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Excel and TI-83 exercises in the text, with corresponding page numbers Excel Choosing Random Numbers Graphical
Representation of Data Measures of Central Tendency Measures of Dispersion Permutations and Combinations Binomial Probabilities Poisson Probabilities Normal Probabilities Finding Values for Given Probabilities Finding Critical Values, Constructing Confidence Intervals Two-Tailed Critical Values for the t-Distribution Finding Critical Values, Constructing Confidence Intervals for Population Proportions Hypothesis Tests for a Single Population Mean Hypothesis Tests for a Single Population Mean (Small Sample) Hypothesis Tests for a Single Population Proportion Hypothesis Test for Two Population Means (Large Samples) Hypothesis Test for Two Population Means (Small Samples) Paired Difference Test Hypothesis Test for Two Population Proportions (Large Samples) Hypothesis Test for Two Population Variances ANOVA Goodness of Fit Test for Independence Correlation Linear Regression Calculations for Inference for Regression Critical Values for the Sign Test KruskalWallis p-Values 10 28 60 87 127 189 204 226 250 278 291 301 327 343 356 380 398 416 431 446 466 486 501 529 550 575 600 632
TI-83 11 30 62 90 127 191 206 229 251 279 292 303 331 347 359 383 401 418 433 448 468 487 502 531 553 * * 632
Excel Explorations The Law of Large Numbers 159 Binomial Probabilities through Simulation 198 Generating Random Normal Data 233 Investigating the Normal Approximation 244 Central Limit Theorem 270 Confidence Intervals 285 Confidence Intervals for Population Proportions 309 The Effect of Sample Size 337 p-Values 363
Mini-Projects
An opportunity to learn about your fellow classmates 16 Fellow classmates: frequency distributions, histograms, and pie charts 45 Fellow classmates: mean, median, first quartile, third quartile, midrange, and mode 73 Fellow classmates: range, interquartile range, and standard deviation 103 Probability that a randomly selected student on your campus is female 162 The mean age of students at your school 285 The mean number of units taken by students at your school this term 296 Proportion of all vehicles in your schools parking lots that are convertibles 308 Testing the claim that the mean height for adult females is 5 5 Testing the claim that the mean height for adult males is 5 9 337 352
Testing the claim that more than 50% of students at your school are female 363 Have college males and females had the same mean number of serious relationships? 393 The amount of money your classmates spent in the last week on food 408 Weight students have gained or lost since attending your school 425 Do students at your school approve of the job the president is doing? 439 Hours per week study time of students at your school 454 Number of calories in one serving of several brands of soda, beer, and juice 477 Eye color in a random sample of at least 100 students 494 Gender and eyeglasses in a random sample of at least 100 students 508 Shoe size and height in a random sample of male and female students 564 Research project: start to finish 649
George Woodbury
College of the Sequoias
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Sponsoring Editor: Carolyn Crockett Editorial Assistant: Jennifer Jenkins Assistant Editor: Ann Day Marketing: Tom Ziolkowski/Samantha Cabaluna Production Editor: Tessa Avila Production Service: Helen Walden Manuscript Editor: Helen Walden COPYRIGHT 2002 Wadsworth Group. Duxbury is an imprint of the Wadsworth Group, a division of Thomson Learning, Inc. Thomson Learning is a trademark used herein under license. ALL RIGHTS RESERVED. No part of this work covered by the copyright hereon may be reproduced or used in any form or by any meansgraphic, electronic, or mechanical, including photocopying, recording, taping, Web distribution, or information storage and retrieval systemswithout the prior written permission of the publisher. Printed in Canada. 10 9 8 7 6 5 4 3 2 1
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Library of Congress Cataloging-in-Publication Data Woodbury, George An introduction to statistics / George Woodbury. p. cm. Includes index. ISBN 0-534-37755-6 1. Statistics. I. Title. QA276.12.W67 2002 519.5dc21
OO B
K IS P RIN
A CID
ER ED E CY C L
To Tina, Dylan, and Alycia
George Woodbury is Chair of the Mathematics and Engineering Division at the College of the Sequoias, a public community college in Visalia, California, where he has taught statistics and mathematics since 1994. His teaching has earned him awards and recognition from students and colleagues alike. He earned a B.S. in Mathematics at University of California, Santa Barbara in 1990 and a M.S. in Mathematics at California State University, Northridge in 1994. George and his wife Tina have two children, Dylan and Alycia. In his free time, George likes to spend time in the garden with his family and on the golf course.
1 DESCRIPTIVE STATISTICS 2
CHAPTER 1 INTRODUCTION TO STATISTICSANALYZING DATA 5 CHAPTER 2 MEASURES OF CENTRAL TENDENCY AND DISPERSION 47
2 PROBABILITY 112
CHAPTER 3 COUNTING AND PROBABILITY 115
3 PROBABILITY DISTRIBUTIONS 166
CHAPTER 4 DISCRETE PROBABILITY DISTRIBUTIONS 169 CHAPTER 5 NORMAL PROBABILITY DISTRIBUTIONS 215
4 ONE-SAMPLE CONFIDENCE INTERVALS
AND HYPOTHESIS TESTS 260
CONFIDENCE INTERVALS 263
CHAPTER 6 THE CENTRAL LIMIT THEOREM AND CHAPTER 7 ONE-SAMPLE HYPOTHESIS TESTS 311
5 TWO-SAMPLE INFERENCES, F-TESTS,
CHI-SQUARE TESTS 370
CHAPTER 8 TWO-SAMPLE HYPOTHESIS TESTS 373 CHAPTER 9 OTHER HYPOTHESIS TESTS 441
6 LINEAR CORRELATION AND REGRESSION 517
CHAPTER 10 LINEAR CORRELATION AND REGRESSION 519
7 NONPARAMETRIC TESTS 588
CHAPTER 11 NONPARAMETRIC TESTS 591
unit 1 Descriptive
one INTRODUCTION TO STATISTICS
ANALYZING DATA 5
1.1 AN INTRODUCTIONDATA AND ITS COLLECTION 6
Data 7 Gathering Data 8 Exercises 1.1 14 1.2 GRAPHICAL REPRESENTATION OF DATA 16 Stem-and-Leaf Display 16 Frequency Distributions 20 Histograms 23 Frequency Polygons 25 Bar Charts 26 Pie Charts 27 Creating a Histogram from a Set of Data 28 Creating a Histogram from a Frequency Distribution 29 Creating a Pie Chart 30 Exercises 1.2 32
two MEASURES OF CENTRAL TENDENCY
AND DISPERSION 47
2.1 MEASURES OF CENTRAL TENDENCY 48
Mean 48 Median 50 Quartiles 52
Midrange 54 Mode 55 Weighted Mean 56 Estimating a Mean Using a Frequency Distribution 58 Exercises 2.1 63 MEASURES OF DISPERSION 73 Range 74 Interquartile Range 74 Boxplot 75 Mean Deviation 76 Variance 78 Standard Deviation 80 Skewness 82 Standard Units 84 Chebyshevs Theorem 85 Estimation of Standard Deviation Using a Frequency Distribution 86 Exercises 2.2 92 EXTRA: Using Samples to Estimate a Population Mean 103
UNIT 1 SUMMARY 107
Probability 112
three COUNTING AND PROBABILITY 115
3.1 COUNTING 116
Tree Diagrams 116 Permutations 122 Combinations 124 Exercises 3.1 128 CLASSICAL PROBABILITY 129 Basic Rules of Probability 133 Exercises 3.2 140 TABLE PROBABILITIES, CONDITIONAL PROBABILITY 141 Conditional Probability 144 Exercises 3.3 148
3.4 INDEPENDENT EVENTS, MULTIPLICATION RULE,
LAW OF LARGE NUMBERS 150 Independent Events 150 Multiplication Rule 153 Multiplication Rule for Dependent Events 154 Subjective Probability 158 Law of Large Numbers 159 Exercises 3.4 161
unit 3 Probability
Distributions 166
four DISCRETE PROBABILITY DISTRIBUTIONS 169
4.1 PROBABILITY DISTRIBUTIONS 170 4.2 4.3
Exercises 4.1 178 BINOMIAL PROBABILITIES 182 Exercises 4.2 194 EXTRA: The Geometric Distribution 196 THE POISSON DISTRIBUTION 199 Exercises 4.3 209 EXTRA: Poisson Distribution as a Limit for the Binomial Distribution 212
five NORMAL PROBABILITY DISTRIBUTIONS 215
5.1 THE NORMAL DISTRIBUTION 216
Exercises 5.1 232
5.2 DISCRETE RANDOM VARIABLES THAT ARE
APPROXIMATELY NORMAL; NORMAL APPROXIMATION TO THE BINOMIAL DISTRIBUTION 233 Normal Approximation to the Binomial Distribution 237 Exercises 5.2 242 FINDING VALUES FOR GIVEN PROBABILITIES 245 Exercises 5.3 252 EXTRA: Is a Set of Data Approximately Normally Distributed? 254
unit 4 One-Sample Confidence Intervals
six THE CENTRAL LIMIT THEOREM AND
6.1 THE SAMPLING DISTRIBUTION OF THE SAMPLE MEAN;
THE CENTRAL LIMIT THEOREM 264 Probabilities Involving Sample Means 266 Exercises 6.1 268 CONFIDENCE INTERVALS FOR A POPULATION MEAN (LARGE SAMPLES) 271 Constructing a Confidence Interval for a Population Mean 272 Determining the Appropriate Sample Size 281 Exercises 6.2 283 EXTRA: One-Sided Confidence Intervals 286 CONFIDENCE INTERVALS FOR A POPULATION MEAN (SMALL SAMPLES) 288 Exercises 6.3 293 CONFIDENCE INTERVALS FOR A POPULATION PROPORTION (LARGE SAMPLES) 296 Appropriate Sample Size 303 Exercises 6.4 305
seven ONE-SAMPLE HYPOTHESIS TESTS 311
7.1 HYPOTHESIS TEST FOR A SINGLE POPULATION MEAN
(LARGE SAMPLES) 312 Step 1: State the Claim and the Hypotheses 313 Step 2: Choose a Level of Significance 315 Step 3: Choose a Test Statistic 315 Step 4: State the Decision Rule 316 Step 5: Do the Calculations and Draw Conclusions 319 P-Values 321 Exercises 7.1 333 HYPOTHESIS TEST FOR A SINGLE POPULATION MEAN (SMALL SAMPLES) 338 Exercises 7.2 349
7.3 HYPOTHESIS TEST FOR A SINGLE POPULATION PROPORTION
(LARGE SAMPLES) 352 Exercises 7.3 359
unit 5 Two-Sample Inferences, F-Tests,
eight TWO-SAMPLE HYPOTHESIS TESTS 373
8.1 HYPOTHESIS TEST FOR TWO POPULATION MEANS
(LARGE SAMPLES) 374 Confidence Intervals 379 Exercises 8.1 386 8.2 HYPOTHESIS TEST FOR TWO POPULATION MEANS (SMALL SAMPLES) 393 Confidence Intervals 397 Exercises 8.2 404 8.3 PAIRED DIFFERENCE TEST FOR DEPENDENT SAMPLES 408 Confidence Intervals 414 Exercises 8.3 420 8.4 HYPOTHESIS TEST FOR TWO POPULATION PROPORTIONS (LARGE SAMPLES) 425 Confidence Intervals 429 Exercises 8.4 434
nine OTHER HYPOTHESIS TESTS 441
9.1 HYPOTHESIS TEST FOR TWO POPULATION VARIANCES 442
Exercises 9.1 450 EXTRA: Hypothesis Test for Two Population Means (Small Samples), Assuming Equal Population Variances 454 9.2 ANALYSIS OF VARIANCE (ANOVA) 459 Exercises 9.2 469 9.3 THE GOODNESS OF FIT TEST 478 Exercises 9.3 489
9.4 CONTINGENCY TABLES AND THE HYPOTHESIS
TEST FOR INDEPENDENCE 494 Exercises 9.4 503
unit 6 Linear Correlation
10.1 CORRELATION 520
Exercises 10.1 533
and Regression 517
ten LINEAR CORRELATION AND REGRESSION 519
10.2 LINEAR REGRESSION 543
Exercises 10.2 554 10.3 INFERENCE FOR REGRESSION 564 Inferences about the Slope 567 Confidence Intervals for 572 Prediction Intervals for y for a Given Value of x Exercises 10.3 577
unit 7 Nonparametric
Tests 588
eleven NONPARAMETRIC TESTS 591
11.1 SIGN TEST 592
Exercises 11.1 601 11.2 WILCOXON SIGNED-RANK TEST 604 Exercises 11.2 612
11.3 WILCOXON RANK-SUM TEST 615
11.4 KRUSKALWALLIS TEST 626
Exercises 11.4 632 11.5 RANK CORRELATION 635 Exercises 11.5 641
UNIT 7 SUMMARY 647 APPENDIX TABLES 652
Table Table Table Table Table Table Table Table A B C D E F G H Binomial Probabilities 652 Poisson Distribution 659 Standard Normal Distribution 660 Students t-Distribution 662 Right-Tailed Critical Values for the F-Distribution 663 Right-Tailed Critical Values for the Chi-Square Distribution 665 Critical Values for the Wilcoxon Signed-Rank Test 666 Critical Values for Spearmans Rank Correlation Coefficient 667
ANSWERS TO ODD-NUMBERED EXERCISES 669 INDEX 693
I believe that students need to understand concepts, rather than mimic steps, to be successful in a math course. This is especially true in an introductory statistics course. One of my main goals in writing this text was to help students understand statistics, to understand the why as well as the how. Toward this end, the many examples in this text are supplemented by clear explanations. Whenever possible, I have tried to show that this is a course of connected material rather than a group of disjointed topics. I try to foreshadow material that will show up in later chapters, while referring back to material covered earlier in the text.
CHAPTER 6 The Central Limit Theorem and Confidence Intervals
M I N I P RO J E C T
What is the mean number of units taken by students at your school this term? Randomly sample fewer than 30 students at your college and find out how many units they are taking this term. Use these sample data to construct a 95% confidence interval for the mean number of units being taken this term by all of the students at your school. In addition to your confidence interval, include
Exercises, Examples, and Mini-Projects
Throughout the book, the exercises and examples describe situations that students can relate to. Some are based on real data found in newspapers, magazines, and government reports, and others are based on projects that students have turned in over the years. In class, I try to inspire my students to apply what they are learning to problems found outside of class, and I believe that this text will do the same. The mini-projects that follow many of the sections are written with this in mind. They allow students to apply the techniques learned in a section to data they can collect on their own. One option is to allow students to work in teams in gathering their data.
Your raw data A histogram for the number of units being taken A pie chart showing the breakdown of units. Set this up in any way that
you feel accurately represents the students in your survey. Finally, answer the following questions.
How did you obtain your sample? Which type of sampling did you use? Was it truly random? What potential biases may show up in your sample?
Confidence Intervals for a Population Proportion (Large Samples)
Quite often, the population parameter that we are interested in is not a mean or average, but a proportion or percentage. In this section, we will learn how to construct a confidence interval for a population proportion. We will use a sample proportion (p) to estimate the population proportion (). Creating a confidence interval is based on our work with the normal approximation to the binomial distribution. The conditions for the normal approximation must be satisfied [np 5 and n(1 p) 5], so we must have at least 5 successes and 5 failures in our n independent trials (sample). The basic idea is the same as it was in the previous two sections. We will calculate a margin of error, and then subtract it from the sample proportion p to get the left endpoint of the confidence interval. We add the margin of error to the sample proportion to find the right endpoint of the confidence interval. The major difference is that the formula for margin of error is different.
E = z/2
(1 p) n
where z/2 = critical value p = sample proportion n = sample size We then construct the confidence interval using
p E p + E
E XAMPLE
A random sample of 400 pregnant women revealed that 60 are smoking during their pregnancy. Construct a 95% confidence interval for the proportion of all pregnant women who smoke while pregnant.
Focus on Choosing the Appropriate Tool
A common problem seen in introductory statistics courses is that students have a hard time deciding which hypothesis test is appropriate for a given situation. At the end of Chapters 7 through 9 is a section titled Choosing the Appropriate Tool. Here students find suggestions for deciding which test is appropriate, as well as summaries of the assumptions for each test. A feature in each section of Chapters 7 through 9 is What Is Wrong with This Picture? Each of these presents a flawed statistical project, and students are asked to find the error and explain how to correct it. Some projects do not meet the necessary assumptions, while others use the wrong test.
Topics that are not covered in all statistics courses appear in Extra sections. For example, a section on the geometric probability distribution follows Section 4.2 (binomial probability distribution), and a section on determining whether a set of data is normally distributed follows Section 5.3 (finding values for given probabilities).
7.3 Hypothesis Test for a Single Population Proportion (Large Samples)
We begin by selecting 100 samples, each of size 100, from a population that has a normal distribution. Pull down the Tools menu and click on Data Analysis. When the dialog box appears, click on Random Number Generation and OK. When the next dialog box appears, enter 100 in the box labeled Number of Variables. (This is the number of samples that we are going to take.) Enter 100 in the box labeled Number of Random Numbers. (This is the sample size.) In the box labeled Distribution, click on Normal. For Parameters, enter 50 for the mean and 5 for the standard deviation. Usually, the population parameter is unknown. Therefore, we try to estimate it using a confidence interval. However, for the purpose of this simulation it is imperative that we know the true mean so that we can find out how many of our confidence intervals contain . Click OK. (This may take a few minutes, so be patient.) In cell A101 type =AVERAGE(A1:A100). After pressing the Enter key, click on cell A101 again. Hold down the Shift key and click on cell CV101. Pull down the Edit Menu; select Fill and then Right. In row 102 we will calculate the lower limit of each confidence interval. In cell A102 type =(A101 - 1.96*(STDEV(A1:A100)/(SQRT(100)))). After pressing the Enter key, click on cell A102 again. Hold down the Shift key and click on cell CV102. Pull down the Edit Menu; click on Fill and then Right. In row 103 we will calculate the upper limit of each confidence interval. In cell A103 type =(A101 + 1.96*(STDEV(A1:A100)/(SQRT(100)))). After pressing the Enter key, click on cell A103 again. Hold down the Shift key and click on cell CV103. Pull down the Edit Menu; click on Fill and then Right. Now we will determine what percentage of our intervals contain the actual population mean (50) in row 104. Type =IF((A102<50)*AND(A103>50),1,0) in cell A104. After pressing the Enter key, the cell will contain a 1 if the interval contains 50 and a 0 if it does not. Click on cell A104. Hold down the Shift key and click on cell CV104. Pull down the Edit Menu; click on Fill and then Right. To determine what percentage of our intervals contain the actual population mean (50), type =SUM(A104:CV104)/100 in cell A105. Did 95% of your intervals contain the population mean? If not, was it close?
40. A random sample of 1954 high school students showed that 186 of them have tried cocaine at least once. At the 0.01 level of significance, test the claim that 10% of all high school students have ever tried cocaine at least once. (Based on the Morbidity and Mortality Weekly Report by the Centers for Disease Control and Prevention.)
The following statistical project contains a major error. Potential errors could be the choice of the wrong hypothesis test or that the assumptions required for this test are not met. Find the error, and explain what could be done to correct it.
A student believes that most college students feel that tuition is too high. He asks 42 randomly selected students on his campus and 38 of them feel that tuition is too high. He uses these results to test the claim that 90% of all college students feel that tuition is too high, using the hypothesis test introduced in this section. What is wrong with this picture?
I claim that more than 50% of the students at your school are female. You are to test my claim at the 0.05 level of significance. Randomly sample at least 100 students and record their gender. Use the sample data to test the claim. In addition to your complete hypothesis test, include
Your raw data A pie chart showing the percentage of students at your school that are
female and the percentage that are male Finally, answer the following questions.
EXTRA One-Sided Confidence Intervals
Occasionally, there is a need for a one-sided confidence interval. For example, we may want to be 95% sure that the population mean is below a certain value, as opposed to being between two values. Or we may want to be 99% sure that the population mean is above a certain value. The procedure is similar to the confidence interval procedure developed in this section. One difference is when we find the critical value. The second difference is that we only find one endpoint, as opposed to two. We begin with an example.
How did you obtain your sample? Which type of sampling did you use? Was the sample truly random? What potential biases may show up in your sample?
EXCEL EXPLORATION
p-Values
The president of a community college claims that the mean age of the students at his school is less than 25. A random sample of 100 of his students had a mean age of 23.1 years, with a standard deviation of 7.10 years. Construct a 99% left-sided confidence interval for the mean age of all students at his school.
We will use Excel to explore the idea of p-values. We will generate several random samples based on a known population proportion. We will then see what proportion of these randomly generated sample proportions are at least as extreme as our given sample proportion. This is the p-value for the given sample. Before we begin we must be sure that the Analysis Tool Pak has been added. Pull down the Tools menu. If you see a choice that says Data Analysis, then the Tool Pak is added and you may continue. If you do not see Data Analysis, click on Add-ins under Tools. When the dialog box opens up, click on the box for Analysis Tool Pak, and click on OK. We will use the following hypothesis test as an example.
3.1 Counting
Use of Technology
The availability of technological resources has had a dramatic effect on the way this course is taught and the types of problems that students can solve. I believe that technology should supplement a students understanding, and have written the technology material in the text with that in mind. I have chosen to incorporate Microsoft Excel as a software package and the Texas Instruments TI-83 as the calculator for this text. Most sections contain directions for using these tools to solve problems found in those sections. Microsoft Excel is used because it is the statistical software most likely to be on a students own computer. A supplement to the text is available for Minitab. Microsoft Excels generation of random numbers makes it easy to run simulations. In several sections you will find Excel Explorations, which are used to illustrate topics. For example, the Excel Exploration after Section 5.2 examines the normal approximation to the binomial distribution.
Microsoft Excel has two built-in functions to help us calculate permutations and combinations. Here is the general form. =PERMUT(n, r) For example, suppose you wanted to find how many different ways you could select a president, a vice president, and a treasurer from a group of 20 people (20 P3). To do this, type the following in any open cell: =PERMUT(20,3) For problems involving combinations, we use =COMBIN(n, r) For example, suppose you wanted to find how many different ways you could select three people from a group of 20 people (20C3). To do this, type the following in any open cell: =COMBIN(20,3)
The TI-83 can help us to calculate permutations and combinations. Suppose you wanted to find how many different ways you could select a president, a vice president, and a treasurer from a group of 20 people (20P3). You first type 20, then press MATH . Use three times to access the PRB menu.
, which should produce the following screen.
Now type 3 and press ENTER .
To the Student
This book is written with you in mind. You will find plenty of examples to help you learn how to do the exercises, as well as clear explanations to show you why we do what we do. The best course of action for you to take is to read each section before it is discussed in class, so that your instructor can help reinforce what you have read. The homework exercises are extremely important. They are designed to help you gain an understanding of the topics in that section. To assist you, a Students Solution Manual is available. Many students dread taking the introductory statistics course (I have heard students refer to it as sadistics), but I believe that you will find it to be one of the more rewarding and useful classes you will take in your college career. Good luck!
Learning and Teaching Aids
Duxbury offers a complete set of tools to help students master the course and to assist instructors in teaching.
Improving Your Grade for An Introduction to Statistics. A student solutions manual containing detailed solutions for the textbooks odd-numbered exercises and tips on how to solve these exercises. (ISBN: 0-534-38926-0) Improving Your Grade Version 2 WebTutor on Web CT (and Blackboard) for An Introduction to Statistics. An electronic resource that offers students real-time access to a full array of study tools, including the student solutions manual (Improving Your Grade for An Introduction to Statistics), practice quizzes, flashcards for each chapter with glossary terms, course goals, and objectives, and chapter learning objectives. (WebCT ISBN: 0-534- 39621-6 and Blackboard ISBN: 0-534- 39622-4)
Instructors Resource Manual. A complete resource containing suggestions on teaching from the book, ideas for classroom lectures, uses of technology, comments on mini-projects, answers to the What Is Wrong with This Picture? sections, answers to all exercises, and discussions of other material in the book. This manual is available through the local sales representative or by submitting a request at info@duxbury.com. (ISBN: 0-534-38921-X)
The publisher also maintains a Web site supporting instruction using An Introduction to Statistics at the address http://www.duxbury.com. At this site, following the Book Companions and Data Library links, are
Data sets Minitab procedures for selected examples from the book Lists of known errata (as needed)
The publishing of this text follows years of hard work, in which I have received an extraordinary amount of support and help from many people. My editor Carolyn Crockett has been most supportive and encouraging throughout the entire process. Liz Gilley-Poulsen provided the early interest and excitement that got this project off the ground. I have enjoyed my relationship with everyone at Duxbury Press, and I thank all who have helped along the way. Thanks to the following reviewers: Michael Allen, Glendale Community College M. C. Bhattacharjee, New Jersey Institute of Technology Michael Butler, College of the Redwoods Guang-Hwa (Andy) Chang, Youngstown State University Pinyuen Chen, Syracuse University Carol Curtis, Fresno City College Gregory J. Davis, University of Wisconsin-Green Bay Janice Dykacz, Community College of Baltimore County Charles Johnson, Collin County Community College, Texas James Lang, Valencia Community College E. D. McCune, Stephen F. Austin State University Ron Persky, Christopher Newport University Geetha Ramachandran, California State University-Sacramento Glenn Weber, Christopher Newport University Mark Wilson, University of Charleston My colleagues at the College of the Sequoias have been most helpful, especially Jared Burch, who I have bounced more ideas off of than I could possibly count. Thanks also to my students, who helped to light the fire in me to be the best possible teacher I can be, for they deserve nothing less than that. This book is dedicated to the memory of four tremendous people: Joy Bishop, John Schaeffer, Claudia Romero, and Lynn Dwyer. From them I have learned that a positive attitude makes all the difference. All of them left us too early, but left their mark on me and many others as well. They are truly missed.
Thanks to family and friends who have all helped me throughout the years, especially my parents, who were always very supportive. A special thanks to my father for all of the early probability lessons! Most important, I thank my wife Tina and our children Dylan and Alycia. They make everything possible and worthwhile. I count them as my greatest blessing, and love them all more than words could possibly say. George Woodbury
INTRODUCTION TO STATISTICS ANALYZING DATA
Section 1.1 An IntroductionData and Its Collection Section 1.2 Graphical Representation of Data
MEASURES OF CENTRAL TENDENCY AND DISPERSION
Section 2.1 Measures of Central Tendency Section 2.2 Measures of Dispersion
elcome to statistics! In this course, we will use statistics to formulate answers to questions such as:
What is the average weight of a newborn baby boy?
What percentage of people between the
ages of 18 and 24 smoke? that plan to attend college greater than the percentage of high school males that plan to attend college? Do students who take a prep course for the SAT have a higher average score than students who do not take such a course? Is there a relationship between the hours a student works each week and his or her score on a final exam?
Is the percentage of high school females
To address such questions, we follow three steps: 1. Collect information. 2. Analyze the information. 3. Draw a conclusion or estimate from our analysis. Clearly, it cannot be as simple as this. However, these are the three main parts of each statistical project. You will learn the finer points as we move along, as well as develop a skeptical eye for what may be going wrong with a project. The first section presents an introduction to the big picture of statistics. You will be introduced to important terms, as well as to the difference between descriptive statistics (the second step above) and inferential statistics (the third step above). After a discussion of what constitutes data, and the different types of data that we will be working with, we will go on to cover the different ways of gathering data. The methods all have advantages and disadvantages, and we shall examine those. In Section 1.2 we will look at the different ways to represent a set of data graphically. A set of graphs can tell us a great deal about a set of data. They can also be our best tools for presenting our case to someone reading our results.
An IntroductionData and Its Collection
A senator is considering running for president in the next election and is interested in what percentage of registered voters consider her to be a serious candidate. How can we go about determining this? Contacting all registered voters would be expensive and impractical, if not impossible. Contacting ten registered voters would not produce reliable results. We will take the middle ground, contacting enough registered voters to make the results reliable without contacting too many to make the project impractical for our resources. How many voters should we contact? This question will be answered in a later chapter, where we will see that a survey of approximately 2000 registered voters should give an estimate to within approximately 2% of the actual percentage. We study two branches of statistics in this course: descriptive statistics and inferential statistics. Descriptive statistics involves methods that are used to describe data that we have collected. We can describe a set of data using numbers and graphs. For instance, if 280 out of 350 people living in Luserna, Italy have the last name Nicolusi, we can go on to say that 80% of the people in Luserna have the last name Nicolusi. The number 80% describes the set of data that we have. If a car salesman sold four cars this weekend, one car the previous weekend, and one car the weekend before that, we can say that the salesman averaged two cars sold for the last three weekends. A descriptive statement must be verifiable from the data provided. A population is a collection of people, objects, or measurements that we are interested in analyzing. For our senator who is considering running for president, the set of registered voters in the United States is the population, because she is interested in how all registered voters feel. Since it is not practical to contact all registered voters, the senator must contact a sample of all registered voters. A sample is a part of the population. If we contacted 2000 registered voters, they would be a sample of all registered voters. Inferential statistics involves methods that are used to make a generalization, or evaluate a claim, about a population based on sample data. For an inference to be reliable, the sample has to be representative of the population. A sample is biased if it does not accurately represent the population. We discuss the selection of a quality sample later in the section, as well as potential biases.
In the last four semesters that an instructor taught intermediate algebra, the following numbers of people passed the class.
1.1 An IntroductionData and Its Collection
Which of the following statements are descriptive, and which statements make an inference? (a) In the last four semesters that the instructor taught intermediate algebra, an average of 15 people passed the class. This statement is descriptive, because we can verify that the average for these four semesters is 15. (b) The next time the instructor teaches intermediate algebra, we can expect approximately 15 people to pass the class. This statement makes an inference, predicting a future event based on the evidence from this sample of four semesters. (c) This instructor will never pass more than 20 people in an intermediate algebra class. This statement is an inference; it makes a prediction based on the fact that the instructor did not pass more than 20 students in these four semesters. (d) In the last four semesters that the instructor taught intermediate algebra, no more than 20 people passed the class. This statement is descriptive, because we can verify that no more than 20 people passed the course in these four semesters. (e) Only four people passed one semester because the instructor was in a bad mood the entire semester. This statement makes an inference, because we are not told that the instructor was in a bad mood that semester. (f) The last time he taught the class, the instructor passed 20 people to keep the administration from criticizing him for poor results. Again, this statement makes an inference, because it is not based on any presented fact.
A number that describes a characteristic of a population is called a parameter, whereas a number that describes a characteristic of a sample is called a statistic. In a statistics class of 40 students, 24 had a credit card with them. If we say that 60% of the students in this class had a credit card with them, the 60% is a parameter because the class of 40 students is the population. However, if we make the inference that approximately 60% of all statistics students bring a credit card to class with them, the 60% is a statistic because the class of 40 students is now a sample.
Data are information that we collect, and this information is not necessarily numerical. Categorical data, or qualitative data, are data that are comprised of names or categories. Numerical data, or quantitative data, are data that are comprised of numerical values. The colors of M&Ms in a bag are an example of categorical data, whereas the number of M&Ms in a bag is an example of numerical data. Categorical data can be broken into two levels: nominal and ordinal. Nominallevel data are categorical data that lack an ordering scheme; ordinal-level data are categorical data that have an ordering scheme. When considering whether a set of data has an ordering scheme, think about whether one value is better or higher
CHAPTER 1 Introduction to StatisticsAnalyzing Data
than another value. For example, household income described as low, middle, or high is ordinal because we can tell that a middle-income household earns more than a lowincome household. The colors of M&Ms in a bag are an example of nominal data. We cannot say whether a red candy is higher or lower than a blue candy. Numerical data can be broken into two levels as well: interval and ratio. Intervallevel data are the next step up from ordinal data, in that we can measure the difference between two values. Interval-level data lack a zero starting point, which means that we cannot construct ratios between two data values. An example of interval-level data would be Celsius temperatures. Suppose we had two Celsius temperatures, 10 and 20. Can we tell which one is hotter? Yes, 20 is hotter than 10. Can we tell how much hotter? Yes, it is 10 hotter. Can we say that 20 is twice as hot as 10? No. This ratio is not valid, because Celsius temperatures do not start at 0. If we convert these temperatures to Fahrenheit temperatures (50 and 68), we can see that the higher temperature is not twice the lower temperature. Numerical data that have a zero starting point are ratio-level data. Cash in pocket is an example of ratio-level data. Suppose two people have $10 and $20 in their wallets. Can we say that the person with $20 has twice as much money as the person who has $10? Yes, because the lowest amount of money that a person could have in his or her wallet is $0. Numerical data can further be classified as either discrete or continuous. A set of data is discrete if there are only a finite number of values possible or if there is a space on the number line between each two possible values. Discrete data are usually associated with some sort of count. If a five-question quiz is given in a math class, the number of correct answers on a students quiz is an example of discrete data. The number of correct answers would have to be one of the following: 0, 1, 2, 3, 4, or 5. There are not an infinite number of values, therefore these data are discrete. Also, if we were to draw a number line and place each possible value on it, we would see a space between each pair of values. For some discrete data there can be an infinite number of possible values, as long as there are gaps between the values. To obtain a taxi license in Las Vegas, a person must pass a written exam regarding different locations in the city. The number of times it would take a person to pass this test is an example of discrete data. A person could take it once, or twice, or three times, or four times, or . . . , so the possible values are 1, 2, 3, . . . . There are infinitely many possible values, but if we were to put them on a number line, we would see a space between each pair of values. Continuous data make up the rest of numerical data. This is a type of data that is usually associated with some sort of physical measurement. For any two possible values, there are infinitely many values between them. Think of a number line. For any two points on a number line, there are infinitely many values between them. The same holds true for continuous data. The height of trees at a nursery is an example of continuous data. It is possible for a tree to be 76" tall, and it is possible for a tree to be 77" tall. All decimal heights between 76" and 77" are also possible. One general way to tell whether data are continuous is to ask yourself whether it is possible for the data to take on values that are fractions or decimals. If your answer is yes, these are probably continuous data.
There are two basic ways to gather data: either through an experiment or an observational study. In an experiment, we apply a treatment and measure its effect. An example of such an experiment is a test of a new drug for lowering blood pressure.
People in the study would be given the new medication, and after using it for a period of time their blood pressure would be checked to see whether the medication was effective in lowering blood pressure. People who know that they are part of this study may change their daily habits, such as increasing the amount of exercise they get or reducing the amount of fatty foods that they eat. This could make it difficult to determine whether the medication was responsible for the drop in blood pressure. Enter the control group. Often an experiment has a group of people that believe they are part of the experiment, but actually are receiving a placebo. If there is a substantial difference between the experimental group and the control group, then this is evidence of the effectiveness of the treatment. Some recent studies suggest that the placebo effect is so strong that perhaps it should be considered as a treatment itself. Another example of an experiment is a taste test. People are given two different brands of the same product, such as chocolate chip cookies, and asked to tell which of the two they prefer. We may obtain different results if we ask the people in the study to simply name which brand they prefer, without tasting. For instance, if we ask Do you prefer Burger King french fries or McDonalds french fries? the results may differ from a taste test of the two types of french fries. A person who has never been to Burger King may answer McDonalds without even knowing what the fries at Burger King taste like. Asking a person to name his or her favorite brand is an example of an observational study. In an observational study, no treatment is applied to measure a response. Instead, we simply observe the members of our study and record data. Such studies can include opinion surveys. We do not try to influence a persons opinion; the persons opinion is already determined. Other examples include trying to determine the percentage of students at your school that are female or trying to determine the average weight of newborn baby boys. Any observational study or experiment in which the sample is not representative of the population may produce poor data and unreliable results. The selection of a sample is a very important step in a statistical project, and lays the foundation that all later work will be built upon. We will employ probability sampling in choosing our sample. In a probability sample, each member of the population is equally likely to be chosen to be included in the sample. One method of sampling is random sampling. The basic idea is to write down each member of the population on a piece of paper, put all of the pieces in a big hat, mix them all up, and then blindly reach in and select one piece of paper at a time. In practice, we arrange the population in an ordered list. Then we randomly select numbers that correspond to members of the population on the list. Some use a table of random numbers to do this, but technology can help us to do this as well. Microsoft Excel and the TI-83 can randomly generate numbers, as can other computer software packages. Can a telephone book for a certain city be used to generate a random sample? A problem with using the phone book is that some people have unlisted numbers and are not in the phone book. These people do not have an equal chance of being included in the sample. Would randomly generating phone numbers work? Although this would address the unlisted number problem, there are some people who do not have a telephone at all. A second sampling technique is systematic sampling. A starting point in the list is randomly picked, and then from there we include every kth member on the list. For example, we could begin with person number 19 and then select every 50th person from there. One potential problem with this type of sampling arises with a list that
follows a certain pattern. For example, if we had a list of the number of customers at a restaurant every day for an entire year and selected every 21st entry, we would always select the same day of the week. Stratified sampling is an excellent way to try and ensure that a sample looks just like the population but smaller, although it requires the most resources and information about the population. The population is broken into different groups, or strata, and then we randomly sample from each group. We select a number from each group that is proportional to the breakdown of these groups in the population. Some different strata for an election poll could include gender, race, age, income, political affiliation, or geographical location. If you desire to sample students at your school about some campus-wide issue, and 55% of the students at your school are female students and 45% are male students, then it makes sense to choose your sample in such a way that 55% of your sample is female. We could break the sample down into freshmen, sophomores, juniors, and seniors as well. Another sampling technique is cluster sampling. We divide the population into groups called clusters, usually having a common location. We then randomly select some of these clusters, and include each member of that cluster in the sample. For example, we could divide up a campus by classrooms that are in session at 10 A.M., randomly select some of those classrooms, and include each student in those classrooms at that time. One problem with this example is that it may only be representative of students that take classes at 10 A.M., leaving out evening students, for example. A convenience sample makes use of data that are convenient to gather. For example, a student could stand at the center of campus, or in front of the library or dining hall, and sample people as they pass by. Keep in mind that samples obtained in such a fashion are not necessarily representative of the population. Another convenience sample could use someone elses data, found in a publication or on the Internet. Always be cautious about using data gathered by another party. The data gatherer may have some bias or hidden agenda that could affect the data. Bias can appear in any set of data, even one that is carefully gathered. One problem is nonresponse. If the people who do not respond have a common characteristic, then that group will be underrepresented. Occasionally, bias is created by poorly worded questions. Dont you believe that this is another instance of big government interfering with our daily lives? may lead a participant to agree and say yes. Sometimes respondents do not tell the truth. If you ask strangers whether they wash their hands after using a restroom, very few if any will say no because it is viewed as poor hygiene. However, we know that some people do not wash their hands after using a restroom. One should never put much faith in the results of a call-in survey, mail-in poll, or on-line poll. The people who make the effort to phone in their opinions or go to a Web site to cast their votes often feel more strongly about the issue. These people can represent the majority in a poll, but the minority in the population.
Choosing Random Numbers
We will be learning to use Microsoft Excel, a spreadsheet software package, throughout this course. In this section, we will learn how to use it to randomly generate numbers
to help us select a random sample. For this example, we will assume that a college has 10,384 students and we want to select a sample of 100 students. After opening Microsoft Excel, you will see a rectangular array of boxes that are called cells. Cells are arranged in columns (denoted by the letters A, B, C, . . . across the top of the screen) and rows (denoted by the numbers 1, 2, 3, . . . down the left side of the screen). Cell A1 is the cell in column A and row 1. In cell A1, type the following command and press Enter on your keyboard: =RANDBETWEEN(1,10384) RANDBETWEEN is Excels built-in function to generate a random integer. The first number in the parentheses is the lowest integer that could be generated, and the second number in the parentheses is the highest integer that could be generated. Since there are 10,384 students, we would arrange the list from 1 to 10,384. The number that appears in cell A1 after you press Enter will tell us the first student to include in the sample. To generate the remaining 99 numbers for our sample, press the F9 key on your keyboard. This will recalculate the function, giving us a new random number in cell A1. Repeat this process until there are 100 different random numbers. A second way to do this is to begin by typing =RANDBETWEEN(1,10384) in cell A1, and pressing Enter on your keyboard. Next, click on cell A1. Select the Edit menu (click on the word Edit at the top of the screen), and choose Copy. This copies the formula to the computers clipboard for later use. While holding down the Shift key on your keyboard, press the down arrow on your keyboard until the first column from cell A1 through A100 is highlighted. Select the Edit menu again, and click on Paste. This puts the formula from cell A1 into all 100 cells. There may be a number that gets repeated, and you will have to select additional random numbers to reach a total of 100 unique random numbers.
Use Excel to generate the following samples. 1. 2. 3. 4. Select Select Select Select 25 40 23 87 people from a population of 700. patients from a group of 295. days from a total of 365 days. people from a list of 3500 people.
We will be learning to use the Texas Instruments TI-83 calculator throughout this course. In this section, we will learn how to use it to randomly generate numbers to help us select a random sample. For this example, we will assume that a college has 10,384 students and we want to select a sample of 100 students. The TI-83 has a built-in function randInt that randomly generates a series of integers. Begin by pressing the button labeled MATH . You should see the screen shown on the right.
Press the right arrow randInt function.
until PRB is highlighted, and press
to select the
You will be taken back to the main screen. Type 1,10384,100) after randInt( and press the ENTER key. After a few seconds, you will see the following screen.
To see all 100 values, we will store these values in a statistical list. Press the key labeled STO , followed by the 2nd key and 1 . This accesses L1, in orange above 1 on your calculator, which is the name of a statistical list. After pressing ENTER , you will see the following screen.
To view this list, press
STAT , which produces the following screen.
Press 1 to select the Edit option. In the column labeled L1, you will find our 100 values. Here is an example of what you should see.
It may be convenient to see these values in ascending order. To do this, press and select option 2 under Edit. Put L1 in parentheses after SortA( and press ENTER . Your screen should look like this.
Go back to view list L1 in the ing order.
editor, and the values will now be in ascend-
Sorting will make it easier to find duplicates in the list. Use the key to move your way down the list. If there are duplicates, you can generate more values on the main screen by pressing 2ND MODE , which selects Quit and returns you from the STAT editor to the main screen. Once there, enter randInt(1,10384) as many times as is necessary until you have 100 unique values.
Use the TI-83 to generate the following samples. 1. 2. 3. 4. Select Select Select Select 25 40 23 87 people from a population of 700. patients from a group of 295. days from a total of 365 days. people from a list of 3500 people.
EXERCISES 1.1
Here are the number of complete games thrown in the American League and the National League for the years 1997 through 1999. (A complete game is a game for which the teams starting pitcher is the only pitcher; he pitches the entire game.)
League American National 1997 123 143 1998 141 161 1999 108 127
Determine whether the following statements are descriptive or inferential. Explain the reasoning behind your choice, in your own words. 1. During each of these three seasons, there were at least 120 complete games in the National League. 2. There will always be at least 120 complete games in a season in the National League. 3. In the season of 2005, there will be more complete games in the National League than in the American League. 4. During these three seasons, the American League had fewer complete games than the National League. 5. The National League has more complete games because the hitters in their league are not as good as the hitters in the American League. 6. The American League has fewer complete games because its relief pitchers are superior to the National Leagues relief pitchers. 7. Based on the data, make a statement of your own that is descriptive. 8. Based on the data, make a statement of your own that is inferential. Here are the number of homicide arrests of juveniles age 18 and under in the United States for the years 1993 through 1997.
Year Arrests 1993 3284 1994 3102 1995 2560 1996 2172 1997 1731
(Source: U.S. Justice Department)
Determine whether the following statements are descriptive or inferential. Explain the reasoning behind your choice, in your own words. 9. 10. 11. 12. The number of homicide arrests of juveniles declined each year. A government program is responsible for the decline in arrests. The number of homicide arrests of juveniles in 1992 was higher than 3284. The average number of homicide arrests of juveniles is above 2500.
Here are the test scores for two students in a statistics class.
Student Alycia Kevin Test 1 85 73 Test 2 88 78 Test 3 83 81 Test 4 86 84 Test 5 85 84
Determine whether the following statements are descriptive or inferential. Explain the reasoning behind your choice, in your own words. 13. Alycias score on the final exam will be in the 80s. 14. Alycia is a better student than Kevin.
15. Alycias score on the final exam will be higher than Kevins. 16. Kevins average for these five tests is 80 points. Determine whether the following data are nominal-level, ordinal-level, interval-level, or ratio-level data. In addition, if the data are numerical, determine whether they are discrete or continuous. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. The colors of the marshmallows in a box of cereal A college students degree (associate, bachelors, masters, etc.) Temperatures of beers from a tap (C) Weight of a college student Letter grades on an exam (A, B, C, D, F) Percentage grade on an exam Points scored by a basketball team Ratings of an instructor (superior, average, poor) Elevations of U.S. National Parks, in feet above/below sea level IQ scores Length of commute to school (distance) Length of commute to school (time) 1999 profit/loss of companies on the New York Stock Exchange Number of classes taken by a student this semester Year in school (freshman, sophomore, etc.) Political affiliation (Democrat, Republican, etc.) Average high temperatures of spring break vacation spots (F) Major in college (mathematics, biology, psychology, etc.) Number of passengers whose baggage was mishandled Height of two-year-old baby boys
For the following scenarios, determine which sampling technique was used. Also list biases that may be present. 37. Average weight of newborn baby boys: Twelve hospitals are selected at random, and the weight of each baby boy born in January is recorded. 38. Percentage of 18- to 25-year-olds who used drugs during the past 30 days: At a shopping mall, people who appear to be in the proper age group are stopped and asked for their age and whether they have used drugs in the past 30 days. 39. Percentage of masters degrees in mathematics earned by females: Several universities are selected at random and asked for the number of masters degrees awarded in mathematics, and how many were awarded to females. 40. Average length (in days) of a sexual harassment trial: The records of a law firm are analyzed, and the lengths of all of their sexual harassment trials are recorded. 41. Effectiveness of a pain reliever against migraine headaches: Patients who have a history of migraines are divided into three groups, using random numbers. The three groups are given a placebo, a half-dose, and a full-dose of the medication. The patients are then asked to rate the effectiveness of the medication on a scale of 1 to 10. 42. Average SAT score for females compared to males: Fifty students are randomly selected from a universitys list of students. Their gender and SAT scores are recorded. 43. Explain how you could use cluster sampling to find out how students in the dormitories feel about the quality of food in the dining hall. 44. Explain how you could use convenience sampling to find out how students in the dormitories feel about the quality of food in the dining hall. 45. Explain how you could use stratified sampling to determine how your candidate for governor is doing. List the different strata that you would use. 46. Explain how you could use stratified sampling to find out how students at your school feel about a proposed tuition hike. List the different strata that you would use.
M I N I P RO J E C T SECTION 1.2
This introductory mini-project will give you the opportunity to learn a little bit about your fellow classmates. The data gathered will be used in later miniprojects. Collect the following data about each of your classmates:
Age Gender Height Eye color Left-handed or right-handed Number of units enrolled in this term Number of hours worked per week (outside of class) Major Smoker or nonsmoker
Collect your data in a table. Numbering the students 1, 2, 3, and so on will help to maintain anonymity.
Graphical Representation of Data
Now that we know how to gather data, we can begin to learn how to analyze data. A picture is worth 1000 words, and in this section we will learn some graphical ways to present our data. There are two ideas that we should keep aware of while analyzing data. First, we need to look for where the data are centered. This will give us an idea of what a typical value is for this set of data. Second, we must note the dispersion of the data. There is a great deal to be learned from how the data are spread out.
Stem-and-Leaf Display
The first graphical tool that we will employ is a stem-and-leaf display. A stem-and-leaf display is a shorthand notation for writing the values in order, from the lowest value to the highest value. From the display we can begin to get a feel for the typical value, or center, of the set of values, as well as how the values are distributed. To construct a stem-and-leaf display, each value is split into two partsa stem and a leaf. The stem is the first part of the number, and the leaf is the last part of the number. Although the values may be split in many different ways, depending on the types of values you are working with, we usually let the leaf be the last digit of the value and the stem be all of the preceding digits. For example, the value 46 has a stem of 4 and a leaf of 6. For the value 192, the stem is 19 and the leaf is 2. (Occasionally, we will choose to use a stem of 1 and a leaf of 92 instead. This is a good idea when the values are spread out from 100 to 900, for instance.) In constructing the actual stem-and-leaf display, we begin by finding the lowest value and the highest value in the set of data. This will give you the first and last stem that is needed in your display. In a vertical column, write down each possible stem from the lowest stem through the highest stem. Then, for each value, write its leaf in the row containing its stem. Once this has been done for each value, each individual row must be sorted from lowest to highest. This step is not necessary if the values have already been sorted. If you are using technology to assist you, such as Microsoft Excel or the TI-83 calculator, it is an excellent idea to sort the values first.
1.2 Graphical Representation of Data
Here are the scores of 32 randomly selected statistics students on an exam on Unit 1. Construct a stem-and-leaf display for these scores.
The lowest value is 45 (stem: 4, leaf: 5), and the highest value is 98 (stem: 9, leaf: 8). This means that our stems will be 4, 5, 6, 7, 8, and 9.
Stem 4 5 6 7 8 9 Leaf
Begin with the first value, 83. Its stem is 8 and its leaf is 3. We will place a 3 in the row containing the leaf 8.
After repeating this for all of the values, your display should look like this.
Stem 4 5 6 7 8 9 7 1 5 6 3 4 8 6 4 5 6 1 5 8 1 8 2 9 6 1 9 6 4 5 6 3 2 2 6 0 2 8 Leaf
The values for each stem need to be sorted, which produces the final stem-andleaf display.
It is extremely important that you keep the leaves aligned vertically. If you do, you can look at the length of each row to determine which stem has the most values. For these test scores, the stem 7 has the most values (9). Being in the middle of the display indicates that a typical value for this set of data may be somewhere in the 70s.
We should also examine the dispersion of these values. Are they evenly distributed? No, there are more values in the upper three stems (23 values) than in the lower three stems (9 values).
Each year, Forbes magazine prints a list of the 400 richest people/families in the world. From the 1999 issue, here are the ages of the 54 richest individuals. All of these individuals have a net worth of at least $5 billion. Construct a stem-and-leaf display for this set of data.
The lowest value in this set is 31 (Pierre OmidyareBay, net worth: $7.8 billion) and the highest value is 88 (Walter HaefnerCareal Holdings (Switzerland, computers/cars), net worth: $7.2 billion). Our stems will be 3 through 8. Here is the stem-and-leaf display, unsorted.
Stem 3 4 5 6 7 8 4 3 9 8 3 4 6 6 5 8 1 2 1 3 4 2 3 8 2 9 2 5 1 3 9 9 1 2 1 1 0 7 Leaf 4 1 0 6 2 9 8 1 8 0 8 2 1 6 2 6 5 9 4 5 2 2
Now here is the display after sorting.
The largest number of these individuals is in their 50s, while most are in their 50s, 60s, and 70s. The typical value is probably somewhere in the low to middle 60s. In the next chapter, we will develop methods to come up with an exact value for the typical value. Sometimes one or more rows will contain too many values. In such a situation, we may break up each row into 2 rows (or more if we wish). To break up a row whose stem is 3, we create two new stems: 3 and 3*. Stem 3 will hold the values from 30 to 34, and 3* will hold the values from 35 to 39. If one row is too long and you want to break it down further, you must break down all of the other rows for consistency. Here is the stem-and-leaf display for the Forbes data, with each stem split in 2.
This display gives us more information about these ages.
Stem 3 3* 4 4* 5 5* 6 6* 7 7* 8 8* 1 6 1 6 0 5 0 6 1 5 1 8 4
2 3 3 3 4 0 6 1 8 1 5 2 1 8 2 8 2 5 2 1 8 2 8 2 6 4 1 9 2 9 3 7 2 4 9 9 9 2 3 4 9
Here are the weights of the 48 players on the 1998 San Francisco Forty-Niners football team, as well as the 25 players on the San Francisco Giants roster on August 11, 1999. Create two stem-andleaf displays for these two sets of data.
Forty-Niners 290 275 237 195 305 200 250 Giants 190 190 210 210 195 205 185 205 220 205 215 202 225 246 187 180 210 190 180 215 192 202 210 218 185 286 310 250 215 241 270 168 220 262 275 223 278 258 305 260 285 234 235 213 295 220 200 275 260 197 223 185 276 320 275 185 215 240 275 205 315 223 300 195 200 217
The stem-and-leaf displays are shown here. As we take a look at these stem-andleaf displays, we can see that the two sets of weights are distributed in a much different fashion. The weights of the Giants are much closer together than the weights of the Forty-Niners. It should be no surprise that the typical weight of a football player is higher than the typical weight of a baseball player.
Stem 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 8 5 5 0 3 0 4 0 0 0 0 5 0 0 0 0 5 5 0 5 0 5 1 0 0 5 6 5 5 5 7 0 5 5 7 3 3 3 7 8 2 5 5 5 5 6 8 5 Forty-Niners Stem 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Giants
Frequency Distributions
Score 4050 5060 6070 7080 8090 90100 Frequency 3 2 4 9 7 7
Another graphical tool we can use to analyze data is a frequency distribution. We divide the data into groups called classes, and count how many times each class is represented. A frequency distribution for the 32 test scores from the first example in this section is shown on the left. The first class, 4050, consists of values from 40 up to, but not including, 50. A value of 50 would be counted in the second class, 5060. In the first class, 40 is referred to as the lower class limit, whereas 50 is the upper class limit. The upper class limit is not actually included in the class, but is the boundary point for beginning the next class. The class size for a given class is the distance from its lower limit to its upper limit.
Class size = upper class limit lower class limit
The class size for each class in the previous example is 10 (50 40 = 10, 60 50 = 10, etc.). The class mark for a given class is the midpoint of the class.
Class mark = lower class limit + upper class limit 2
In the previous frequency distribution, the first class mark is 45.
40 + 50 = 45 2
Score 4055 5065 6075 7085 85100
The other class marks are 55, 65, 75, 85, and 95. Now that we have discussed the terminology for frequency distributions Frequency and seen what one looks like, we can focus on the guidelines for creating one. There are two rules that must be followed. First, the classes must be mutually exhaustive. This means that each value in the set of data must belong to one of the classes. Consider the classes in the frequency distribution on the left. There are several values between 40 and 99 that are not accounted for. Suppose you had a value of 50 in your set of data. Which class would that belong in? It does not belong in any of the classes, and this creates a problem. There should be no gaps between your classes. The second rule is that the classes must be mutually exclusive. This Frequency means that two classes may not overlap. Consider the classes in the frequency distribution shown on the left. Several of these classes overlap. Suppose you had a value of 52 in your set of data. That value fits into the first class, 4055, as well as the second class, 5065. Where should we count it? It would be wrong to arbitrarily place it in one of these classes, and it would be wrong to count it twice. The best way to handle this is to prevent overlapping classes to begin with. The best way to satisfy both of these rules is to verify that the lowest value fits into the first class, the highest value fits into the last class, and the upper limit of one class is the same as the lower limit of the next class. This prevents overlaps as well as missing values. There are other suggestions that you may not be able to follow, depending on the set of data. Try to incorporate as many as you can. The first suggestion is about the number of classes. Using too many classes may break down the data too much,
and using too few classes may not break down the data enough. A good genScore eral rule of thumb is to use somewhere between 5 and 20 classes. 4050 Another good idea is to use classes that are of equal size. When we com5060 pare the frequency of one class to another, the classes must be of the same 6075 size for the comparison to be fair. In our first frequency distribution, each 7585 class size was 10. Consider this frequency distribution for the 32 test scores, 85100 shown on the right. These classes do not have equal sizes. The first, second, and fourth classes have a class size of 10, and the third and fifth classes have a class size of 15. This distorts the picture that we got from the first frequency distribution. Although the frequency of the third class seems large, you must keep in mind that it is based on a class size that is 50% bigger than the first, second, and fourth classes. Another suggestion to consider is to avoid using open-ended classes. An openended class is a class that is missing one of its limits. For example, if we were analyzing household incomes, $100,000 and up would be an example of an open-ended class. However, below $20,000 is not an open-ended class because a households income cannot be lower than $0. Therefore, $0 is really the lower limit for that class. The class should be written as $0$20,000. Classes that have class sizes of 5 or 10 are often convenient to work with. It is also convenient to use lower class limits that are multiples of the class size. These suggestions will help you at the beginning, but you will find that creating frequency distributions is very intuitive. Often, there is a natural breakdown for classes; the data will dictate what your classes look like.
Frequency 3 2 7 10 10
Lets look again at the Forbes magazine list of the 400 richest people/families in the world. From the 1999 issue, here are the ages of the 54 richest individuals. All of these individuals have a net worth of at least $5 billion. Construct a frequency distribution for this set of data.
I recommend that you begin with the stem-and-leaf display, which was done earlier in this section. Here it is.
The row whose stem is 3 would correspond to a class of 3040. If it can be done, it is an excellent idea to match your classes to rows from your stem-and-leaf display,
because it then becomes a matter of counting how many values are in each row. There are three values in the first row (3040), seven values in the second row (4050), and so on. Here is the frequency distribution.
Age 3040 4050 5060 6070 7080 8090 Frequency 3 7 15 11 13 5
We could have used the stem-and-leaf display whose stems were split into two rows. Here is that stem-and-leaf display, along with the corresponding frequency distribution.
Stem 3 3* 4 4* 5 5* 6 6* 7 7* 8 8* 1 6 1 6 0 5 0 6 1 5 1 8 4 2 3 3 3 4 0 6 1 8 1 5 2 1 8 2 8 2 5 2 1 8 2 8 2 6 4 1 9 2 9 3 7 4 2 4 9 9 9 2 3 4 9 Leaf Age 3035 3540 4045 4550 5055 5560 6065 6570 7075 7580 8085 8590 Frequency 2 1 6 1 7 8 6 5 7 6 4 1
Score 4050 5060 6070 7080 8090 90100
Relative Frequency (%) 9.4 6.3 12.5 28.1 21.9 21.9
Frequency 4 3 15 26 20 8
One variation on a frequency distribution that can be made is called a relative frequency distribution. The frequencies are converted to a percentage of the total. This allows a comparison of two sets of data that are similar but are of substantially different sizes. It is impossible in such a case to compare the actual frequencies and gain knowledge about the two sets, but comparing percentages levels the playing field. On the left (top) is an example of a relative frequency distribution for the 32 test scores. Note that the relative frequencies do not total to 100% due to rounding. If we had another set of similar scores, we could then compare their distributions. On the left (bottom) is another frequency distribution, this time consisting of 76 scores on the first unit examination in a precalculus class. The relative frequency distribution for both sets of scores follow. Note again that the relative frequencies do not total to 100% due to rounding.
PRECALCULUS SCORES Score 4050 5060 6070 7080 8090 90100 Relative Frequency (%) 5.3 3.9 19.7 34.2 26.3 10.5
STATISTICS SCORES Score 4050 5060 6070 7080 8090 90100 Relative Frequency (%) 9.4 6.3 12.5 28.1 21.9 21.9
We see that the percentage of scores from 60 to 90 is higher for the precalculus exam than the statistics exam, whereas the percentage of scores from 40 to 60 and 90 to 100 is lower. Based on this evidence, is there a major difference between the scores on these two exams? We can only make an educated guess at this point, but later we will be able to make such an inference. Another variation of the frequency distribution is the cumulative frequency distribution. In such a distribution we Score Frequency build a cumulative count of the frequencies. In a cumulative 4050 3 less-than frequency distribution we list how many values are 5060 2 in that class or lower. On the right is an example of a cumu6070 4 lative less-than frequency distribution for the 32 test scores. 7080 9 To obtain the cumulative frequencies, we add our way 8090 7 down the frequency column. For example, the third cumula90100 7 tive frequency is the sum of the first three frequencies. If we look down the right column containing the cumulative frequencies, the 3 represents the number of scores that are in the first class, or in other words, less than 50. The 5 tells us that there are five values in the first two classes, or that there are five values that are less than 60. If a score of 70 is required to pass the test, the value 9 tells us how many students failed to reach that score. If we were looking for the middle of the data set so we could split the data into two equal groups, we would be looking for where the 16th and 17th values are. This cumulative frequency distribution tells us that those values are in the class labeled 7080. We could then estimate that the typical value is somewhere in the upper 70s. We can also create a cumulative greater-than frequency distribution in a similar fashion. This time we add the freScore Frequency quencies from the bottom of the distribution up. Each cumulative frequency will tell us how many values are in that class 4050 3 5060 2 or higher, or how many values are at least as big as the lower 6070 4 class limit of that class. On the right is an example of a cumu7080 9 lative greater-than frequency distribution for the 32 test scores. 8090 7 Taking a look at the cumulative frequency of 27, we see 90100 7 that there are 27 values in the last four classes. In other words, there are 27 test scores that are at least 60. Cumulative frequency distributions can also be done as relative frequency distributions. This will tell us what percentage of the values fit before or after certain values.
Cumulative Less-Than 3 5 9 18 25 32
Cumulative Greater-Than 32 29 27 23 14 7
A histogram is essentially a graph of a frequency distribution. An example of a histogram for the 32 test scores is shown here. We begin by drawing a set of two axes. On the horizontal axis, we label 10 the lower class limits for each class as well as the upper 8 class limit of the last class. These should be scaled precisely, so that none of the bars become too narrow or too 6 wide, which can lead to a deceiving picture. It is not nec4 4 3 essary to put the values below the lower class limit of the 2 2 first class on the horizontal axis, but they have been included here because they are possible test scores. On the 0 vertical axis, we label the frequencies. Since the highest 0 20 40 60 frequency is 9, you must label at least up to 9. Score Frequency
Instead of actual frequencies, you can create a histogram from a relative frequency distribution. Here is the relative frequency distribution (left) for the 32 test scores and the corresponding histogram (right). Relative Frequency (percent)
40 60 Score
This time we had to label the vertical axis past 28.1% instead of 9 as with the previous histogram. Note that the picture looks essentially the same; we are really just looking at a different vertical scale. The picture essentially looks the same as the stemand-leaf display that was used to create the frequency distribution. Here are the stemand-leaf display (left) and the first histogram (right) side by side, with the histogram rotated 90 to show the similarity.
Stem 4 5 6 7 8 9 5 1 4 1 2 0 7 6 5 1 2 1 8 8 4 3 2
40 60 Score 80 100
Now we will take a look at creating a histogram from the beginning.
Once again we use the Forbes magazine list of the 400 richest people/families in the world from the 1999 issue. Here is a frequency distribution of the ages of the 54 richest individuals. All of these individuals have a net worth of at least $5 billion. Construct a histogram for this set of data.
First we must set up the axes. The horizontal axis should have the values 30, 40, 50, 60, 70, 80, and 90, because they are the class limits. On the vertical axis, we must go to a height of 15, because that is the largest frequency.
15 12 Frequency 9 6 3 30 40 50 60 Age 70 80 90
We draw the first bar from 30 to 40, to a height of 3, because its frequency is 3. The next bar, from 40 to 50, is drawn to a height of 7. We draw the next four bars to heights of 15, 11, 13, and 5, respectively. Here is the completed histogram, with the frequencies labeled above each bar.
15 12 Frequency 9
60 Age
Frequency Polygons
A frequency polygon is another way to graph what we see in a frequency distribution. We set up the vertical axis in the same way that we would for a histogram. For the horizontal axis, instead of dealing with the class limits, we begin by finding the class marks. Above each class mark put a point at the height of the frequency for that class. We also put a point on the horizontal axis where the class mark would be if there were one class below the first class, as well as the class mark that would be one class above the last class. At right is an example for the ages of the 54 richest individuals in the world. Note that the graph follows the same shape as the histogram.
12 9 6 3 25 35 45 55 65 75 85 95 Age
Here are the scores of 32 randomly selected statistics students on an exam on Unit 1, arranged in a frequency distribution. Construct a frequency polygon for these data.
10 8 Frequency 6 4 2
30 40 50 60 70 80 90 100 Score
Grades on Exam 1
First we make note of the class marks for this frequency distribution: 45, 55, 65, 75, 85, and 95. Then, on the horizontal axis, we place a point at 35 (one class below the first class) and a point at 105 (one class above the last class). We find the first class mark, 45, on the horizontal axis, and go above it to a height of the first class frequency, 3. We place a point there, and then repeat this for the rest of the class marks. We connect each of the points, from left to right, with line segments to produce the frequency polygon shown on the left.
A bar chart is similar to a histogram, but is used for categorical data. There is no scale to worry about on the horizontal axis, but it is important that each bar is of the same width. Also, the bars do not touch each other; there is space between them. On the left is an example of a bar chart for the grades on the exam on Unit 1, created using Microsoft Excel.
C D Letter Grade
On the San Francisco Giants current roster there are 12 pitchers, 3 catchers, 5 infielders, and 5 outfielders. Create a bar chart for these data.
There will be four bars in this chart: one for pitchers, a second for catchers, a third for infielders, and a fourth for outfielders. Since the highest total is 12, we must be sure to label the vertical axis to a height of at least 12. At left is the completed chart.
The last graphical way to represent data in this section is the pie chart. This type of chart is used for categorical data to show what percentage of the total each category makes up. Here is an example of a pie chart for the letter grade breakdown for 32 students on exam 1 (created using Microsoft Excel).
Grades on Exam 1 F 16%
B 22% C 27%
When creating a pie chart by hand, begin with a circle. The circle represents 100%. Divide the circle into four equal parts, each representing 25%. You can then break down each quarter into five pieces of 5% each. You will end up with a circle that looks like this. 100%
50% The first wedge of the pie, corresponding to your first category, should begin at 0%. Begin the second wedge where the first wedge ends, and so on.
In 1999, according to CommerceNet, 62% of on-line shoppers were men and 38% were women. This is a significant change from 1995, when 81% were men and 19% were women. Create two pie charts to show the breakdown of on-line shoppers by gender.
Gender Breakdown of On-line Shoppers, 1995 Women 19%
Men 81%
We can use Microsoft Excel to assist us with representing a set of data graphically. We will use the Forbes magazine richest individuals data to walk through all of the techniques. Here are the data. 43 73 68 31 75 61 51 68 71 54 60 79 56 58 46 73 77 82 51 66 72 43 75 76 88 62 75 62 34 71 62 68 41 52 72 59 84 59 59 50 59 50 55 43 62 81 74 44 42 36 69 51 82 58
We must put all of the values in a single column to analyze them. Type the first value, 43, in cell A1, the second value, 68, in cell A2, and so on until the last value, 50, is in cell A54. We will begin with a stem-and-leaf display. Excel will not prepare this display for us, but it can sort the values to make our work easier. Choose the Data menu from the toolbar. Then choose Sort, and a dialog box will appear on the screen. Excel asks which column you will be sorting by, and whether you want to sort the values in that column in an ascending or descending order. For the stem-and-leaf display, it is best to sort your values in an ascending order. In the dialog box you have the opportunity to tell Excel whether your column has a row at the top of the column that serves as a title for that column (header row). This will prevent you from sorting the title with your values. Before continuing, note that when sorting with Microsoft Excel, Excel will sort the other columns in your worksheet based on the column you are sorting. This keeps data together for each member of the sample or population that you are analyzing. Once the values are sorted, it is easy to determine what your highest and lowest stem values are. Once you write the stems in place, you may read down the worksheet to fill in the leaves.
Creating a Histogram from a Set of Data
Before creating a histogram, you must decide what you want to use as upper class limits. Since these values range from 31 through 88, a good choice would be to use 40,
50, 60, 70, 80, and 90 as upper class limits. Excel requires us to put the highest possible value for the class as the upper class limit. So the class 3040 has an Excel upper limit of 39. For this set of data, type these upper class limits in column B, putting 39 in cell B1, 49 in B2, . . . , and 89 in B6. To create the histogram, we will use the Analysis ToolPak, which is an add-in to Microsoft Excel. To see if it has already been added, select the Tools menu. If you do not see Data Analysis as one of the choices in this menu, then you have to add in the Analysis ToolPak. From the Tools menu, select Add-Ins. When the dialog box appears, click on the box to the left of Analysis ToolPak so that a checkmark appears, then click OK. Select Data Analysis from the Tools menu. When the dialog box appears, select Histogram and click on OK. Here is how you need to fill out that dialog box.
Input Range A1:A54 Bin Range B1:B6 Click on Chart Output at the bottom of the box.
Once the dialog box is complete, click on OK. Your histogram will be created on a new worksheet. Note that Excel leaves gaps between the bars. We can take care of that by clicking on any of the bars with the right mouse button (right-clicking), and then selecting Format Data Series. When the dialog box appears, click on the Options tab. Change the Gap Width to 0 and click OK. Here is an example of what you should see. Ages of 54 Richest Individuals 16 14 12 10 8 6 4 2 0
69 79 Age
Creating a Histogram from a Frequency Distribution
If you have created a frequency distribution, Excel can assist you in drawing a Age histogram. We will work with a frequency distribution for the Forbes data, 3040 shown on the right. 4050 Go back to the original worksheet by clicking on Worksheet 1 at the bot5060 tom left of the screen. Type the classes in column C, and the frequencies in col6070 umn D. If you include the header row of Age and Frequency, the last frequency 7080 should be in cell D7. Click in any open cell, and then choose the Insert menu 8090 from the toolbar. Then choose Chart, and a dialog box will appear for the Chart Wizard. Select Column as the chart type, and then click on the button labeled Next> to continue to Step 2. For the data range, type A1:B7, and be sure that Columns is selected under Series. Step 3 allows you to put a title on your graph, as well as label the horizontal and vertical axes. When you have completed Step 3, advance to Step 4, where you will be asked whether to include the chart on the same worksheet or
somewhere else. Click on Finish to see your chart. Readjust the Gap Width in the same fashion as for the previous histogram. Here is an example of what you should see. Ages of 54 Richest Individuals 20 15 10 5 0 30 to 40 to 50 to 60 to 70 to 80 to 40 50 60 70 80 90 Age
Creating a Pie Chart
To create a pie chart in Microsoft Excel, we begin by typing the categories in one column and the corresponding frequencies in the next column. We then choose the Insert menu from the toolbar. Then choose Chart, and a dialog box will appear for the Chart Wizard. Select Pie as the chart type, and then click on the button labeled Next> to continue to Step 2. The range of cells in Step 2 begins at the upper cell in the first column and extends to the lower cell in the second column. While working on Step 3, click on the tab labeled Legend, and then unselect Show legend. Click on the Labels tab, and select Show value and percent. Finish Step 4 by deciding where your pie chart should be displayed. Here is what it should look like.
Ages of 54 Richest Individuals 80 to 90 9% 30 to 40 6% 40 to 50 13%
70 to 80 24%
We can use the TI-83 calculator to assist us with representing a set of data graphically. We will use the Forbes magazine richest individuals data to walk through all of the techniques. Here are the data.
We must put all of the values in a single list to analyze them. We begin by clearing list L1. Press STAT , and choose option 4 under Edit, which is ClrList. This will take us to the main screen, where we need to enter the name of the list (L1) that we 1 and then ENTER . Now, to enter want to clear. We do this by pressing 2nd the values in the list L1, press STAT and select option 1 under Edit. In the column underneath L1, begin typing the values. Type the first value, 43, and then press the key. Type the second value, 68, and press the key. Repeat this until you have entered the last value, 50. The TI-83 does not create a stem-and-leaf display, but it can help us by sorting the values. To do this, press STAT and select option 2 under Edit. Put L1 in parentheses after SortA( and press ENTER . To view L1, press STAT , and then press 1 under Edit. Use the key to move your way down the list and write down the leaves next to the appropriate stems. Y= to activate the STAT PLOT To create a histogram for the data, press 2nd window. Here is the screen that you will see.
, which leads to the following screen.
On the first line of the display, Plot1 should be highlighted. On the second line, be sure that the option On is highlighted. To move from one line to the next, simply use the key. To highlight an option, use the and keys to move back and forth and press ENTER when you reach the option that you want to highlight. . Xlist For the type option, highlight the histogram, which is denoted by
should have L1 next to it, and Freq should have 1 next to it. To see the histogram, press ZOOM and then option 9 (ZoomStat). Here is what your histogram should look like.
To examine the classes that the TI-83 has selected, and the frequencies of those classes, press TRACE . The screen should now look like this.
The first class is activated. The lower class limit is 31, the upper class limit is 39.142857, and the frequency is 3. Use the and keys to move from one class to the next.
Construct stem-and-leaf displays for the following sets of data. 1. Number of Republicans in the House of Representatives, from the 194951 session through the 199799 session:
Year 194951 195153 195355 195557 195759 195961 196163 196365 196567 Number of Republicans 171 199 221 203 201 153 175 176 140 Year 196769 196971 197173 197375 197577 197779 197981 198183 Number of Republicans 187 192 180 192 144 143 158 192 Year 198385 198587 198789 198991 199193 199395 199597 199799 Number of Republicans 167 182 177 175 167 176 230 227
(Source: Vital Statistics on Congress 199798, reprinted in the New York Times.)
2. First downs gained by the 31 NFL teams in the first two weeks of the 1999 season:
(continues)
3. 1999 average SAT verbal scores for the 50 states and the District of Columbia:
State AL AK AZ AR CA CO CT DE DC FL GA HI ID IL IN IA KS Verbal Score 561 516 524 563 497 536 510 503 494 499 487 482 542 569 496 594 578 State KY LA ME MD MA MI MN MS MO MT NE NV NH NJ NM NY NC Verbal Score 547 561 507 507 511 557 586 563 572 545 568 512 520 498 549 495 493 State ND OH OK OR PA RI SC SD TN TX UT VT VA WA WV WI WY Verbal Score 594 534 567 525 498 504 479 585 559 494 570 514 508 525 527 584 546
(Source: The College Board.)
4. 1999 average SAT math scores for the 50 states and the District of Columbia:
State AL AK AZ AR CA CO CT DE DC FL GA HI ID IL IN IA KS Math Score 555 514 525 556 514 540 509 497 478 498 482 513 540 585 498 598 576 State KY LA ME MD MA MI MN MS MO MT NE NV NH NJ NM NY NC Math Score 547 558 503 507 511 565 598 548 572 546 571 517 518 510 542 502 493 State ND OH OK OR PA RI SC SD TN TX UT VT VA WA WV WI WY Math Score 605 538 560 525 495 499 475 588 553 499 565 506 499 526 512 595 551
5. Daily volume of the New York Stock Exchange for August 1999 (millions of shares): 649 745 732 739 692 864 789 583 719 859 691 569 699 683 605 684 684 860 836 661 793 682
6. Percentage of high school graduates that took the 1999 SAT for the 50 states and the District of Columbia:
State AL AK AZ AR CA CO CT DE DC FL GA HI ID IL IN IA KS
Percentage of Graduates 9 50 34 6 49 32 80 67 77 53 63 52 16 12 60 5 9
State KY LA ME MD MA MI MN MS MO MT NE NV NH NJ NM NY NC
Percentage of Graduates 12 8 68 65 78 11 9 4 8 21 8 34 72 80 12 76 61
State ND OH OK OR PA RI SC SD TN TX UT VT VA WA WV WI WY
Percentage of Graduates 5 25 8 53 70 70 61 4 13 50 5 70 65 52 18 7 10
7. Time (in minutes) to complete 20 National League baseball games: 166 156 191 149 153 176 161 154 153 172 165 154 141 153 183 166 134 159 176 194 183 196 194 200 167 131 218 193 240 181 162 145 210 188 165 147
8. Time (in minutes) to complete 16 American League baseball games:
Compare this stem-and-leaf display to the previous one for the National League. Can you make any inferences about the length of time that it takes to play a game in the two different leagues? (Compare the typical values and the spread of the data.) For Exercises 9 and 10, explain what is wrong with the stem-and-leaf display. Also explain why the mistake may lead to improper conclusions about the data. 9. Stem
3 6 7 8 9 1 1 0 0 0 3 2 1 1 2 Leaf 7 5 7 8 8 9 4 5 5 5 6 7 7 9 1 4 8 9
10. Stem
3 3* 4 4* 5 6 6* 7 8 0 5 1 5 0 0 5 0 2 1 5 1 5 0 1 8 3 4 7 2 6 1 3 8 8
Leaf 8 3 4 4 6 6 7 9 9 1 2 4 5 6 8 9 9 3
Construct frequency distributions for the following sets of data. For Exercises 1114, you may want to use the stem-and-leaf displays that you created in Exercises 14.
11. Number of Republicans in the House of Representatives, from the 194951 session through the 199799 session:
12. First downs gained by the 31 NFL teams in the first two weeks of the 1999 season: 36 30 42 42 35 22 38 30 28 22 14 42 42 23 44 52 47 32 43 14 28 31 45 35 46 53 34 46 38 29 36
13. 1999 average SAT verbal scores for the 50 states and the District of Columbia:
14. 1999 average SAT math scores for the 50 states and the District of Columbia:
State AL AK AZ AR CA CO CT DE DC Math Score 555 514 525 556 514 540 509 497 478 State FL GA HI ID IL IN IA KS KY Math Score 498 482 513 540 585 498 598 576 547 State LA ME MD MA MI MN MS MO MT Math Score 558 503 507 511 565 598 548 572 546 (continues)
(continued) State Math Score NE NV NH NJ NM NY NC ND 571 517 518 510 542 502 493 605
State OH OK OR PA RI SC SD TN
Math Score 538 560 525 495 499 475 588 553
State TX UT VT VA WA WV WI WY
Compare this frequency distribution to the previous one for the SAT verbal scores. Can you make any inferences concerning the scores on these two different tests? (Compare the typical values and the spread of the data.) 15. Here are the ages of the presidents of the United States at inauguration (from George Washington through Bill Clinton). Construct a frequency distribution and a relative frequency distribution for these ages. 57 50 54 56 61 48 42 61 57 65 51 52 57 52 56 69 58 56 55 64 57 46 51 46 61 54 54 54 49 51 68 51 60 51 47 62 49 55 43 64 55 55
16. Here are the numbers of home runs hit by each major league baseball team during the 1999 season. Construct a frequency distribution and a relative frequency distribution for these values. 230 105 188 209 212 168 151 235 197 193 158 187 212 223 163 203 179 128 176 216 194 162 161 171 145 209 189 244 165 153
17. Here are the numbers of earthquakes of magnitude 7.0 or greater for each year from 1900 through 1998. Construct a frequency distribution and a relative frequency distribution for the number of earthquakes per year. 1900 1901 1902 1903 1904 1905 1906 1907 1908 1909 1910 1911 1912 1913 1914 1915 1916 1917 1918 1919 13 14 8 10 16 26 32 27 18 32 36 24 22 23 22 18 25 21 21 14 1920 1921 1922 1923 1924 1925 1926 1927 1928 1929 1930 1931 1932 1933 1934 1935 1936 1937 1938 1939 8 11 14 23 18 17 19 20 22 19 13 26 13 14 22 24 21 22 26 21 1940 1941 1942 1943 1944 1945 1946 1947 1948 1949 1950 1951 1952 1953 1954 1955 1956 1957 1958 1959 23 24 27 41 31 27 35 26 28 36 39 21 17 22 17 19 15 34 10 15 1960 1961 1962 1963 1964 1965 1966 1967 1968 1969 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 22 18 15 20 15 22 19 16 30 27 29 23 20 16 21 21 25 16 18 15 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 18 14 10 15 8 15 6 11 8 7 13 10 23 16 15 25 22 20 16
(Source: U.S. Geological Survey.)
18. Here are the daily volumes (in millions of shares) of the New York Stock Exchange for the first three months of 1999. Construct a frequency distribution and a relative frequency distribution for these volumes. 892 797 893 705 700 752 727 784 784 789 852 735 718 771 752 707 987 785 916 721 781 835 752 748 863 905 799 815 782 714 831 729 937 871 845 691 740 804 922 930 818 786 876 653 785 845 662 800 728 856 735 669 909 819 934 896 871 742 754 826 762
19. Here are the publishers prices of 50 books, rounded to the nearest dollar, on USA Todays list of best-selling books (8/26/99). Construct a frequency distribution for these prices, as well as cumulative less-than and greater-than frequency distributions. 17 14 14 7 7 7 10 20 22 8 18 8 8 13 22 9 14 15 5 7 26 23 23 24 18 8 5 25 12 22 8 14 13 5 12 24 7 8 7 7 4 28 5 11 4 8 13 8 6 14
20. Here are the high temperatures, in degrees Fahrenheit, reported in the New York Times for 68 foreign cities on August 15, 1999. Construct a frequency distribution for these temperatures, as well as cumulative less-than and greater-than frequency distributions. 93 42 70 73 74 72 89 68 99 76 93 72 64 89 92 87 77 69 70 111 75 60 81 90 88 91 83 89 93 62 65 71 99 91 59 91 87 89 97 85 94 73 71 103 86 89 64 86 70 84 65 89 87 86 85 77 68 65 90 71 69 66 81 68 73 99 73 57
21. Here are the median incomes, in dollars, for the 50 states and the District of Columbia for the years 199798. Construct a frequency distribution for these incomes, as well as cumulative less-than and greater-than frequency distributions.
State AL AK AZ AR CA CO CT DE DC FL GA HI ID IL IN IA KS Income ($) 34,351 49,717 35,170 27,117 40,623 45,253 45,589 42,581 32,895 33,935 37,950 41,199 35,302 42,552 39,613 35,664 36,875 State KY LA ME MD MA MI MN MS MO MT NE NV NH NJ NM NY NC Income ($) 35,113 32,757 34,461 48,714 42,511 40,583 45,576 29,031 38,662 30,622 35,823 39,608 43,297 49,297 31,049 36,875 36,118 State ND OH OK OR PA RI SC SD TN TX UT VT VA WA WV WI WY Income ($) 31,229 37,811 32,783 38,447 38,558 38,012 34,031 31,471 32,602 35,702 43,870 37,485 43,490 46,339 27,310 40,769 34,597
(Source: Census Bureau.)
22. Here are the average number of students per elementary school for the 50 states and the District of Columbia for the years 199697. Construct a frequency distribution for these averages, as well as cumulative less-than and greater-than frequency distributions.
Number of Students 489 357 560 394 629 422 465 567 413 783 661 623 365 427 445 288 290 Number of Students 423 494 262 551 437 433 434 522 392 187 172 595 329 456 399 614 548 Number of Students 190 425 340 382 488 399 544 178 515 547 548 237 533 452 296 371 201
(Source: U.S. Department of Education National Center for Educational Statistics Common Core of Data Survey.)
23. The following relative frequency distribution is based on a sample of 400 mothers who smoked while pregnant in 1994. Use this to create a standard frequency distribution for the number of cigarettes smoked per day.
Number of Cigarettes 16 611 1116 1621 2131 3141 Relative Frequency 0.237 0.403 0.063 0.250 0.033 0.014
(Source: Centers for Disease Control and PreventionMonthly Vital Statistics Report.)
24. The following relative frequency distribution is based on a total of 195 pro golfers, and shows the number of rounds of golf that they played in 1997. Use this to create a standard frequency distribution for the number of rounds of golf played that year. (Note that relative frequencies do not add to 1 due to rounding.)
Number of Rounds 4555 5565 6575 7585 8595 95105 105115 115125 Relative Frequency 0.062 0.108 0.169 0.221 0.205 0.205 0.021 0.010
25. The following is a cumulative less-than frequency distribution of SAT verbal scores of 200 students. Use this to create a standard frequency distribution for the SAT verbal scores of these students.
SAT Verbal Score 200300 300400 400500 500600 600700 700800
Cumulative Frequency 6 32 94 158 190 200
26. The following is a cumulative greater-than frequency distribution of the IQ scores of 368 college students. Use this to create a standard frequency distribution for the IQ scores of these students.
IQ Score 8595 95105 105115 115125 125135
Cumulative Frequency 368 298 207 97 30
Construct histograms for the following sets of data. For Exercises 2730, you may want to use the frequency distributions that you created in Exercises 1114. 27. Number of Republicans in the House of Representatives, from the 194951 session through the 199799 session:
Year 194951 195153 195355 195557 195759 195961 196163 196365 196567
Number of Republicans 171 199 221 203 201 153 175 176 140
Year 196769 196971 197173 197375 197577 197779 197981 198183
Number of Republicans 187 192 180 192 144 143 158 192
Verbal Score 561 516 524 563 497 536 510 503 494 499 487 482 542 569 496 594 578
Math Score 555 514 525 556 514 540 509 497 478 498 482 513 540 585 498 598 576
31. Here are the weights of 40 men before participating in a weight-loss trial. Construct a histogram for these weights. 157 181 174 174 176 169 178 161 179 188 182 175 183 162 175 175 170 155 178 167 184 171 172 180 182 197 178 186 170 170 174 173 160 155 166 168 180 188 180 174
32. Here are the scores of 82 students on the exam for Unit 1. Construct a histogram for these scores.
33. Here are the times, in seconds, of the 66 songs on Bruce Springsteens Tracks. Construct a histogram for these times. 267 168 211 266 138 124 277 172 231 223 190 126 211 291 158 197 240 109 280 245 268 118 156 257 231 306 206 179 258 227 291 253 199 167 274 275 269 268 264 213 227 295 213 180 219 286 117 236 356 224 164 442 204 158 264 195 201 165 174 185 505 231 201 185 314 259
34. Here are the number of complaints, by state, about nursing homes received in 1997. Construct a histogram for the number of complaints that each state received.
Number of Complaints 998 114 621 1,625 17,764 7,669 231 1,043 5,199 5,029 3,274 3,952 6,638 4,731 1,792 254 4,552
Number of Complaints 5,314 1,992 499 2,260 10,438 2,674 2,953 661 7,324 1,174 1,453 7,158 593 2,192 9,171 5,483 2,457
Number of Complaints 546 4,016 2,164 2,950 4,641 686 1,896 541 1,332 9,495 1,958 330 406 1,670 1,043 3,314 733
(Source: 1997 preliminary figures, U.S. Department of Health and Human Services, Administration on Aging, USA Today.)
35. Here are the 1999 average SAT verbal scores for the 50 states and the District of Columbia. Use the histogram created in Exercise 29 to create a frequency polygon for these scores.
36. Here are the 1999 average SAT math scores for the 50 states and the District of Columbia. Use the histogram created in Exercise 30 to create a frequency polygon for these scores.
37. Here are the number of home runs allowed by each of the 30 major league baseball teams during the 1999 major league season. Construct a frequency polygon for these totals.
AL Team Anaheim Baltimore Boston Chicago Cleveland Detroit Kansas City Minnesota Home Runs 177 198 160 210 197 209 202 208 NL Team Arizona Atlanta Chicago Cincinnati Colorado Florida Houston Los Angeles Home Runs 176 142 220 188 237 171 128 192 (continues)
(continued) AL Team New York Oakland Seattle Tampa Bay Texas Toronto
Home Runs 158 160 191 172 186 191
NL Team Milwaukee Montreal New York Philadelphia Pittsburgh San Diego San Francisco St. Louis
Home Runs 213 152 167 212 160 193 194 161
38. Here are amounts spent by each state per welfare case in 1998. Construct a frequency polygon for these amounts.
State AL AK AZ AR CA CO CT DE DC FL GA HI ID IL IN IA KS Spending ($) 4,034 6,393 5,637 4,206 5,279 6,574 5,654 4,267 4,355 5,190 4,344 5,807 17,624 3,423 5,212 5,226 7,326 State KY LA ME MD MA MI MN MS MO MT NE NV NH NJ NM NY NC Spending ($) 3,444 3,508 5,096 4,817 6,917 6,268 5,530 3,764 3,613 6,415 4,339 4,322 6,129 4,946 6,054 7,252 4,073 State ND OH OK OR PA RI SC SD TN TX UT VT VA WA WV WI WY Spending ($) 8,061 5,189 6,126 8,826 5,330 4,942 3,952 5,535 3,440 3,436 7,329 6,429 3,705 5,200 6,350 21,674 17,272
(Source: New York Times analysis of data from the Department of Health and Human Services.)
39. The Atlanta Braves made the playoffs eight times in the 1990s, the Cleveland Indians and the New York Yankees each made the playoffs five times, and the Boston Red Sox made the playoffs four times. Draw a bar chart to display these data. 40. Here are the letter grades of 82 students on the exam for Unit 1. Draw a bar chart to display these data.
A 15 B 27 C 22 D 8 F 10
41. A community college offered the following number of sections of math classes this semester. Draw a bar graph to display these data.
Beginning Algebra 21 Intermediate Algebra 23
Prealgebra 28
42. Here are the average annual salaries of five occupations in 1998, from the American Federation of Teachers. Draw a bar chart to display these data.
Attorney $72,000 Engineer $64,000 Systems Analyst $63,000 Accountant $46,000 Teacher $39,000
43. According to the International Labor Organization, here are the annual number of hours worked per person in five selected countries. Draw a bar chart to display these data.
United States 1966 Japan 1889 Canada 1732 Britain 1731 Norway 1399
44. Here are the market shares of leading personal computer makers for the third quarter of 1998 and 1999, according to Dataquest (New York Times). Construct two bar charts to show the distribution of the PC market share for these companies.
Market Share (%) (3rd Quarter 1998) 13.4 8.2 8.4 5.9 3.8 60.3 Market Share (%) (3rd Quarter 1999) 12.8 10.8 7.6 6.2 4.3 58.3
Company Compaq Dell IBM Hewlett-Packard Gateway Others
45. According to a Bureau of Justice statistics study in 1999, 49% of civil trials decided by a jury are won by the plaintiff and 51% are won by the defendant. In contrast, when a judge decides a civil case, 62% are won by the plaintiff and 38% are won by the defendant. Construct a pie chart showing the breakdown for cases decided by a jury. Construct a second pie chart showing the breakdown for cases decided by a judge. 46. According to The College Board, during the 199899 school year 70% of high school graduates in Rhode Island took the SAT college admissions test, and 30% did not. For the same school year only 10% of the high school graduates in Wyoming took the test. Construct a pie chart showing the percentage of students in Rhode Island who took the test. Construct a second pie chart showing the breakdown for Wyoming students. 47. A USA Today/CNN/Gallup poll asked What bothers you the most on network TV? Here are the results.
Response Violence Sexual situations Lewd and profane language Do not watch TV No opinion Percentage 44 22 23 7 4
Construct a pie chart showing the percentages for each response. 48. A USA Today/CNN/Gallup poll asked Have you ever been shocked by something you saw on TV? Here are the results.
Response Yes No No opinion Percentage 52 45 3
Construct a pie chart showing the percentages for each response. 49. According to the National Restaurant Association, Saturday is the busiest day for restaurants. For each day of the week, here is a percentage breakdown of business for restaurants.
Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday
Percentage 14.2 11.3 11.9 12.6 12.6 18.1 19.2
Construct a pie chart showing the percentages for each day of the week. 50. In a study of women age 2549 by Yankelovich Partners for Leggs, women were asked how rested they are on waking up. Here are their responses.
Response Extremely refreshed and well rested Fairly well rested Somewhat well rested Not very well rested Not rested at all Percentage 8 42 32 13 5
Construct a pie chart showing the percentages for each response. 51. A 1999 CNN/Time poll of 516 potential Democratic primary voters in New Hampshire revealed that 227 preferred Bill Bradley for president, 212 preferred Al Gore, and 77 were undecided. Construct a pie chart showing the percentages for each candidate, as well as the percentage undecided. 52. A survey of 150 fourth- and fifth-graders about their favorite breakfast food produced the following results.
Breakfast Food Cereal Eggs Pancakes Toast Doughnuts Toaster pastries Other Number of Kids 63 33 21 12 9 9 3
Construct a pie chart showing the percentages that favor each breakfast food.
In this mini-project, we will use the data gathered in the mini-project of Section 1.1. Construct frequency distributions and histograms for
Age Male height Female height Number of units enrolled in this term Number of hours worked per week (outside of class)
Use these to make an estimate for a typical value for each set of values. Create pie charts showing the percentage breakdown for the following.
Gender Eye color Left-handed or right-handed Major Smoker or nonsmoker
n this chapter we will examine two different ways to describe a set of data numerically. First we will describe a typical value for a set of data using measures of central tendency. These measures will tell us where the set of data is centered. The most important measure is the mean, which is the arithmetic average. We will also describe how the data values are spread out using measures of dispersion. This is important because two sets of data may have the same typical value, but are completely different sets due to the way their values are distributed.
In this section we will develop ways to find numerical values that we can say are typical values for a set of data. These values are called measures of central tendency, because they tell us where the data are centered. It makes sense that a typical value for a set of data will be somewhere in the middle of the set.
The first measure that we examine is the mean. The mean is the arithmetic average for a set of data. Recall that to find the arithmetic average for a set of values we add up all the values and then divide that total by how many values there are. We denote the mean of a sample with the symbol x, which is read x-bar. We denote the mean of a population with the Greek letter , or mu (pronounced mew, the sound you would get if you crossed a cat with a cow). This is the lower case m in the Greek alphabet (m for mean). We will use our alphabet whenever we refer to a sample statistic, and we will use the Greek alphabet whenever we refer to a population parameter. Recall that the field of inferential statistics makes inferences about population param eters based on sample statistics. For example, we will be using x to estimate . and is the same: total the values and divide by how The way that we calculate x many values there are. Here is the formula for x . Sample Mean
x x = n
The symbol is the capital Greek letter sigma and is used to represent a summation, so x is the sum of all values. The letter n is the number of values in the sample and is referred to as the sample size. Here is the formula for . Population Mean
= x N
These formulas are the same, with the exception that we use N for the size of a population.
A manager of a local restaurant is interested in the number of people who eat there on Fridays. Here are the totals for nine randomly selected Fridays:
Find the mean for this set of data.
2.1 Measures of Central Tendency
Does this set of data represent a population or a sample? The manager is interested in the number of patrons on Fridays in general, not only on these nine Fridays, so these data represent a sample. We sum the values and get a total of 5571, and divide by 9 to find the mean.
x x = n = 5571 9
The mean for this set of data is 619 people.
It is a good idea to verify that the mean is a typical value. Does 619 customers seem like a typical value for this set of data? It may help to look at the values after they have been put in order from smallest to largest. 526 532 596 600 626 642 655 682 712
The value 619 seems to fit nicely in the center of these data.
Here are the rushing totals (in yards) for Barry Sanders in the years 1989 through 1998. Find the mean rushing total for these ten seasons.
This time we are looking for a population mean.
= = x n 15,269 10
= 1,526.9
Barry Sanders mean rushing total for these ten seasons is 1526.9 yards.
It is important to learn how to use technology, whether it is your calculator or a computer software package, to enter a set of data and use the built-in functions to calculate the mean. Other measures that we will need to calculate later in this chapter and in the next chapter can be easily found once all of the data have been entered. As the sets of data get larger in size, many of the calculations become very tedious, and technology will be your friend. The mean is the most frequently used measure of central tendency. It is the most reliable measure for inferential purposes. Later in the text, we will be using a sample mean x to make an inference about a population mean . If many samples are taken from a population, the sample means vary from the population mean less than other measures of central tendency. Another strength of the mean is that it uses every value in the set of data in its calculation. It should make sense that if we are looking for a value to represent an entire set of data, it should be based on all of the values. We shall see later that other measures of central tendency do not have this property. Other strengths of the mean
CHAPTER 2 Measures of Central Tendency and Dispersion
that other measures do not have include the fact that the mean always exists (we can always calculate the mean of a set of data) and that the mean is unique (a set of data cannot have more than one mean). The mean does have a weakness in that it is sensitive to outliers. An outlier is a value that is located far away from the rest of the set of data. Here is an example that shows this weakness.
Tracy, a real estate agent, sold six homes last week. Here are their sale prices. Find the mean sale price for these six homes.
We are looking for a population mean this time, because there is no suggestion that we are interested in anything other than these six houses.
= = x n 3,000,000 6
= 500,000
The mean sale price is $500,000.
Is the above value typical for the set of sale prices? No, it is off by approximately $400,000 for five of the houses and it is off by $2,000,000 for the other house. If each homeowner was given $500,000 for his or her house there would be five very happy people, and one person would be quite upset. The problem here is the outlier of $2,500,000. The other five prices are grouped closely together, but this one price is substantially higher. The one high value raises the sum of the values and consequently inflates the mean. An outlier that is significantly lower than the other values would make the mean too small to be typical for the set of data. If that value was omitted from the list, the remaining homes would have a mean sale price of $100,000, which would be a typical value for those homes. We cannot simply omit values that are not convenient. There is a measure that works well in cases where outliers are likely, and it is called the median.
The median of a set of data is a value that divides the set of data into two equal groups, after the values have been put in order from lowest to highest. Where do we find the median of a road? In the center of the road. Two scenarios could occur when finding a median. If there are an odd number of values in the set, there will be one value in the center of the data, and this value is the median. However, if there are an even number of values in the set of data, there is not a single value in the center. In this case, we take the mean of the two values in the center.
Find the median for this set of data. The first step is to arrange the values in ascending order:
Now we need to divide the set into two equally sized groups. Since there are nine values in the set, we have two groups of four values, with one value left over in the middle.
That one value left over in the middle, 626, is the median for this set. Recall that the mean for this set of data was 619. Both values seem typical for this set of data.
Here are the rushing totals (in yards) for Barry Sanders in the years 1989 through 1998. Find the median rushing total for these ten seasons.
Again, we begin by placing the values in ascending order. Since there are ten values, we will have two groups of five values, with no values left over.
To find the median, we need to find the mean of the two values in the middle, 1491 and 1500.
Median = 1491 + 1500 2
= 1495.5
The median for this set of data is 1495.5 yards. This is fairly close to the mean, which was 1526.9 yards.
Now we will reexamine the example involving home sale prices, which included an outlier.
Tracy, a real estate agent, sold six homes last week. Here are their sale prices. Find the median sale price for these six homes.
We begin by putting the values in ascending order. We get two groups of three values, with no values left over in the center.
The median is equal to the mean of $99,000 and $106,000.
Median =
99,000 + 106,000 2
The median sale price is $102,500. This value is typical of all of the values except for the outlier, and thus describes the set well.
The median handles the problem that outliers create. It also always exists, and it is unique like the mean. It does not use all of the values in its calculation, but instead uses only one or two of the values. The median is used often in situations that produce outliers, such as home prices. Another situation that usually involves the median is household income, which lends itself to high outliers.
The median breaks a set of data into two halves. If we desire to break the set of data into quarters, the appropriate measures are quartiles. The first quartile, denoted Q1, is the value that separates the first quarter of a data set from the rest. The third quartile, denoted Q3, is the value that separates the last quarter of a data set from the rest. The median divides a set of data into two equal groups of values. The first quartile is the median of the first group, while the third quartile is the median of the second group.
Here are the SAT math scores for 19 randomly selected students. Find the median, as well as the first quartile and third quartile.
The first step is to put the values in ascending order, and find the median as before. Since there are 19 values, we will have two groups of nine values with one value left in the middle. That middle value is the median.
First Group: Second Group:
400 Median: 540
The median is 470. Now to find the first quartile, we need to find the median of the first group of nine values. This first group will be broken into two groups of four values with one value left in the middle. That middle value is the first quartile, Q1.
The first quartile is 430. This is the score that separates the first quarter of the values from the rest. To find the third quartile, Q3, we repeat the same procedure with the second group of nine values.
The third quartile is 570. This is the score that separates the last quarter of the data from the rest.
Here are the heights in inches of 12 randomly selected college females. Find the median, as well as the first quartile and third quartile.
Here are the heights in ascending order.
We have two groups of six heights, with no value left in the middle. The median will be the mean of the middle two values, 64 and 65.
Median = 64 + 65 2
= 64.5
The median is 64.5. To find the first quartile, we work with the first group of six values.
The first quartile is found by taking the mean of the two middle values, which is 63. We repeat this procedure to find the third quartile by using the second group of six values.
The third quartile is found by taking the mean of the two middle values: 66 and 67. The third quartile is 66.5.
Each year, Forbes magazine prints a list of the 400 richest people/families in the world. From their 1999 issue, here are the ages of the 54 richest individuals. All of these individuals have a net worth of at least $5 billion. Find the median, the first quartile, and the third quartile.
We begin with the stem-and-leaf display, which was done in the previous chapter:
The advantage of using the stem-and-leaf display is that the values are already in ascending order. Since there are 54 values, we will have two groups of 27 values, with no value left between them. We count until we find the 27th value, which ends the first group. This value is 61. The value that begins the second group is 62. The median for this set is the mean of those two values, which is 61.5. To find the first quartile, we use the first group of 27 values. This group can be broken into two groups of 13 values, with one value left in the middle, which will be the first quartile. We count until we find the 14th value, which is the second 51 in the list. The first quartile is 51. To find the third quartile, we use the second group of 27 values, which begins at 62. In a similar fashion, we count until we find the 14th value of that group, which will be the third quartile. The 14th value is the second 73 in the list. The third quartile is 73.
The midrange is a measure of central tendency that is quite easy to calculate. It is found by finding the mean of the lowest value and highest value in the set. The midrange is exactly halfway between the lowest value and the highest value, which often places it in the center of the set of data. However, if there is an outlier the midrange will be pulled away from the center toward that outlier.
Find the midrange for this set of data. The lowest value for this set of data is 526, and the highest value is 712. To find the midrange, we find the mean of these two values.
Midrange = 526 + 712 2
The midrange is 619.
For the previous set of data, the midrange happens to be the same as the mean for this set of data as well. Thus this midrange could be considered a good measure of central tendency. This is not always the case. Consider the following.
Tracy, a real estate agent, sold six homes last week. Here are their sale prices. Find the midrange sale price for the six homes.
The midrange is the mean of the two extreme values: $88,000 and $2,500,000.
Midrange =
88,000 + 2,500,000 2
= 1,294,000
The midrange for this set of data is $1,294,000. This is far from what we would call a central value. The outlier ($2,500,000) has pulled the midrange out of the center toward the outlier.
The midrange for a set of data is sensitive to outliers, just like the mean. However, unlike the mean it does not use all of the values in its calculation. The use of the midrange should be limited to providing a quick estimate of the center of the data. Further analysis of the data is required to get a better idea of the center of the data.
The mode of a set of data is the value(s) that occur most frequently in the set.
Here are the ages of the 42 presidents of the United States at inauguration. Find the mode for this set of data.
The mode for this set of data is 51, because it appears five times in the list.
The mode has a few weaknesses for quantitative data. The first problem is that the mode may not be typical for a set of data. Consider the set of data below. 1 13 2 14 4 15 5 16 6 17 7 19 9 20 10 100 11 100
The mode for this set of data is 100 because it is the only value that appears twice, but it is hardly typical. Simply because a value is repeated the most number of times, there is no guarantee that it is in the center of the values. In the preceding example, the value 100 would be considered an outlier. Another weakness of the mode is that it may not exist. If no value in the set is repeated, then the set of data does not have a mode.
Here are the amounts of vitamin C in a serving of 14 different fruits and vegetables, in milligrams. Find the mode for this set of data.
Since no value is repeated in this set of data, there is no mode.
There is sometimes more than one mode for a set of data. If there are two or more values that are repeated the same number of times, then each value could be a mode.
Each year, Forbes magazine prints a list of the 400 richest people/families in the world. From their 1999 issue, here are the ages of the 54 richest individuals. All of these individuals have a net worth of at least $5 billion. Find the mode age for this set of values.
Again, we begin with the stem-and-leaf display. Since the values are in order, it is easy to see which values are repeating, as well as how many times they repeat.
Stem 3 4 5 6 7 8 1 1 0 0 1 1 4 2 0 1 1 2 6 3 1 2 2 2 Leaf
6 4 5 6 8 8 9 9 9 9 6 8 8 8 9 4 5 5 5 6 7 9
The values 59 and 62 both appear four times, which is more than any other value. There are two modes for this set of data: 59 and 62.
We should use the mode cautiously with numerical data. For nominal-level data, the mode is a good measure because it shows us which category is represented the greatest number of times. For ordinal-level data, the median is a better measure to use because the values can be put in order. The mode will still be helpful for ordinal-level data, but it will not necessarily show us where the center of the data is.
Consider the following example.
A college cross-country team has 12 members. During the offseason, seven of the members run 20 miles per week, four members run 25 miles per week, and one runs 30 miles per week. Find the mean weekly distance for the team during the off-season.
It would be incorrect to simply add 20, 25, and 30, and then divide by 3 to find the mean because there are twelve values, not three. The value 20 needs to contribute to the sum seven times, and the value 25 needs to contribute to the sum four times. Here is the correct calculation.
x x = n = = 20 + 20 + 20 + 20 + 20 + 20 + 20 + 25 + 25 + 25 + 25 + 30 12 270 12
The mean distance for these runners is 22.5 miles per week.
In the previous example, the value 20 has a greater relative importance than the other two values when calculating the mean because it shows up seven times. The weighted mean takes the relative importance of each value into consideration. The value 20 should count seven times, so we say that the weight for the value 20 is 7. The value 25 has a weight of 4, and the value 30 has a weight of 1. Here is the formula for the weighted mean.
w x xw = w
We use a subscript w to denote the weighted mean. In the formula, w represents the weight for the value x. To calculate the weighted mean, we multiply each value by its weight and total those products. We then divide by the sum of the weights.
This time we will use the weighted mean formula.
w x xw = w = = 7(20) + 4(25) + 1(30) 7+4+1 270 12
The mean distance for the runners is 22.5 miles per week.
The weighted mean is useful when we want to find the combined mean of two or more sets of data that are of different sizes. The size of each set is the weight.
A college math department gives a common final in its precalculus classes. There are five different versions of the exam. Here are the results from last semester for each version. Find the combined mean for the five versions.
Version Number of Students Mean
A 37 74.0
B 47 72.7
C 40 72.0
D 37 80.1
E 51 84.2
The weights for this problem are the number of students that took each version.
w x xw = w = = 37(74.0) + 47(72.7) + 40(72.0) + 37(80.1) + 51(84.2) 37 + 47 + 40 + 37 + 51 16,292.8 212
= 76.85
The combined mean for all these exams is 76.85.
The weighted mean can be used to find the mean of two or more percentages as well.
At a community college, 70% of the students who took statistics last year passed the class. At the nearest university, 90% of the students who took statistics last year passed the class. Find the combined percentage of students who passed the class last year at the two schools if 300 students took it at the community college and 700 students took it at the university.
This time the weights are the number of students at the two schools.
w x xw = w = = 300(70) + 700(90) 300 + 700 84,000 1,000
At the two schools, 84% of the students who took statistics passed the class.
Note that the percentage in the previous example was not simply the mean of 70% and 90%. The percentage that passed was closer to 90% than it was to 70% because the weight was much higher for the percentage of students who passed at the university.
Estimating a Mean Using a Frequency Distribution
We can use the idea of the weighted mean to estimate the mean of a set of data, even if we do not have the actual values, but instead have a frequency distribution. Consider the frequency distribution for the test scores of 32 statistics students on Exam 1, shown on the next page in the margin.
There are three scores in the first class. What would you estimate the mean of those three values to be? A safe estimate would be the midpoint, or class mark, of that class, which is 45. If the values in a class are evenly distributed throughout the class, then their mean will be close to the class mark. We will estimate the mean of the next two values to be 55, the mean of the next four values to be 65, and so on. We then find the weighted mean for these individual means. Here is a formula.
frequency class mark xg = frequency
Frequency 3 2 4 9 7 7
The frequency for each class will be the weight, and the class mark will be used as the mean of that class. After finding each class mark, multiply the frequency for each class by its class mark. Total those products, and divide that sum by the total of the frequencies. Here are the calculations necessary to estimate the mean using the frequency distribution for the 32 test scores.
Score 4050 5060 6070 7080 8090 90100 Totals Frequency 3 2 4 9 7 7 32 Class Mark 45 55 65 75 85 95 Frequency Class Mark
135 110 260 675 595 665 2440
2440 xg = 32 = 76.25
We estimate the mean to be 76.25. The actual mean is 75.91, so this is an accurate estimate.
Each year, Forbes magazine prints a list of the 400 richest people/families in the world. From their 1999 issue, here are the ages of the 54 richest individuals arranged in a frequency distribution. Use the frequency distribution to estimate the mean for this set of data.
The class marks are 35, 45, 55, 65, 75, and 85. We multiply each class mark by the frequency for that class, and total those products. To calculate the estimate of the mean, we divide that total by 54, which is the total of the frequencies.
Age 3040 4050 5060 6070 7080 8090 Totals 3360 xg = 54 = 62.22
Frequency 3 7 15 11 13 5 54
Class Mark 35 45 55 65 75 85
Class Mark
Using this frequency distribution, we estimate the mean to be 62.22 years. (The actual mean is 61.35 years.)
We can use Excel to help us calculate quickly measures of central tendency for sets of data. We will walk through an example that demonstrates how to calculate certain measures of central tendency. If you follow the directions carefully, you will reproduce the results shown. You can then adapt these instructions for other sets of data. Following is the set of data that will be used for the example. These are the scores of 57 randomly selected students on a statistics exam. 73 98 92 75 44 78 91 66 64 83 50 99 97 64 71 86 72 87 75 77 99 65 66 90 85 75 81 90 66 91 99 89 95 46 69 63 87 77 92 89 79 94 88 94 84 81 71 63 93 84 56 77 79 85 97 90 76
To begin, type Scores in cell A1. Below that, type the values in cells A2 through A58. It is important that all of the values end up in the same column for the upcoming operations. These cells are the cell range that contains the data we wish to work with, and to refer to them in Excel we will be typing A2:A58. MEAN In cell B2 type Mean, and next to it (C2) type =AVERAGE(A2:A58) AVERAGE is the built-in Excel function to calculate the mean of a list of numbers. It is important to type the equal sign (=) before the word AVERAGE. Inside the parentheses we type A2:A58 because those cells contain the scores. MEDIAN In cell B3 type Median, and next to it (C3) type
=MEDIAN(A2:A58) MEDIAN is the built-in Excel function to calculate the median of a list of numbers. FIRST QUARTILE In cell B4 type First Quartile, and next to it (C4) type =QUARTILE(A2:A58,1) QUARTILE is the built-in Excel function to calculate a quartile of a list of numbers. We will use it again to calculate the third quartile. THIRD QUARTILE In cell B5 type Third Quartile, and next to it (C5) type =QUARTILE(A2:A58,3). SMALLEST VALUE In cell B6 type Minimum, and next to it (C6) type =MIN(A2:A58) MIN is the built-in Excel function to find the smallest value in a list of numbers. LARGEST VALUE In cell B7 type Maximum, and next to it (C7) type =MAX(A2:A58) MAX is the built-in Excel function to find the largest value in a list of numbers. MIDRANGE In cell B8 type Midrange, and next to it (C8) type =(C6+C7)/2 This adds the values in cell C6 and C7, and divides that total by 2. Here is what the top of your worksheet should look like.
Scores 73 91 97 75 85 99 87 88 63 79 98 66 Mean Median First Quartile Third Quartile Minimum Maximum Midrange 79.77 81 71 90 44 99 71.5
ESTIMATING A MEAN USING A FREQUENCY DISTRIBUTION Excel can also help us in estimating the mean of a set of data from a frequency distribution. Here is a frequency distribution for these test scores.
Frequency 2 2 9 13 14 17
Using the same worksheet, type Score in cell D1. Below it, in cell D2, type 4050. Type the remainder of the classes in cell D3 through D7. In the next column, type Frequency in cell E1. Type the respective frequencies in cells E2 through E7, next to their classes. To estimate the mean using a frequency distribution, we multiply the class mark for each class by the frequency for that class, so we will need a column containing the class marks. In cell F1, type Class Marks. Type 45 in cell F2, 55 in F3, . . . , and 95 in cell F7. Now for the multiplication. In cell G1, type Product. In cell G2, type =E2*F2. This multiplies the frequency of the first class by its class mark. In G3 type = E3*F3. Repeat these steps to fill column G through cell G7. To estimate the mean, we need to divide the sum of these products by the sum of the frequencies. In cell D9 type Estimate of Mean, and in E9 type =(SUM(G2:G7))/(SUM(E2:E7)). That portion of the worksheet should look like this:
Score 4050 5060 6070 7080 8090 90100 Frequency 2 2 9 13 14 17 Class Mark 45 55 65 75 85 95 Product 90 110 585 975 1190 1615
Estimate of Mean 80.0877193
The actual mean for this data is 79.77, so our estimate of approximately 80.09 is a very good estimate.
We can use the TI-83 to help us calculate quickly measures of central tendency for sets of data. We will walk through an example that demonstrates how to calculate certain measures of central tendency. The set of data that will be used for the example is the scores of 57 randomly selected students on a statistics exam: 73 98 92 75 44 78 91 66 64 83 50 99 97 64 71 86 72 87 75 77 99 65 66 90 85 75 81 90 66 91 99 89 95 46 69 63 87 77 92 89 79 94 88 94 84 81 71 63 93 84 56 77 79 85 97 90 76
We must put all of the values in a single list to analyze them. We begin by clearing list L1. Press STAT , and choose option 4 under Edit, which is ClrList. This will
take you to the main screen, where we need to enter the name of the list (L1) that 1 and then ENTER . Now, to we want to clear. We do this by pressing 2ND STAT and select option 1 under Edit. In the enter the values in the list L1, press column underneath L1, begin typing the values. Type the first value, 73, and then press the key. Type the second value, 91, and press the key. Repeat this until the last value, 94, has been entered. Be sure that you have entered 57 values before proceeding. We can calculate several measures at the same time. Press STAT and use the key to access the CALC menu, which looks like this.
Press 1 to select the 1-Var Stats option. On the main screen, enter L1 after 1-Var Stats by using the 2ND and 1 keys and press ENTER . The results take up more than one window. The TI-83 will put an arrow pointing downward in the lower left corner of the window to let us know when this happens. The following two screens show all of the information.
We now see that
The The The The The The
mean is approximately 79.77. median is 81. first quartile is 71. third quartile is 90.5. smallest value is 44. largest value is 99. 44 + 99 or 71.5. 2
From this information, we can calculate that the midrange is
Find the mean of the following sets of values. 1. 32 2. 47 57 42 39 46 85 45 74 41 14 49 45 49 62 112 21 42 19 44
3. 84 72 4. 19 50 49 5. 45 50 6. 149 147 7. 60 30 15 98 8. 204 111 181
93 87 18 23 18 58 55 135 127 56 49 55 47 161 107 213
77 72 52 52 51 60 46 144 149 74 86 104 74 67 76 231
89 68 52 53 20 40 56 142 142 63 101 57 33 145 146 239
78 80 20 18 56 43 46 135 127 55 91 57 107 110 194
Find the median and quartiles for the following sets of values.
127 116 105 70 17 245 116 166 23 76 107 182 243 185
9. A sample of five college females produced the following heights (in inches). Find the midrange for these heights. 64 67 65 70 62 10. Here are the scores of six randomly selected professional football teams on a certain Sunday. Find the midrange for these scores. 31 14 25 17 20 6 11. Here are the math exam scores of eight randomly selected students. Find the midrange for these scores. 62 75 72 95 82 86 80 71 12. Here are the speeds of seven pitches thrown by a professional baseball pitcher, in miles per hour. Find the midrange of these speeds. 95 79 88 98 96 94 72 13. Here are the widths of five model year 2000 compact sports utility vehicles (in inches). Find the mode and midrange for these widths.
Make/Model Chevy Tracker (4-door) Isuzu Amigo Jeep Wrangler Suzuki Vitara (2-door) Toyota RAV4 Width (in.) 67.3 71.4 66.7 67.3 66.7
(Source: Consumer Guide Car & Truck Test.)
14. Here are the widths of five model year 2000 midsize sports utility vehicles (in inches). Find the mode and midrange for these widths.
Make/Model Chevy Blazer (4-door) Dodge Durango Jeep Cherokee (4-door) Nissan Pathfinder Toyota 4Runner Width (in.) 67.8 71.5 69.4 69.7 66.5
15. A class for volunteer Braille transcribers has eight students, whose ages are given here. Find the mean and median for these ages. 55 65 59 30 54 41 60 62 16. A study group has six members, who have the following IQs. Find the mean and median for these IQs. 108 104 110 104 92 106 17. Here are the math exam scores of eight randomly selected students. Find the median and mean for these scores. 62 75 72 95 82 86 80 71 18. Here are the speeds of seven pitches thrown by a professional baseball pitcher, in miles per hour. Find the median and mean of these speeds. 95 79 88 98 96 94 72 19. For the taping of a game show, ten potential contestants are present. Here are the ages of these potential contestants. Find the mean and mode for these ages. 46 35 37 38 28 42 58 38 32 39 20. The Registrars Office employs 12 student workers. Here are the number of units each is taking this semester. Find the mean and mode for these amounts. 9 11 14 10 16 12 12 15 12 12 18 11
For the sets of data in Exercises 2134, calculate (a) the mean, (b) the median, (c) the first quartile, (d) the third quartile, (e) the midrange, and (f) the mode. 21. Daily volumes of the New York Stock Exchange for August 1999 (millions of shares): 649 745 732 739 692 864 789 583 719 859 691 569 699 683 605 684 684 860 836 661 793 682
22. Times (in minutes) to complete 20 randomly selected National League baseball games: 166 156 153 176 153 172 141 153 134 159 183 196 167 131 240 181 210 188 165 147
23. Times (in minutes) to complete 16 randomly selected American League baseball games: 191 149 161 154 165 154 183 166 176 194 194 200 218 193 162 145
Year 194951 195153 195355 195557 195759 195961 196163 196365 196567 196769 196971 197173 197375 Number of Republicans 171 199 221 203 201 153 175 176 140 187 192 180 192 Year 197577 197779 197981 198183 198385 198587 198789 198991 199193 199395 199597 199799 Number of Republicans 144 143 158 192 167 182 177 175 167 176 230 227
25. Ages of the presidents of the United States at inauguration (George Washington through Bill Clinton): 57 64 55 62 61 50 55 43 57 48 54 55 57 65 42 56 58 52 51 61 57 56 56 52 61 46 55 69 54 54 51 64 68 49 54 46 51 51 51 49 47 60
26. Market values, in millions of dollars, of the 20 companies making IPOs (initial public offerings) during the week of October 25, 1999: 1538 2275 130 50 1044 621 2131 408 449 51 202 911 346 1157 566 129 283 189 1883 62
(Source: Renaissance Capital in USA Today.)
27. Scores of 20 randomly selected high school sophomores on the PSAT verbal exam: 29 37 23 300 390 530 520 426 524 53 52 60 410 530 700 520 391 483 36 59 42 510 460 530 550 495 531
Company AOL Yahoo Microsoft Lycos Go Network Excite@Home Amazon Time Warner Go2Net Altavista
28. Scores of eight randomly selected high school juniors on the PSAT math exam: 29. Scores of 12 randomly selected high school students on the SAT I verbal exam:
30. Scores of 15 randomly selected high school students on the SAT I math exam: 470 540 355 463 560
31. Point totals of 14 algebra students before the final exam (out of 600 points possible):
32. Number of visitors to ten companies Web sites in August 1999:
Visitors (in millions) 53.43 40.24 35.02 29.38 20.26 16.06 12.54 11.75 11.10 10.22
(Source: Media Metrix; Jupiter Communications New York Times.)
33. Number of shares (in millions) being offered by the 19 companies making initial public offerings (IPOs) during the week of October 18, 1999: 6.0 7.8 4.0 4.6 100.0 42.9 2.9 6.0 3.5 1.5 6.5 6.5 9.0 2.0 3.4 10.0 7.2 3.8 5.5
34. Average number of wet days in September for 34 cities from around the world:
City Athens Beijing Wet Days 4 7 City Bermuda Boston Wet Days 10 9 (continues)
(continued) City Budapest Buenos Aires Cairo Chicago Delhi Dublin Frankfurt Geneva Hong Kong Houston Jerusalem Johannesburg London Los Angeles Madrid
Wet Days 7 8 0 9 4 12 13 10 12 8 0 2 13 1 6
City Moscow New York Paris Phoenix Rio de Janeiro Rome San Francisco San Juan Stockholm Sydney Tokyo Toronto Washington Sydney Tokyo
Wet Days 13 9 13 3 11 5 2 18 14 12 12 12 8 12 12
(Source: Times Books World Weather Guide, New York Times.)
35. Given here are the amounts of calcium (in milligrams) in certain foods.
Dairy Products Plain fat-free yogurt Swiss cheese Chocolate milk shake American cheese Milk Chocolate milk Cheddar cheese Ricotta cheese Ice cream or frozen yogurt Fruits and vegetables Collard greens Soybeans Broccoli, cooked or raw Orange Calcium-fortified orange juice Fish Sardines Oysters, raw Canned pink salmon Miscellaneous Cheese pizza Macaroni and cheese Pancakes made with milk Taco salad Small salad Caffe latte 1 slice 1 2 cup 2 1 1 12 oz. 220 180 72 280 280 412 3 oz. 1 cup 31 oz. 3 370 220 200 1 cup cooked 1 2 cup cooked 1 cup 1 1 cup 300 130 118 52 300 Amount 1 cup 11 oz. 2 10 oz. 2 oz. 8 oz. 8 oz. 11 oz. 2 1 2 cup 1 cup Calcium (mg) 400 408 374 348 300 280 300 300 200
(Source: Rockford Register Star, in Visalia Times Delta)
(a) Find the mean amount of calcium in these 23 items. (b) Find the mean amount of calcium in each of the four subgroups: dairy products, fruits and vegetables, fish, and miscellaneous. (c) If you add up the four means found in (b) and divide by 4, will this produce the same answer as you found in (a)? Why or why not? If not, explain what needs to be done with these four means to produce the same mean found in (a).
36. Here are the amounts of caffeine (in milligrams) from various sources.
Product Coffee Brewed (8 oz.) Instant (8 oz.) Decaffeinated Tea Tea (8 oz., leaf or bag) Arizona Iced Tea (16 oz.) Snapple Iced Tea (16 oz.) Decaffeinated Soft Drinks (12 oz. can) Barqs Root Beer Coca-Cola (classic and diet) Dr. Pepper (regular and diet) Jolt Josta Mellow Yellow Mountain Dew Pepsi (regular and diet) RC Cola Chocolate Hot Cocoa (8 oz.) Hershey Bar (1.5 oz.) Hersheys Special Dark (1.5 oz.) Over-the-counter medications Anacin (2 tablets) Excedrin (2 tablets) NoDoz (1 tablet, regular strength) Caffeine (mg) 135 95 5 50 30 48 5 23 46 41 71 58 52 55 36 36 5 10 31 64 130 100
(Source: National Coffee Drinking Trends Survey, National Coffee Association of USA Inc.; Nutrition Action Healthletter, Center for Science in the Public Interestin USA Today magazine.)
(a) Find the mean amount of caffeine in these 22 items. (b) Find the mean amount of caffeine in each of the five subgroups: coffee, tea, soft drinks, chocolate, and over-the-counter medications. (c) If you add up the five means found in (b) and divide by 5, will this produce the same answer as you found in (a)? Why or why not? If not, explain what needs to be done with these five means to produce the same mean found in (a). 37. Make up a set of data that has at least six distinct values, for which the mean is greater than the median. 38. Make up a set of data that has at least six distinct values, for which the mean is less than the median. 39. Make up a set of data that has at least six distinct values, for which the mean is 75. 40. Make up a set of data that has at least six distinct values, for which the mean is 68.2. 41. Make up a set of data that has at least six distinct values, for which the median is 15. 42. Make up a set of data that has at least six distinct values, for which the median is 17.5. 43. Make up a set of data that has no mode. 44. Make up a set of data that has three modes. Wilt Chamberlain played center for 14 NBA seasons in Philadelphia, San Francisco, and Los Angeles. Among his notable achievements were scoring 100 points in a single NBA
game, averaging over 50 points per game over an entire season, and never fouling out of a basketball game during his entire career. Here are his point totals and rebound totals for those 14 seasons.
Season 195960 196061 196162 196263 196364 196465 196566 196667 196768 196869 196970 197071 197172 197273
Team Philadelphia Philadelphia Philadelphia San Francisco San Francisco San Francisco Philadelphia Philadelphia Philadelphia Los Angeles Los Angeles Los Angeles Los Angeles Los Angeles
Points 2707 3033 4029 3586 2948 2534 2649 1956 1992 1664 328 1696 1213 1084
Rebounds 1941 2149 2052 1946 1787 1673 1943 1957 1952 1712 221 1493 1572 1526
45. (a) Find the mean number of points per season and the median number of points per season. (b) An injury during the 196970 season cut Wilts season dramatically short. If we omit that outlier, find the mean number of points per season, and the median number of points per season for the remaining 13 seasons. (c) Did the mean number of points change significantly from (a) to (b)? Did the median number of points change significantly from (a) to (b)? Explain these results. 46. (a) Find the mean number of rebounds per season and the median number of rebounds per season. (b) An injury during the 196970 season cut Wilts season dramatically short. If we omit that outlier, find the mean number of rebounds per season, and the median number of rebounds per season for the remaining 13 seasons. (c) Did the mean number of rebounds change significantly from (a) to (b)? Did the median number of rebounds change significantly from (a) to (b)? Explain these results. 47. During the 46 years from 1950 through 1995, the state of Utah was not struck by a tornado in 19 of those years. Here is a table that shows the frequency of tornadoes in Utah for these 46 years.
Number of Tornadoes 0 1 2 3 4 5 6
Number of Years 19 8 6 4 3 3 3
Use the weighted mean to calculate the mean number of tornadoes per year in Utah.
48. Here are the number of absences for 20 classes at an elementary school.
Absences 0 1 2 3 4 Number of Classes 11 4 3 0 2
Use the weighted mean to calculate the mean number of students absent. 49. A statistics class had four exams. On exam 1, 32 students had a mean score of 81.3. On exam 2, 30 students had a mean score of 68.6. On exam 3, 24 students had a mean score of 74.2. On exam 4, 23 students had a mean score of 83.5. Use the weighted mean to find the mean score of all four exams combined. 50. Here are the number of rushing attempts and average yards per carry for six NFL teams. Use the weighted mean to find the mean number of yards per carry for the six teams combined.
Team Kansas City Washington Jacksonville Arizona Denver Carolina Number of Attempts 39 24 34 19 30 18 Yards per Carry 2.5 6.8 5.4 4.0 4.5 3.5
51. Here is a frequency distribution showing the scores of 40 students on an algebra midterm exam.
Score 3040 4050 5060 6070 7080 8090 90100 Frequency 4 2 5 7 14 5 3
Use this frequency distribution to estimate the mean score on this exam. 52. Here is a frequency distribution showing the income level of 98 randomly selected 25-year-old males.
Income $0$5,000 $5,000$10,000 $10,000$15,000 $15,000$25,000 $25,000$35,000 $35,000$50,000 $50,000$75,000 $75,000$100,000 Frequency 4 9 11 20 17 17 12 8
Use this frequency distribution to estimate the mean income of these 25-year-old males.
53. Here is a frequency distribution showing the ages of 121 randomly selected people who have a bachelors degree or higher.
Ages 1825 2535 3545 4555 5565 6585 85105 Frequency 8 32 35 23 11 9 3
(Source: Based on the results of a study by the Census Bureau.)
Use this frequency distribution to estimate the mean age of these 121 people. 54. According to the Centers for Disease Control and Prevention, there were 30,903 suicides in the United States in 1996. Here is a frequency distribution for the ages of the 30,879 people whose ages were known. (Twenty-four of the ages were unknown.)
Age 010 1020 2030 3040 4050 5060 6070 7080 80100 Number 4 2115 5357 6521 6050 3667 2729 2667 1769
(Source: USA Today.)
Use this frequency distribution to estimate the mean age of a person who commits suicide. 55. Here are the SAT I composite scores (verbal + math) of 40 high school students. 580 1040 880 930 1060 1040 860 740 1030 1260 920 1010 1130 1000 1240 1200 1230 1120 1160 1030 1240 950 770 810 1030 1110 1110 1200 1170 1440 890 680 1120 910 1170 1000 940 1010 1100 820
(a) Calculate the mean score for these 40 students. (b) Create a frequency distribution for these scores. (c) Use the frequency distribution in (b) to estimate the mean score for these students. Compare your result to (a). Does it seem reasonable? 56. Here are the number of electoral college votes that the 50 states and the District of Columbia had in the 1996 presidential election.
State AL AK AZ AR CA CO CT Votes 9 3 8 6 54 8 8 State DE DC FL GA HI ID IL Votes 3 3 25 13 4 4 22 State IN IA KS KY LA ME MD Votes 12 7 6 8 9 4 10 (continues)
(continued) State Votes MA MI MN MS MO MT NE NV NH NJ 12 18 10 7 11 3 5 4 4 15
State NM NY NC ND OH OK OR PA RI SC
Votes 5 33 14 3 21 8 7 23 4 8
State SD TN TX UT VT VA WA WV WI WY
Votes 3 11 32 5 3 13 11 5 11 3
(a) Calculate the mean number of electoral votes for the 50 states and Washington, D.C. (b) Create a frequency distribution for these scores. (c) Use the frequency distribution in (b) to estimate the mean score for these students. Compare your result to (a). Does it seem reasonable? 57. Here are the number of earned runs allowed by each of the 30 major league teams during the 1999 baseball season. The teams are divided into American League teams and National League teams.
Al Team Anaheim Baltimore Boston Chicago Cleveland Detroit Kansas City Minnesota New York Oakland Seattle Tampa Bay Texas Toronto Earned Runs 762 761 638 787 792 824 845 795 665 760 836 806 809 788 NL Team Arizona Atlanta Chicago Cincinnati Colorado Florida Houston Los Angeles Milwaukee Montreal New York Philadelphia Pittsburgh San Diego San Francisco St. Louis Earned Runs 615 596 835 643 957 781 622 718 815 749 691 788 692 705 764 764
(a) Calculate the mean number of earned runs for the American League teams, and then calculate the mean number of earned runs for the National League teams. (b) Calculate the median number of earned runs for the American League teams, and then calculate the median number of earned runs for the National League teams. (c) Based on your results from (a) and (b), do you feel that the number of earned runs allowed in the American League and in the National League are substantially different? 58. Here are the number of home runs hit by each of the 30 major league teams during the 1999 baseball season. The teams are divided into American League teams and National League teams.
AL Team Anaheim Baltimore Boston Home Runs 158 203 176 NL Team Arizona Atlanta Chicago Home Runs 216 197 189 (continues)
2.2 Measures of Dispersion
(continued) AL Team Chicago Cleveland Detroit Kansas City Minnesota New York Oakland Seattle Tampa Bay Texas Toronto
Home Runs 162 209 212 151 105 193 235 244 145 230 212
NL Team Cincinnati Colorado Florida Houston Los Angeles Milwaukee Montreal New York Philadelphia Pittsburgh San Diego San Francisco St. Louis
Home Runs 209 223 128 168 187 165 163 179 161 171 153 188 194
(a) Calculate the mean number of home runs for the American League teams, and then calculate the mean number of home runs for the National League teams. (b) Calculate the median number of home runs for the American League teams, and then calculate the median number of home runs for the National League teams. (c) Based on your results from (a) and (b), do you feel that the number of home runs hit in the American League and in the National League are substantially different?
Again, use the data collected in the mini-project in Section 1.1. For each of the following, calculate (a) the mean, (b) the median, (c) the first quartile, (d) the third quartile, (e) the midrange, and (f) the mode.
Measures of Dispersion
Although measures of central tendency are important when we are trying to describe a set of data, they are not enough. Two sets of data may be centered in the same location, but still be totally different sets of data. Another important characteristic to consider is how the data are dispersed or spread out. In some sets the values are closely grouped together, while in others they are far apart from each other. Here are the five test scores of two statistics students. Jack Jill 85 72 70 68 55 70 41 65 99 75
The mean score for each student is 70, but the two sets of scores are different. Jills scores are close together; no score is more than 5 points away from her mean score. Jacks scores are more spread out. Two of his scores are 29 points away from his mean score. We can expect Jill to score closer to 70 on the next test than Jack.
The first measure of dispersion that we will discuss is the range. To find the range for a set of data, we subtract the lowest value from the highest value.
Range = highest lowest
Here are the rushing totals (in yards) for Barry Sanders in the years 1989 through 1998. Find the range for these ten seasons.
The highest value is 2053 yards, and the lowest is 1115 yards.
Range = 2053 1115 = 938
The range for this set of data is 938 yards.
The range tells us how far apart the lowest and highest values are. One problem with the range is that it is sensitive to outliers. If a set of data has an outlier, the range uses it in its calculation. Another problem with the range is that two sets of data can be spread out in totally different fashions, but have the same range. Here are the scores of two golfers from last months matches. Jack Greg 71 75 72 73 73 77 71 78 73 78 82 81 70 74 72 71 68 85
The range for both golfers is 14 strokes, but the scores are dispersed in a much different fashion. Jacks scores are closely grouped together between 68 and 73, with an outlier at 82. Gregs scores are evenly dispersed from 71 to 85. The range should be used only as a first step in investigating the dispersion for a set of data. Although it can give us an idea about how spread out the values are, it cannot paint the whole picture for us.
Another measure of dispersion is the interquartile range, the distance between the first and third quartiles.
Interquartile range = Q3 Q1
Instead of telling us how far apart the two extreme values are, it tells us the range in which we can find the middle 50% of the values. It is not sensitive to outliers.
Here are the SAT math scores for 19 randomly selected students. Find the interquartile range.
Recall that the first step is to put the values in ascending order, and find the median. Then to find the first quartile we find the median of the first group. The third quartile is the median of the second group.
First Group Second Group
Median: 470
Recall from the last section that the first quartile is 430, and the third quartile is 570.
570 430 = 140
The interquartile range for these scores is 140 points.
Boxplot
One more way to represent a set of data graphically is with a boxplot. A boxplot needs five values for its construction: the lowest value, the first quartile, the median, the third quartile, and the highest value. These five values are often referred to as the five-number summary for a set of data. Above a horizontal axis, we draw a box from the first quartile to the third quartile. We put a dashed line in the box at the median. Finally, extend out line segments from the box to the lowest and highest values. For the previous 19 test scores, here is the five-number summary.
Lowest 320 Q1 430 Median 470 Q3 570 Highest 710
Here is a boxplot for this set of data.
The range is the distance from the far left to the far right, and the interquartile range is the width of the box.
Each year, Forbes magazine prints a list of the 400 richest people/families in the world. From their 1999 issue, here are the ages of the 54 richest individuals. All of these individuals have a net worth of at least $5 billion. Construct a boxplot for these data.
We will begin with the stem-and-leaf display.
Here is the five-number summary; verify it with the work from the previous section.
Lowest 31 Q1 51 Median 61.5 Q3 73 Highest 88
The range is 57 years, and the interquartile range is 22 years. Here is the boxplot.
Although the range and interquartile range can give us a good idea of a set of datas dispersion, they do have their drawbacks. They do not use all of the data values in their calculation. They measure dispersion from one end to the other, measuring the width between two values. Another way to look at dispersion is by examining how far the values are from the center of the set. For instance, an important characteristic of a new home is its distance from the center of town. We will continue now by examining three measures that determine dispersion in this fashion.
Mean Deviation
To find the mean deviation for a set of data, we begin by finding the distance from each value to the mean. We then find the mean of these distances. This is the mean deviation. Roughly interpreted, it tells us how far from the center of the data the values are on average. A measure of distance is nonnegative, so to calculate each distance we take the absolute value of the difference between the value and the mean. Here is the formula.
Mean deviation = |x x | n
This formula uses sample notation, but the procedure is exactly the same for the mean deviation of a population. Simply replace x by , and n by N. Here are the steps for this calculation. 1. 2. 3. 4. 5. Find the mean. Subtract the mean from each value. Take the absolute value of each difference. Total these distances. Divide the total by the number of values in the set.
A taxi dispatcher is interested in the number of fares for his drivers on Fridays. He randomly selects seven drivers, and then randomly selects one Friday for each of the drivers. Here are the number of fares for each. Find the mean deviation for these totals.
For this calculation it is best to use a column approach.
x 32 27 30 41 29 38 34 x = 33 28 7 x x 1 6 3 8 4 5 1 |x x | 1 6 3 8 4 5 1 28
Mean deviation =
The mean deviation is 4 fares. We can say that on average, the values are 4 fares away from the mean.
We should note one more thing from the previous example. Look at the middle column, labeled x x . The sum of that column is 0. This shows us that the sum of the distances to the right of the mean (positive values) is equal to the sum of the distances to the left of the mean (negative values). This is evidence that the mean truly is in the center of the data. In practice, that middle column is not necessary. Simply find the distance between each value and the mean, and place that in the right column.
Here is a list of the New York Stock Exchanges daily volumes for one week of trading, in millions of shares. Find the mean deviation for these values.
Since there is no suggestion that we are interested in anything except these five values, we will treat this set of data as a population.
x 669 754 752 771 835 = 756.2 187.2 5 x 87.2 2.2 4.2 14.8 78.8 |x | 87.2 2.2 4.2 14.8 78.8 187.2
The mean deviation is 37.44 million shares. On the average, the values are 37.44 million shares away from the mean.
Another measure of dispersion that measures from the inside out is variance. Variance is similar to mean deviation, with two exceptions. The first difference between these two measures is that we square the difference between each value and the mean, rather than taking the absolute value. The other difference is that there are two formulas, depending on whether we are finding the variance of a sample or the variance of a population. Here are the two formulas. Sample Variance
s2 (x x )2 = n 1
Population Variance
2 = (x )2 N
We use the symbol s 2 to represent sample variance, and the symbol 2 to represent population variance. is the lowercase Greek letter sigma, which is the Greek letter s. Note that there is a significant difference in the two formulas. The denominator in the sample variance formula calls for us to subtract 1 from the sample size, whereas the denominator in the population variance uses the population size, without subtracting 1. Why the difference? Although this is beyond the scope of this course, subtracting 1 from the sample size makes the sample variance an unbiased estimator of the population variance. It is crucial that you are able to identify whether a set of data is a sample or a population. Using the sample formula by mistake will produce a variance that is too large. Using the population formula by mistake will produce a variance that is too small.
A taxi dispatcher is interested in the number of fares for his drivers on Fridays. He randomly selects seven drivers, and then randomly selects one Friday for each of the drivers. Here are the number of fares for each. Find the variance for these totals.
Since the dispatcher is interested in all Fridays, these data are a sample. For this calculation, just as with mean deviation, it is best to use a column approach.
x 32 27 30 41 29 38 34 x = 33 s2 = = 152 7 1 152 6 x x 1 6 3 8 4 5 1 (x x )2 1 36 9 64 16 25 1 152
The variance is 25.33.
A drawback for variance is that it lacks the interpretation that mean deviation has. The idea to keep in mind is that the bigger the variance is, the more spread out the values are. An advantage of variance is that its formula does not involve absolute values, which are difficult to manipulate algebraically.
Here is a list of the New York Stock Exchanges daily volumes for one week of trading, in millions of shares. Find the variance for these values.
x 669 754 752 771 835 = 756.2 (x )2 N 14,054.8 5 x 87.2 2.2 4.2 14.8 78.8 (x )2 7,603.84 4.84 17.64 219.04 6,209.44 14,054.8
2 = =
= 2,810.96
The variance is 2,810.96.
The measure of dispersion that we will use most often in this course is the standard deviation. The standard deviation of a set of data is the square root of the variance.
Standard deviation = variance
We use the letter s to represent sample standard deviation, and to represent population standard deviation. Sample Standard Deviation
s = (x x )2 n 1
Population Standard Deviation
= (x )2 N
A taxi dispatcher is interested in the number of fares for his drivers on Fridays. He randomly selects seven drivers, and then randomly selects one Friday for each of the drivers. Here are the number of fares for each. Find the standard deviation for these totals.
In a previous example, we calculated the sample variance for these values to be 25.33.
s = 25.33
The sample standard deviation is 5.03 fares.
In a previous example, we calculated the population variance for these values to be 2810.96.
= 2810.96
The standard deviation is 53.02 million shares.
A student is interested in how old women are when they first get married. To estimate the mean age, she goes into 11 randomly selected chat rooms, and asks randomly selected women how old they were at their first marriage until she gets a response in each room. Here are the 11 ages. Find the standard deviation of these ages.
Since the student is interested in the mean age for all women, these 11 values represent a sample.
x 21 20 19 16 22 21 21 19 18 24 18 x = 19.9 s = (x x )2 n 1 48.91 10 x x 1.1 0.1 0.9 3.9 2.1 1.1 1.1 0.9 1.9 4.1 1.9 (x x )2 1.21 0.01 0.81 15.21 4.41 1.21 1.21 0.81 3.61 16.81 3.61 48.91
The standard deviation is 2.21 years.
On July 15, 1999, Bruce Springsteen and the E Street Band began a series of 15 concerts in Bruces home state, New Jersey. Fans who attended several of the concerts raved about how different the shows were each night. Here is a list of the number of songs played at each of the 15 concerts. Find the standard deviation.
Since there is no suggestion that we are interested in anything beyond these 15 concerts, we will treat this as a population.
x 26 26 24 23 23 23 25 23 22 22 23 22 23 25 24 = 23.6 x 2.4 2.4 0.4 0.6 0.6 0.6 1.4 0.6 1.6 1.6 0.6 1.6 0.6 1.4 0.4 (x )2 5.76 5.76 0.16 0.36 0.36 0.36 1.96 0.36 2.56 2.56 0.36 2.56 0.36 1.96 0.16 25.6
(x )2 N 25.6 15
The standard deviation is 1.31 songs.
As mentioned before, we should try to take advantage of technology (computer, calculator) when calculating standard deviation. When using the calculator, the process is the reverse of what we have shown by hand. We first calculate the standard deviation, and then if the variance is needed we square the standard deviation.
Skewness
Standard deviation is a measure of dispersion that is used often in inferential statistics. We introduce three of its uses here. The first has to do with skewness. A set of data is said to be skewed if it is not symmetrical. Here is a histogram from a set of data that is roughly symmetrical.
Histogram 160 140 120
100 80 60 40 20 0
The peak of the previous histogram is located in the center. A set of data is skewed if we notice that the histogram is stretched to the left or stretched to the right, such as in the following two histograms. Histogram Frequency Frequency 150 100 50 0 Classes 40 30 20 10 0 Classes
A set of data that is stretched to the left is negatively skewed. A set of data is negatively skewed if the values to the left of the median are more spread out than the values on the right side of the median. A low outlier(s) can cause a set of data to be negatively skewed. In such a situation, the mean will be less than the median. (Why?) If a set of data is stretched to the right, we say that it is positively skewed. We have a measure that calculates how skewed a set of data isthe coefficient of skewness. Here is the formula.
sk = 3(mean median) standard deviation
If a set of data is positively skewed, its mean will be greater than its median, causing the coefficient of skewness to be positive. A negative coefficient of skewness indicates that a set of data is negatively skewed. The coefficient of skewness will be a number between 3 and 3. The closer it is to 0, the more symmetric the data are. The farther away from 0 the coefficient of skewness is, the more skewed the data are.
Here are the ages of ten randomly selected women at a college orientation. Find the coefficient of skewness for these data.
The data represent a sample. To calculate the coefficient of skewness we need to find the mean, median, and standard deviation. First the mean.
x x = n = 225 10
Next, find the median.
21 + 22 2
Finally, find the standard deviation. You could use your calculator for this; however, you should also be able to calculate standard deviation through the use of the formula.
x 18 18 19 19 21 22 25 25 27 31 x = 22.5 x x 4.5 4.5 3.5 3.5 1.5 0.5 2.5 2.5 4.5 8.5 (x x )2 20.25 20.25 12.25 12.25 2.25 0.25 6.25 6.25 20.25 72.25 172.5
(x x )2 n 1 172.5 9
Now we can calculate the coefficient of skewness.
sk = = 3(mean median) standard deviation 3(22.5 21.5) 4.38
These data have a coefficient of skewness of 0.68. They are moderately positively skewed. Here is a boxplot to demonstrate the skewness.
Note that the side to the right of the median is more stretched out than the side to the left of the median.
Standard Units
Standard deviation can be used to help us compare two values from two different sets of data. For instance, suppose a man is 6 4 tall, and his IQ is 130. Both of these values are above the mean, but which stands out the most? The mean height of adult males is 69.0 inches (5 9 ), so the man is 7 inches above the mean height. The mean IQ is 100 points, so the mans IQ is 30 points above the mean IQ. We cannot compare the 7 inches to the 30 points, because these are based on two different scales. A fair comparison would be to compare how many standard deviations above the mean these two values are. Finding how many standard deviations away from the mean a value is converts the value to standard units, or its z-score. To convert a value to standard units, we subtract the mean from the value, and then divide by the standard deviation.
z = value mean standard deviation
Adult male heights have a mean of 69.0 inches, with a standard deviation of 2.8 inches. Here is the z-score that corresponds to 6 4 :
z = = value mean standard deviation 76 69.0 2.8
This tells us that the mans height is 2.5 standard deviations above the mean. IQ scores have a mean of 100 points and a standard deviation of 15 points. Here is the standard deviation that is associated with an IQ of 130 points.
z = = value mean standard deviation 130 100 15
This tells us that the mans IQ is 2 standard deviations above the mean. Relatively speaking, the mans height is greater than his IQ. Standard units, or z-scores, are positive for values that are to the right of the mean and negative for values that are to the left of the mean. A z-score of 0 indicates that a value is the mean. The bigger the z-score, the farther away from the mean that value is. We will use z-scores a great deal later in the course.
Here is a list of the New York Stock Exchanges daily volumes for one week of trading, in millions of shares. Convert each volume to its z-score.
Recall that earlier the mean for these values was 756.2 million shares, and the standard deviation was 53.02 million shares.
z = 669 756.2 53.02 z=
754 756.2 53.02 z=
752 756.2 53.02
771 756.2 z = 53.02 = 0.28
In the previous example, the volume of 835 million shares is the farthest from the mean, and therefore has the greatest z-value. Note also that if we total all of the z-values, they add to 0.01. If these values had not been rounded, they would have totaled to 0. This is because the numerators represent the deviations from the mean, which we know total to 0.
Chebyshevs Theorem
Standard deviation can be used to tell us what percentage of the values lie in a certain interval. Our first introduction to this is Chebyshevs Theorem. For example, if k = 2, then CHEBYSHEVS THEOREM For any set of data and any 1 at least 1 22 or 3 of the data must lie within 1 4 constant k > 1, at least 1 k2 of the data must lie within k 2 standard deviations of the mean. What perstandard deviations of the mean on either side. centage of the data must lie within 2.5 standard
deviations of the mean? We apply Chebyshevs Theorem with k = 2.5. At least 1 2.1 2 5 or 84% of the data must lie within 2.5 standard deviations of the mean. Chebyshevs Theorem provides us with a minimum guarantee, but we will see later that if a set of data follows a bell-shaped distribution, then the percentage of values in the interval is much higher than Chebyshevs Theorem tells us. For example, we saw that Chebyshevs Theorem tells us that at least 75% of the data lie within 2 standard deviations of the mean when bell-shaped distributions have approximately 95% of their data within 2 standard deviations of the mean.
The time required to complete a mathematics placement test has a mean of 40 minutes, with a standard deviation of 8 minutes. At least what percentage of students take from 30 minutes to 50 minutes to finish the exam?
This is a problem requiring Chebyshevs Theorem. We begin by finding the value for k. We must figure out how many standard deviations away from the mean our two values are, so we convert them to their z-scores first.
z = 30 40 8
Since each value is 1.25 standard deviations away from the mean, we can use 1.25 for k. At least 1 1.215 2 or 36% of all students take from 30 minutes to 50 minutes to complete the exam.
Estimation of Standard Deviation Using a Frequency Distribution
We conclude this section by learning how to estimate the standard deviation of a set of data if we are given only a frequency distribution. This process is similar to the calculations required to estimate a mean from a frequency distribution. We will estimate the mean first, subtract it from each class mark, square those differences, and then weight these squares according to the class frequencies. We finish by totaling these products, dividing by the appropriate denominator, and taking the square root. Here is the formula for estimating sample standard deviation.
sg = frequency(class mark x g)2 n 1
Given here is a frequency distribution for the scores of 32 randomly selected statistics students on an exam on Unit 1.
We will now estimate the standard deviation for these scores. Keep in mind that this is a sample.
Frequency (f ) 3 2 4 9 7 7 Class Mark (xm) 45 55 65 75 85 95
2440 xg = 32 = 76.25 Frequency (f) 3 2 4 9 7 7 Class Mark (xm) 45 55 65 75 85 95
xm x g 31.25 21.25 11.25 1.25 8.75 18.75
(xm x g)2 976.5625 451.5625 126.5625 1.5625 76.5625 351.5625
(xm x g)2
2929.6875 903.125 506.25 14.0625 535.9375 2460.9375 7350
sg =
7350 31 = 15.40
We estimate the standard deviation to be 15.40. The actual standard deviation of those 32 scores is approximately 14.7, so this is not a bad estimate.
Excel provides numerous built-in functions for calculating measures of dispersion. In each of the following definitions, cell range refers to the range of cells containing data. For example, if you have values in the first 50 cells of column A, type A1:A50 for the cell range. RANGE Excel does not have a built-in function to calculate the range, but we can find it using the built-in functions MAX and MIN. MAX returns the largest value in the range of cells, and MIN returns the smallest value. Type the following in any empty cell.
=MAX(cell range)-MIN(cell range)
If you have already sorted the values in ascending order, the range will be equal to the last value minus the first value.
INTERQUARTILE RANGE Again, Excel does not have a built-in function for calculating interquartile range, but instead we will use QUARTILE to calculate the third and first quartiles and then subtract them. Type the following in any empty cell.
=QUARTILE(cell range,3)-QUARTILE(cell range,1)
MEAN DEVIATION Excel does have a built-in function for this one, AVEDEV. Type the following in any empty cell.
=AVEDEV(cell range)
VARIANCE Excel has two different built-in functions for calculating variance: one for sample variance (VAR) and one for population variance (VARP). To calculate sample variance, type the following in any empty cell.
=VAR(cell range)
To calculate population variance, type the following in any empty cell.
=VARP(cell range)
STANDARD DEVIATION Excel also has two different built-in functions for calculating standard deviation, one for sample standard deviation (STDEV) and one for population standard deviation (STDEVP). To calculate sample standard deviation, type the following in any empty cell.
=STDEV(cell range)
To calculate population standard deviation, type the following in any empty cell.
=STDEVP(cell range)
STANDARD UNITS (z-SCORE) To calculate the z-score using Excel, we use the built-in function STANDARDIZE. This function requires three arguments: the value, the mean, and the standard deviation. If you know the mean and standard deviation, you may type the numbers. If there is a set of data, you may use the built-in functions AVERAGE and STDEV. Type one of the following in any empty cell.
=STANDARDIZE(value, mean, standard deviation)
=STANDARDIZE(cell, AVERAGE(cell range), STDEV(cell range))
COEFFICIENT OF SKEWNESS Excel does not have a built-in function to calculate our coefficient of skewness, but we can use the built-in functions AVERAGE, MEDIAN, and STDEV to calculate it. Type the following in any empty cell.
=3*(AVERAGE(cell range)-MEDIAN(cell range))/STDEV(cell range)
There are different coefficients of skewness. Excels built-in function SKEW calculates the coefficient of skewness defined by the formula
n z 3 (n 1)(n 2)
To use this coefficient of skewness, type the following in any empty cell.
=SKEW(cell range)
ESTIMATING STANDARD DEVIATION USING A FREQUENCY DISTRIBUTION We can also use Excel to estimate the standard deviation of a set of data from a frequency distribution. We will now estimate the standard deviation of a set of test scores based on the following frequency distribution. We will assume that these test scores represent a sample of all students.
Score 4050 5060 6070 7080 8090 90100 Frequency 2 2 9 13 14 17
Before we begin, recall that the formula for estimating sample standard deviation is
Type Score in cell A1. Below it, in cell A2, type 40 to 50. Type the remainder of the classes in cell A3 through A7. In the next column, type Frequency in cell B1. Type the respective frequencies in cells B2 through B7, next to their classes. To estimate the standard deviation using a frequency distribution, we must first estimate the mean. To do that we multiply the class mark for each class by the frequency for that class. We will need a column containing the class marks. In cell C1, type Class Marks. Type 45 in cell C2, 55 in C3, . . . , and 95 in cell C7. Now we want to multiply each class mark by its class frequency. In cell D1, type Product. In cell D2, type =B2*C2. This multiplies the frequency of the first class by its class mark. In D3, type =B3*C3. Repeat these steps to fill column D through cell D7. To estimate the mean, we need to divide the sum of these products by the sum of the frequencies. In cell C9, type Estimate of Mean, and, in D9, type =(SUM(D2:D7))/(SUM(B2:B7)). The next step is to subtract the estimate of the mean from each class mark. In cell E1, type Mark-Mean. In cell E2, type =C2-D9. This subtracts the mean from the class mark of the first class. Continue by typing =C3-D9 in cell E3 to find the difference between the mark of the second class and the mean. Repeat this in cells E4 through E7. We follow this by squaring each of these differences. In cell F2, type =E2*E2. In cell F3, type =E3*E3. Repeat this in cells F4 through F7. The next step is to multiply each of these squared differences by their respective frequencies. In cell G2, type =B2*F2. In cell G3, type B3*F3. Repeat this in cells G4 through G7. Now we will divide the sum of column G by one less than the sum of column B, and take the square root of that. In cell C10, type Standard Deviation. In cell D10,
type =SQRT(SUM(G2:G7)/(SUM(B2:B7)-1)). Here is what the Excel worksheet should look like.
Score 40 to 50 50 to 60 60 to 70 70 to 80 80 to 90 90 to 100 Frequency 2 2 9 13 14 17 Class Marks 45 55 65 75 85 95 Estimate of Mean Standard Deviation Product 90 110 585 975 1190 1615 80.08772 13.51227 Mark - Mean -35.0877193 -25.0877193 -15.0877193 -5.087719298 4.912280702 14.9122807 Squared 1231.148046 629.3936596 227.6392736 25.88488766 24.13050169 222.3761157 Freq* Sq 2462.296091 1258.787319 2048.753463 336.5035396 337.8270237 3780.393967
We can use the TI-83 to help us quickly calculate measures of dispersion for sets of data. This is a continuation of the material that began in Section 2.1. We calculate standard deviation in the same way that we calculated the mean, median, and other measures of central tendency. Here is the set of data that will be used for the example. These are the scores of 57 randomly selected students on a statistics exam. 73 98 92 75 44 78 91 66 64 83 50 99 97 64 71 86 72 87 75 77 99 65 66 90 85 75 81 90 66 91 99 89 95 46 69 63 87 77 92 89 79 94 88 94 84 81 71 63 93 84 56 77 79 85 97 90 76
We must put all of the values in a single list to analyze them. We begin by clearing list L1. Press STAT , and choose option 4 under Edit, which is ClrList. This will take us to the main screen, and we need to enter the name of the list (L1) that we want to clear. 1 and then ENTER . Now, to enter the values in the We do this by pressing 2ND list L1, press STAT and select option 1 under Edit. In the column underneath L1 begin typing the values. Type the first value, 73, and then press the key. Type the second value, 91, and press the key. Repeat this until the last value, 94, has been entered. Be sure that you have entered 57 values before proceeding. We can calculate several measures at the same time. Press STAT and use the key to access the CALC menu, which looks like this.
the sample standard deviation is approximately 13.63, and the population standard deviation is approximately 13.51.
Also, the interquartile range is 90.5 71 or 19.5, and the range is 99 44 or 55. 1 . y= The TI-83 can also create a boxplot for us. Begin by pressing 2ND For Type: select . Your screen should look like this.
Now press ZOOM
and see the following boxplot.
Press TRACE , and you will be able to see the values of the minimum, Q1, median, Q3, and maximum. The following screen shows the median.
14. Here are the scores of six randomly selected professional football teams on a certain Sunday. Find the mean deviation for these scores. 31 14 25 17 20 6 15. Here are the scores of eight randomly selected students on a math exam. Find the mean deviation for these scores. 62 75 72 95 82 86 80 71 16. Here are the speeds of seven pitches thrown by a professional baseball pitcher, in miles per hour. Find the mean deviation of these speeds. 95 79 88 98 96 94 72 17. A sample of five college females produced the following heights (in inches). Find the variance and standard deviation for these heights. 64 67 65 70 62 18. Here are the scores of six randomly selected professional football teams on a certain Sunday. Find the variance and standard deviation for these scores. 31 14 25 17 20 6 19. A class for volunteer Braille transcribers has eight students. Here are their ages. Find the variance and standard deviation for these ages. (Treat the data as a population.) 55 65 59 30 54 41 60 62 20. A study group has six members, with the following IQs. Find the variance and standard deviation for these IQs. (Treat the data as a population.) 108 104 110 104 92 106 21. Here are the scores of eight randomly selected students on a math exam. Find the variance and standard deviation for these scores. 62 75 72 95 82 86 80 71 22. Here are the speeds of seven pitches thrown by a professional baseball pitcher, in miles per hour. Find the variance and standard deviation of these speeds. 95 79 88 98 96 94 72 23. For the taping of a game show, ten potential contestants are present. Following are
26. Amounts of vitamin C (in milligrams) in 16 food sources:
Food Orange juice, fresh Green pepper, raw Grapefruit juice Papaya Brussels sprouts Broccoli, raw Orange Cantaloupe Cauliflower, raw Strawberries Tomato juice Cabbage, raw Blackberries Spinach, raw Blueberries Cherries, sweet
(Source: C for Yourself.)
Serving 1 cup 1 2 cup 1 cup 1 2 medium 4 medium 1 2 cup 1 medium 1 4 medium 1 2 cup 1 2 cup 1 cup 1 2 cup 1 2 cup 1 2 cup 1 2 cup 1 2 cup
Vitamin C (mg) 124 96 93 85 73 70 66 45 45 44 39 21 15 14 10 8
27. Number of goals scored during each season from 197980 through 199899 by Wayne Gretzky (The Great One played 20 NHL seasons in Edmonton, Los Angeles, St. Louis, and New York):
Season 7980 8081 8182 8283 8384 8485 8586 8687 8788 8889 Team Edmonton Edmonton Edmonton Edmonton Edmonton Edmonton Edmonton Edmonton Edmonton Los Angeles Goals 51 55 92 71 87 73 52 62 40 54 Season 8990 9091 9192 9293 9394 9495 9596 9697 9798 9899 Team Los Angeles Los Angeles Los Angeles Los Angeles Los Angeles Los Angeles Los AngelesSt. Louis New York New York New York Goals 40 41 31 16 38 11 23 25 23 9
28. Number of assists in each season from 197980 through 199899 by Wayne Gretzky:
Season 7980 8081 8182 8283 8384 8485 8586 8687 8788 8889 Team Edmonton Edmonton Edmonton Edmonton Edmonton Edmonton Edmonton Edmonton Edmonton Los Angeles Assists 86 109 120 125 118 135 163 121 109 114 Season 8990 9091 9192 9293 9394 9495 9596 9697 9798 9899 Team Los Angeles Los Angeles Los Angeles Los Angeles Los Angeles Los Angeles Los AngelesSt. Louis New York New York New York Assists 102 122 90 49 92 37 79 72 67 53
29. Number of pass receptions by Jerry Rice each season from 1985 through 1996 (Jerry Rice began his NFL career as a wide receiver with the San Francisco FortyNiners in 1985, and currently owns practically every NFL receiving record.)
Receptions 49 86 65 64 82 100 80 84 98 112 122 108
30. Number of yards receiving by Jerry Rice each season from 1985 through 1996
Year 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 Yards 927 1570 1078 1306 1483 1502 1206 1201 1503 1499 1848 1254
31. Per-site averages for the top ten movies during the first weekend of October 1999:
Film Double Jeopardy Three Kings American Beauty Blue Streak The Sixth Sense Drive Me Crazy For Love of the Game Elmo in Grouchland Mystery, Alaska Stigmata
(Source: AC Nielsen EDI Inc. in USA Today.)
Average per Site ($) 7,451 5,387 11,599 2,928 2,490 3,084 1,210 2,690 1,854 1,154
32. Gross value of milk produced in Tulare County, California, per year, for 1989 through 1998 (in millions of dollars):
1989 $287 1990 $363 1991 $413 1992 $411 1993 $455 1994 $477 1995 $547 1996 $569 1997 $712 1998 $718
(Source: Visalia Times-Delta.)
33. Winning mutual payouts for 16 races at Santa Anita Park (for a $2 wager):
34. Number of horses running in 13 races at Santa Anita Park that were restricted to female horses (fillies and mares):
35. Number of complaints about nursing homes, by state, received in 1997:
State AL AK AZ AR CA CO CT DE DC FL GA HI ID IL IN IA KS Complaints 998 114 621 1,625 17,764 7,669 231 1,043 5,199 5,029 3,274 3,952 6,638 4,731 1,792 254 4,552 State KY LA ME MD MA MI MN MS MO MT NE NV NH NJ NM NY NC Complaints 5,314 1,992 499 2,260 10,438 2,674 2,953 661 7,324 1,174 1,453 7,158 593 2,192 9,171 5,483 2,457 State ND OH OK OR PA RI SC SD TN TX UT VT VA WA WV WI WY Complaints 546 4,016 2,164 2,950 4,641 686 1,896 541 1,332 9,495 1,958 330 406 1,670 1,043 3,314 733
(Source: 1997 preliminary figures, U.S. Department of Health and Human Services, Administration on Aging, in USA Today.)
36. Number of earthquakes per year worldwide that were magnitude 7.0 or higher:
Year 1900 1901 1902 1903 1904 1905 1906 1907 1908 1909 1910 1911 1912 1913 1914 1915 1916 1917 1918 1919 Number 13 14 8 10 16 26 32 27 18 32 36 24 22 23 22 18 25 21 21 14 Year 1920 1921 1922 1923 1924 1925 1926 1927 1928 1929 1930 1931 1932 1933 1934 1935 1936 1937 1938 1939 Number 8 11 14 23 18 17 19 20 22 19 13 26 13 14 22 24 21 22 26 21 Year 1940 1941 1942 1943 1944 1945 1946 1947 1948 1949 1950 1951 1952 1953 1954 1955 1956 1957 1958 1959 Number 23 24 27 41 31 27 35 26 28 36 39 21 17 22 17 19 15 34 10 15 Year 1960 1961 1962 1963 1964 1965 1966 1967 1968 1969 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 Number 22 18 15 20 15 22 19 16 30 27 29 23 20 16 21 21 25 16 18 15 Year 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 Number 18 14 10 15 8 15 6 11 8 7 13 10 23 16 15 25 22 20 16
37. Here are the ages of the 16 full-time faculty members in the math department at a community college. Calculate the coefficient of skewness for these ages. Is the skewness positive or negative? Is the skewness moderate or severe?
38. Here are the weights of nine 3-year-old girls in a dance class. Calculate the coefficient of skewness for these weights. Is the skewness positive or negative? Is the skewness moderate or severe?
39. Barry Sanders was an NFL running back with the Detroit Lions for 10 NFL seasons. Here are the number of rushing touchdowns that he scored each season for the 1989 through 1998 seasons. Calculate the coefficient of skewness for the number of touchdowns. Is the skewness positive or negative? Is the skewness moderate or severe?
1989 14 1990 13 1991 16 1992 9 1993 3 1994 7 1995 11 1996 11 1997 11 1998 4
40. Here are the number of shares (in millions) being offered by 19 companies through initial public offerings (IPOs) during the week of October 18, 1999. Calculate the coefficient of skewness for the number of shares. Is the skewness positive or negative? Is the skewness moderate or severe?
(Renaissance Capital in USA Today.)
41. Here are the number of electoral college votes that the 50 states and the District of Columbia had in the 1996 presidential election. Calculate the coefficient of skewness for the number of electoral votes. Is the skewness positive or negative? Is the skewness moderate or severe?
State AL AK AZ AR CA CO CT DE DC FL GA HI ID IL IN IA KS Votes 9 3 8 6 54 8 8 3 3 25 13 4 4 22 12 7 6 State KY LA ME MD MA MI MN MS MO MT NE NV NH NJ NM NY NC Votes 8 9 4 10 12 18 10 7 11 3 5 4 4 15 5 33 14 State ND OH OK OR PA RI SC SD TN TX UT VT VA WA WV WI WY Votes 3 21 8 7 23 4 8 3 11 32 5 3 13 11 5 11 3
42. Here are the median household incomes for the 50 states and the District of Columbia for 199798. Calculate the coefficient of skewness for these median incomes. Is the skewness positive or negative? Is the skewness moderate or severe?
Median Income ($) 34,351 49,717 35,170 27,117 40,623 45,253 45,589 42,581 32,895 33,935 37,950 41,199 35,302 42,552 39,613 35,664 36,875
(Source: Census BureauUSA Today.)
43. Explain, in your own words, why home sale prices would most likely be positively skewed. 44. Explain, in your own words, why household incomes would most likely be positively skewed. 45. If a math instructor gives an easy exam, are the scores more likely to be positively skewed or negatively skewed? Explain your reasoning. 46. If men whose first marriage began 20 years ago were surveyed about the length of their first marriage, are the lengths more likely to be positively skewed or negatively skewed? Explain your reasoning. 47. In fall 1992, a study of part-time instructional faculty in institutions of higher education with regard to the number of students they taught was conducted by the U.S. Department of Education. The following table summarizes the results.
Total Students Taught Less than 25 2549 5074 7599 100149 150 or more Percentage of Part-Time Faculty 36.5 33.5 16.2 6.2 5.1 2.5
Are the numbers of students taught by part-time faculty positively skewed, negatively skewed, or symmetric? Explain your reasoning. 48. A study by the U.S. Department of Education of the ages of college students produced the following table showing the distribution of the ages of full-time students at a two-year institution.
Age Under 18 18 and 19 20 and 21 2224 2529 3034 Percentage of Students 2.1 35.3 21.4 13.1 10.6 6.4 (continues)
(continued) Age 3539 4049 5064 65 and over Age unknown
Percentage of Students 4.6 4.8 1.1 0.2 0.6
Are the ages of full-time students at two-year colleges positively skewed, negatively skewed, or symmetric? Explain your reasoning.
49. Here is the five-number summary for a set of data. Draw a boxplot, and use that boxplot to determine whether the data are positively skewed, negatively skewed, or symmetric. Explain your reasoning.
Lowest 37 Q1 66 Median 71 Q3 74 Highest 85
50. Here are the number of public two-year colleges in 199697 by state, according to the U.S. Department of Education. Construct a boxplot for the data, and use the boxplot to determine whether the data are positively skewed, negatively skewed, or symmetric. Explain your reasoning.
Number of Two-Year Colleges 34 1 20 24 108 15 12 3 0 28 34 7 3 49 14 17 23 Number of Two-Year Colleges 14 43 7 20 18 29 46 22 18 13 9 4 7 19 20 47 58 Number of Two-Year Colleges 9 39 15 17 20 1 21 6 14 67 4 1 24 32 5 19 7
51. Here are the number of colleges and universities in the six New England states. Convert each number to a z-score. Treat the data as a population.
State Connecticut Maine Massachusetts New Hampshire Rhode Island Vermont
(Source: U.S. Department of Education.)
Number of Colleges and Universities 42 34 124 29 12 25
52. Following are the number of home runs hit by Mark McGwire for the 199599 seasons. Convert each total to a z-score. Treat the data as a population.
Year Home Runs 1995 39 1996 52 1997 58 1998 70 1999 65
53. Here are the heights of ten Giant Sequoia trees, in feet. Convert each height to a z-score.
54. Here are the number of doctors degrees awarded to women in 199596 by field for certain fields. Convert each total to a z-score.
Field Agriculture Business Communications Computer and Information Science Education English Mathematics Psychology
Doctors Degrees to Women 336 394 155 126 4151 945 247 2452
What percentage of all z-values lie within 3 standard deviations of the mean? What percentage of all z-values lie within 4 standard deviations of the mean? What percentage of all z-values lie within 1.5 standard deviations of the mean? What percentage of all z-values lie within 1.6 standard deviations of the mean? The heights of 12-month-old boys have a mean of 29.8 inches and a standard deviation of 1.2 inches. What percentage of 12-month-old boys have a height that is between 27.7 inches and 31.9 inches? 60. IQ scores have a mean of 100 points, with a standard deviation of 15 points. What percentage of people have an IQ that is between 80 points and 120 points? 61. The heights of adult males have a mean of 69.0 inches and a standard deviation of 2.8 inches. What percentage of adult males are between 5 6 tall and 6 tall? 62. Scores for high school sophomores on the mathematics portion of the PSAT (Preliminary SAT/National Merit Scholarship Qualifying Test) have a mean of 46.1 and a standard deviation of 10.5. (Scores on the PSAT are whole numbers, with a low of 20 and a high of 80.) What percentage of sophomores score at least 30 points on this test? For Exercises 63 and 64, use the frequency distribution to estimate the standard deviation of the actual values. 63. These are the daily volumes (in millions of shares) of the New York Stock Exchange for the first three months of 1999. (Treat the data as a population.)
Millions of Shares 650700 700750 750800 800850 850900 900950 9501000 Frequency 4 14 16 10 9 7 1
55. 56. 57. 58. 59.
64. These are the high temperatures, in degrees Fahrenheit, reported in the New York Times for 68 foreign cities on August 15. (Treat the data as a sample.)
Temperature (F) 3545 4555 5565 6575 7585 8595 95105 105115 Frequency 1 0 6 22 8 25 5 1
65. These are the number of complaints about nursing homes, by state, received in 1997.
(a) Create a frequency distribution for the number of complaints by state. (b) Use the frequency distribution to estimate the standard deviation of this population. (c) Compare your result in (b) to the standard deviation that you calculated in Exercise 35. Is this estimate a good one? 66. Here are the number of earthquakes worldwide per year that were magnitude 7.0 or higher.
Year 1900 1901 1902 1903 1904 1905 1906 1907 1908 Number 13 14 8 10 16 26 32 27 18 Year 1909 1910 1911 1912 1913 1914 1915 1916 1917 Number 32 36 24 22 23 22 18 25 21 Year 1918 1919 1920 1921 1922 1923 1924 1925 1926 Number 21 14 8 11 14 23 18 17 19 Year 1927 1928 1929 1930 1931 1932 1933 1934 1935 Number 20 22 19 13 26 13 14 22 24 Year 1936 1937 1938 1939 1940 1941 1942 1943 1944 Number 21 22 26 21 23 24 27 41 31 (continues)
(continued) Year Number 1945 1946 1947 1948 1949 1950 1951 1952 1953 1954 1955 27 35 26 28 36 39 21 17 22 17 19
Number 15 34 10 15 22 18 15 20 15 22 19
Number 18 15 18 14 10 15 8 15 6 11 8
Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998
Number 7 13 10 23 16 15 25 22 20 16
(a) Create a frequency distribution for the number of earthquakes per year. (b) Use the frequency distribution to estimate the standard deviation of this population. (c) Compare your result in (b) to the standard deviation that you calculated in Exercise 36. Is this estimate a good one? Dan Marino became a quarterback for the Miami Dolphins in 1983. Here are the number of passes he completed each year, as well as the number of touchdown passes that he threw, for the 19831998 seasons.
Completions 173 362 336 378 263 354 308 306 318 330 91 385 309 221 319 310
Touchdowns 20 48 30 44 26 28 24 21 25 24 8 30 24 17 16 23
67. (a) Calculate the range, interquartile range, and standard deviation for the number of completions that Dan Marino made in these 16 seasons. (b) During the 1993 season, Dan Marino missed a majority of the games due to injury. Remove the 91 completions in 1993 from the list, and recalculate the range, interquartile range, and standard deviation for the number of completions that he made in the remaining 15 seasons. (c) Which of your calculations were greatly affected by deleting that one value, and which ones were not greatly affected? Explain why. 68. (a) Calculate the range, interquartile range, and standard deviation for the number of touchdown passes Dan Marino threw in these 16 seasons.
(b) During the 1993 season, Dan Marino missed a majority of the games due to injury. Remove the 91 completions in 1993 from the list, and recalculate the range, interquartile range, and standard deviation for the number of touchdown passes he threw in the remaining 15 seasons. (c) Which of your calculations were greatly affected by deleting that one value, and which ones were not greatly affected? Explain why.
Again, we will use data collected in the mini-project in Section 1.1. For each of the following, calculate (a) the range, (b) the interquartile range, and (c) the standard deviation.
Using Samples to Estimate a Population Mean EXTRA
On the following pages you will find 500 values, numbered from 1 through 500. The mean of these values is 500, and the standard deviation is 46.899. We will draw samples from this population and examine how close our sample means are to the mean of the population. Here are 12 randomly selected values.
Number 322 51 460 285 55 441 139 446 244 466 184 77 Value 503 485 563 491 542 499 432 450 485 505 560 518
The mean of this sample is 502.75, which is fairly close to the population mean. Will this always be the case? Here are the means of ten samples that were generated in the same fashion.
Sample Number 1 2 3 4 5 Mean 502.75 482.42 494.50 485.58 499.67 Sample Number 6 7 8 9 10 Mean 500.42 485.42 522.83 498.92 522.25
You will notice that some of these means (samples 1, 5, 6, 9) are very close to the population mean, while others (samples 2, 4, 7, 8, 10) are somewhat far away. Thus, an individual sample may vary from the population mean, but look at what happens when we find the mean of these ten sample means. Their mean is 499.476, which is close to 500, and their standard deviation is 14.073. Repeat this procedure for ten samples of size 30, and answer the following questions.
1. List the ten sample means. 2. Does this set of means appear to be closer to or farther away from the population mean than the sample means above? Offer an explanation why. 3. Find the mean of these sample means. Does this appear to be as effective for estimating a population mean? Is the mean approximately close to the mean of the ten previous samples or significantly different? 4. Find the standard deviation of these sample means. Is it lower or higher than the standard deviation for the previous ten samples? Does this suggest that as the sample size increases, the sample means as a group get closer to the population mean?
Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 Value 400 533 499 451 562 498 589 501 452 407 457 448 477 432 469 548 525 485 469 584 511 591 507 495 528 514 518 523 538 513 412 518 527 522 Number 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 Value 496 549 487 488 550 430 485 554 498 481 478 528 555 470 538 523 492 443 604 499 542 488 486 472 527 513 498 510 474 585 504 517 599 396 Number 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 Value 433 516 464 556 526 449 429 506 518 515 433 484 453 438 586 435 457 418 496 493 520 510 533 446 533 524 537 549 436 425 594 430 439 595 Number 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 Value 595 540 514 525 560 469 563 439 474 481 501 421 572 516 516 458 532 478 503 530 474 565 602 571 524 544 420 473 525 505 444 523 494 490 Number 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 Value
516 450 432 459 471 562 515 426 492 545 530 56 463 543 507 494 400 553 459 521 444 527 426 431 424 509 522 434 482 484 452 622 475 450 (continues)
(continued) Number Value 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 449 459 501 428 583 416 518 510 397 442 495 469 518 560 462 407 508 479 523 553 529 500 547 412 527 508 504 437 456 487 465 455 549 469 544 455 633 495 549 494 534 536 477 477 502 480 462 458 536 390 516 511 460 508 428 534
Number 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282
Value 560 587 567 577 490 455 407 519 451 507 601 446 462 448 536 492 449 485 481 514 495 457 449 461 498 572 503 505 570 492 509 523 509 490 473 560 506 506 478 552 494 484 475 470 558 513 461 494 530 526 514 457 508 496 453 553
465 541 603 501 467 544 551 544 531 417 545 564 470 487 500 515 634 462 468 588 588 510 524 500 449 589 493 458 438 412 460 496 548 499 476 535 462 488 520 488 515 471 538 427 538 461 499 506 520 469 571 450 472 498 493 514 (continues)
(continued) Number Value 451 452 453 454 455 456 457 458 459 460 479 533 484 504 521 553 554 497 449 563
Number 461 462 463 464 465 466 467 468 469 470
Value 542 528 507 498 474 505 516 556 501 490
We study two types of statistics, inferential statistics and descriptive statistics. Inferential statistics are techniques and methods that are used to make a generalization, or evaluate a claim, about a population based on sample data. A population is the complete collection of people, objects, or measurements that we are interested in analyzing. A sample is a part of the population that we are interested in. Descriptive statistics are techniques and methods that are used to describe a set of values, both graphically and numerically. Data are information that we have collected. Categorical (qualitative) data are comprised of values that are names or categories. Nominal-level data are a type of categorical data that lack an ordering scheme, while ordinal-level data are a type of categorical data that have an ordering scheme to them (such as low, middle, high). Numerical (quantitative) data can also be broken into two levels: interval-level data and ratio-level data. The difference between these two levels is that interval-level data lack a zero starting point, so ratios between two data values are meaningless. Such ratios are possible only with ratio-level data. Numerical data can also be classified as discrete or continuous. For discrete data, there are gaps between possible data values. Discrete data often involve a count of some type. For continuous data, there are infinitely many possible data values. Continuous data are usually associated with physical measurements. We gather data through experiments and observational studies. In an experiment, we apply a treatment and measure its effect. In an observational study, we simply observe and record data. We select our samples using probability sampling. Different types of sampling include random sampling, systematic sampling, stratified sampling, cluster sampling, and convenience sampling. We can represent a set of data graphically and numerically. One graphic tool that we have is the stem-and-leaf display. We break each number into a stem (first part) and a leaf (last part), and then group the values that have the same stem together. This gives us a rough idea about the center and spread of the data. This display should be the first step in investigating most sets of numerical data. Another graphical tool is the frequency distribution. For a frequency distribution, we divide the data into classes, and count the number of times each class is represented. A histogram is roughly speaking a graph of a frequency distribution. The classes are placed on the horizontal axis, and bars are drawn above them to the height of each class frequency. A pie chart is of great assistance for picturing categorical data. A circle is divided into wedges for each category, to show the percentage breakdown represented by each category. For a set of data, we are often interested in the location of the center of the data, as well as how the data are spread out. Measures of central tendency help us investigate the location of the center of the data. The most important measure of central tendency is the mean. The mean of a sample is the most reliable measure for predicting or estimating the corresponding population parameter. The mean of a set of values is found by dividing the sum of the values by how many values there are. The median is another measure of central tendency. The median is the value for which half of all the values are above it and the other half of the values are below it. The median is especially useful for data that may have outliers, values that are far removed from the majority of values. The first quartile (Q1) and the third quartile (Q3) separate the set of data into four equal parts. Other measures of central tendency include the midrange and the mode. Another important measure for a set of data is a measure of dispersion. Such a measure tells us how spread out the values are. The measure of dispersion that is used the most in this class is standard deviation. There are two different formulas for standard deviation, depending on whether the data represent a sample or a population. Standard deviation is the square root of the variance. Other measures that we employ are the mean deviation, range, and interquartile range.
One simple way to describe a set of data is to use the five-number summary. These five numbers are the lowest value, the first quartile, the median, the third quartile, and the highest value. We can use these five values to construct a boxplot. A boxplot can show where the data are centered, as well as giving a rough idea of where the data are centered. Some uses of standard deviation include z-scores and Chebychevs Theorem. We also use standard deviation in our calculation of the coefficient of skewness. If we are given a frequency distribution, we have the tools to estimate the mean and standard deviation from that frequency distribution.
Unit 1 Review Exercises
1. In a poll of 125 high school seniors, 25 of them felt that having children was "very important. Identify the following statements as descriptive or inferential. (a) Eighty percent of students polled did not feel that having children was very important to them. (b) Twenty percent of all high school seniors feel that having children is very important. (c) In the future, only 20% of people will have children. (d) Many high school seniors will change their feelings about having children before they are 30 years old. 2. You survey the students in each of your classes and find out the amount that each student spent on textbooks this semester. (a) Give an example for which these data would be considered a sample. (b) Give an example for which these data would be considered a population. 3. Identify the following data as categorical or numerical. If the data are categorical, identify whether they are nominal-level data or ordinal-level data. If the data are numerical, identify whether they are continuous or discrete. (a) Number of push-ups that a person can do (b) Year in school (freshman, sophomore, junior, senior) (c) Race/ethnicity (d) Time required to complete an exam 4. You are asked to estimate the number of hours that a student spends studying per week at your school. Explain how you would use the following types of sampling to gather your data. (a) Random sampling (b) Systematic sampling (c) Convenience sampling (d) Stratified sampling (e) Cluster sampling (f) Identify potential biases that may affect the data. Cal Ripken set a major league record by playing in 2424 consecutive games for the Baltimore Orioles between 1982 and 1998. Here are his at-bats (AB), runs (R), hits (H), home runs (HR), runs batted in (RBI), walks (BB), and strike outs (SO). Year 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 AB 598 663 641 642 627 624 575 646 600 650 637 641 R 90 121 103 116 98 97 87 80 78 99 73 87 H 158 211 195 181 177 157 152 166 150 210 160 165 HR 28 27 27 26 25 27 23 21 21 34 14 24 RBI 93 102 86 110 81 98 81 93 84 114 72 90 BB 46 58 71 67 70 81 102 57 82 53 64 65 SO 95 97 89 68 60 77 69 72 66 46 50 58 (continues)
(continued) Year 1994 1995 1996 1997 1998 AB 444 550 640 615 601 R 71 71 94 79 65 H 140 144 178 166 163 HR 13 17 26 17 14 RBI 75 88 102 84 61 BB 32 52 59 56 51 SO 41 59 78 73 68
5. For the number of hits: (a) Construct a stem-and-leaf display. (b) Create a frequency distribution. (c) Make a histogram. 6. For the number of home runs: (a) Calculate the mean. (b) Find the median. (c) Find the first quartile. (d) Find the third quartile. (e) Calculate the midrange. (f) Find the mode. (g) Calculate the range. (h) Calculate the interquartile range. (i) Construct a boxplot. (j) Calculate the standard deviation. (k) Calculate the coefficient of skewness. 7. Here are the market values of oranges grown in Tulare County, California, for the years from 1990 through 1998, in millions of dollars. 1990 $387 1991 $93 1992 $359 1993 $356 1994 $375 1995 $472 1996 $396 1997 $454 1998 $478
(Source: Visalia Times Delta.)
(a) Calculate the mean deviation. (b) Calculate the standard deviation. (c) Calculate the variance. (d) Convert each value to a z-score. 8. SAT I verbal scores have a mean of 505 points and a standard deviation of 111 points. What percentage of students have a score between 330 points and 680 points? 9. The following frequency distribution shows the ages of 480 mathematics instructors at various colleges. Age 2535 3545 4555 5565 6575 Frequency 44 99 197 115 25
(a) Use this frequency distribution to estimate the mean age of those instructors surveyed. (b) Use this frequency distribution to estimate the standard deviation of the ages of those instructors surveyed.
Tying It All Together
A math instructor wrote two versions of the same test. He believed them to be of equal difficulty. He gave the first version to 36 students, and the second version to 41 students. We will consider the two groups to be random samples from the population of all community college statistics
students. Your job is to help the instructor decide whether the two tests were of equal difficulty, or whether one of the exams was harder than the other. Here are the scores of the two versions.
69 71 87 94 82 85 (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) Construct a stem and leaf display for each set of scores. Construct a frequency distribution for each set of scores. Draw a histogram for each set of scores. Construct a pie chart showing the letter grade breakdown for each set of scores. (90100: A; 8089: B; 7079: C; 6069: D; 059: F) Construct a pie chart showing the percentage of students who passed and failed the exam for both sets of scores. (069: Fail; 70100: Pass) Calculate the mean for each set of scores. Calculate the median for each set of scores. Calculate the first quartile and third quartile for each set of scores. Draw a boxplot for each set of scores. Calculate the mode for each set of scores. Estimate the mean for each set of scores using the frequency distribution that you created above. Calculate the range for each set of scores. Calculate the interquartile range for each set of scores. Calculate the standard deviation for each set of scores. Calculate the variance for each set of scores. 84 94 87 80 89 79 70 92 69 76 94 86 69 98 95 85 82 74 96 95 96 91 95 87 72 94 64 77 76 82 79 86 95 91 82
Write a brief essay to answer the question "Were the two exams of equal difficulty?" Use as much of the evidence from above as possible to support your case.
x Sample mean: x = n Population mean: = Midrange: Midrange = x N lowest + highest 2
w x Weighted mean: x w = w
frequency class mark Estimate of a mean using a frequency distribution: x g = frequency Range: Range = highest lowest
Interquartile range: Interquartile range = Q 3 Q1 x x Mean deviation: Mean deviation = n (x x )2 2 Sample variance: s = n 1 Population variance: 2 = (x )2 N (x x )2 n 1 (x )2 N
Sample standard deviation: s =
Population standard deviation: = Coefficient of skewness: sk = Standard units: z =
3(mean median) standard deviation
value mean standard deviation 1 k2
Chebyshevs Theorem: 1
Estimate of a sample standard deviation using a frequency distribution: sg = frequency(class mark x g)2 n 1
unit two
COUNTING AND PROBABILITY
Section 3.1 Counting Section 3.2 Classical Probability Section 3.3 Table Probabilities, Conditional Probability Section 3.4 Independent Events, Multiplication Rule, Law of Large Numbers
nferential statistics is a very important branch of statistics. It involves using information from a sample to draw conclusions about a population. The principle at work behind the scenes of inferential statistics is probability, and that is the main focus of this chapter. The probability of an event occurring is a measure of how likely it is that the event occurs. We have often heard the weatherman say, There is a 40% chance of rain tomorrow. That is a probability. We will begin by studying counting, a tool that is essential to understanding probability.
This section covers the first major area of this chapterfiguring out how many different ways something could happen. Here are some examples of such problems.
In how many different ways could a statistics instructor choose two students from a class of 38 students to give a presentation on measures of dispersion? In how many different ways could a field of 30 figure skaters finish first, second, and third? If a pair of dice is rolled, how many different outcomes are possible? How many pairs will have a sum of 9 or more?
We will use four tools to help us: tree diagrams, the Multiplication Principle, permutations, and combinations. Tree diagrams are introduced to develop intuition and lead to the Multiplication Principle.
A tree diagram is a systematic, graphical way to list all possible outcomes. It looks similar to a family tree. Each level or step in the problem is similar to each generation in a family tree. Following is a tree diagram showing the possible results of three flips of a coin. Each flip could be either heads (H) or tails (T). First Flip Second Flip Third Flip H H T H H T T H H T H T T T
The H and T in the top row represent the possible outcomes of the first flip. If the first flip is heads, there are two possible outcomes for the second flip: heads or tails. These are represented by the two branches on the left side that lead to the second row. Scanning across the bottom row, we count eight branches. This tells us that there are eight different possible outcomes. The first possibility is found by tracing the first branch from top to bottom, and it is heads, heads, heads. The second possibility, from the second branch, is heads, tails, heads. Here are the eight different possible outcomes.
Outcome Number 1 2 3 4 5 6 7 8
First Flip H H H H T T T T
Second Flip H H T T H H T T
Third Flip H T H T H T H T
Number of Heads 3 2 2 1 2 1 1 0
We can now answer questions like How many different ways are there to get exactly two heads? and How many different ways are there to get at least one head?
A tutor can schedule anywhere from zero to two appointments each day. Draw a tree diagram to show in how many different ways she can schedule the next three days. 0 1 2 0 1 2 0 0 1 2 1 0 1 2 2 0 1 2 0 0 1 2 2 1 0 1 2 2 0 1 2
Day 1 Day 2 Day 3 0 0 1 2
There are 27 different ways that she could schedule her next three days.
0 on Day 1 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 0 1 2 0 1 2 0 1 2 Total 0 1 2 1 2 3 2 3 4 1 on Day 1 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 2, 2, 2, 0 1 2 0 1 2 0 1 2 Total 1 2 3 2 3 4 3 4 5 2 on Day 1 2, 2, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0, 1, 1, 1, 2, 2, 2, 0 1 2 0 1 2 0 1 2 Total 2 3 4 3 4 5 4 5 6
We can now answer questions like In how many different ways can she schedule at least four appointments over the next three days? The answer would be 10, because there are 10 totals that are 4 or greater.
If a pair of six-sided dice are rolled, how many different outcomes are possible? The key to this problem is to establish a difference between the two dicefor instance, suppose one of the dice is white and the other is yellow. Then a 2 on the white die and a 3 on the yellow die is a different outcome than a 3 on the white die and a 2 on the yellow die. From here on we will simply refer to the two dice as the first die and the second die. We could draw a tree diagram listing the six different possible outcomes of the first die (16), and then the six different possible outcomes of the second die for
CHAPTER 3 Counting and Probability
each of those six branches. However, the following chart does the trick. Each roll is represented as an ordered pair (x, y), where x is the result of the first die and y is the result of the second die.
Possible Rolls of a Pair of Dice
SECOND DIE 1 FIRST DIE 1 2 3 4 5 6 (1, (2, (3, (4, (5, (6, 1) 1) 1) 1) 1) 1) 2 (1, (2, (3, (4, (5, (6, 2) 2) 2) 2) 2) 2) 3 (1, (2, (3, (4, (5, (6, 3) 3) 3) 3) 3) 3) 4 (1, (2, (3, (4, (5, (6, 4) 4) 4) 4) 4) 4) 5 (1, (2, (3, (4, (5, (6, 5) 5) 5) 5) 5) 5) 6 (1, (2, (3, (4, (5, (6, 6) 6) 6) 6) 6) 6)
There are 36 possible rolls. Often we are interested in the sum of the dice. Here is a chart that helps with the sums.
Possible Sums of a Pair of Dice
SECOND DIE 1 FIRST DIE 1 2 3 4 5 6 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12
How many different ways are there to obtain a sum that is 9 or higher? By counting, we see that there are 10 different sums that are 9 or higher.
If a pair of fair six-sided dice are rolled, how many different ways are there to obtain a sum that is 10 or lower?
From the above chart, we can count that there are 33 different ways to obtain a sum of 10 or lower. Note that the only sums that are not 10 or lower are the two rolls of 11 and the one roll of 12. Subtracting these three rolls from the total of 36 leaves us again with 33 different ways to obtain a sum of 10 or lower.
If a pair of fair six-sided dice are rolled, how many different ways are there to obtain a sum that is between 4 and 10, inclusive?
There are three 4s, four 5s, five 6s, six 7s, five 8s, four 9s, and three 10s. Totaling these, we see that there are 30 different ways to obtain a sum that is between 4 and 10, inclusive.
If a pair of fair six-sided dice are rolled, how many different ways are there to roll doubles?
A roll is doubles if both dice are the same number. There are six different ways to do this: double 1s, double 2s, double 3s, double 4s, double 5s, and double 6s.
If a pair of fair six-sided dice are rolled, how many different ways are there to obtain a sum that is 9 or higher or doubles (or both)?
From the first dice example, we know that there are 10 different ways to obtain a sum that is 9 or higher. Two of those 10 rolls are already doubles (double 5s and double 6s), so we need to add four rolls to the total for the four different doubles that have not been counted yet (double 1s, double 2s, double 3s, and double 4s). This gives us a total of 14 different ways to obtain a sum that is 9 or higher or doubles.
It is rumored that you can get anything you want at Alices Restaurant. If Alices has five different types of coffee (regular, decaffeinated, espresso, cappuccino, and latte) and three different types of pie (apple, blueberry, and cherry), in how many different ways could you order coffee and a slice of pie?
We will use A for apple pie, B for blueberry pie, and C for cherry pie. Here is the tree diagram. Coffee Regular Decaffeinated Espresso Cappuccino Latte
There are 15 different ways to order coffee and a slice of pie at Alices.
One nice thing about tree diagrams is that they list each different possible way something could occur. There are times that we do not need all of that detail, but simply need to THE MULTIPLICATION PRINCIPLE If a choice conknow the total number of different ways. Is there sists of two steps, the first of which can be made in m a quicker, easier way to get the total of 15 differdifferent ways and the second can be made in n different ways, then the whole choice can be made in m n different ent ways out of five types of coffee and three types ways. of pie without using a tree diagram? Here is where the Multiplication Principle comes in. Lets go back to Alices Restaurant to apply the Multiplication Principle.
This is a problem that requires the Multiplication Principle with m = 5 and n = 3.
m n = 5 = 15
There are 15 different ways to order coffee and a slice of pie at Alices Restaurant.
If a problem consists of more than two steps, we can extend the Multiplication Principle. Simply multiply the number of ways for the first step by the number of ways for the second step, then continue on for the third step, and so on.
A community college has 12 math instructors, 14 English instructors, and 5 history instructors. In how many different ways could the president of the college choose a math instructor, an English instructor, and a history instructor to attend an articulation meeting?
There are three steps to this problem. There are 12 different ways to select a math instructor, 14 different ways to select an English instructor, and 5 different ways to select a history instructor. Thus, the president can select the three instructors in 12 14 5 = 840 different ways.
A statistics instructor gives a ten-question multiple choice exam. There are five different choices for each problem, of which only one is correct.
1. In how many different ways could a student answer the ten questions? How many steps are there to this problem? There are ten steps, one for each question that needs to be answered. In how many different ways could a student answer the first question? Since each question offers five possible answers, there are five different ways to answer the first question. This holds true for the other nine questions as well.
5 = 510 = 9,765,625
There are 9,765,625 different ways to answer the ten questions. 2. In how many different ways could a student answer all ten questions correctly? There are still ten steps to this problem, but now there is only one way to answer each question correctly.
1 = 110 = 1
There is only one way to answer all ten questions correctly (and Im pretty sure the one way involves a great deal of studying and homework). 3. In how many different ways could a student answer all ten questions incorrectly? There are ten steps to this problem, and each question can be answered incorrectly in four different ways.
There are 1,048,576 different ways to get all ten questions incorrect.
The next example is different because the number of ways changes from step to step.
If there are 30 skaters competing at the Olympic Games, in how many different ways can the gold, silver, and bronze medals be given?
There are three steps to this problem: determining the number of ways the gold medal could be awarded, determining the number of ways the silver medal could be awarded, and determining the number of ways the bronze medal could be awarded. How many different skaters could win the gold medal? Any of the skaters could win the gold medal, so there are 30. Once we have selected the winner of the gold medal, how many skaters could win the silver medal? Any of the skaters except the gold medal winner could take the silver medal, so that leaves 29 skaters. Finally, now that two medal winners have been determined, how many different ways could the bronze medal be awarded? There are 28 different skaters left, so there are 28 ways.
28 = 24,360
There are 24,360 different ways to give the gold, silver, and bronze medals.
The reason that the number of ways went down by one for each step is that once a person was selected for one of the medals, that person was not eligible for any of the other medals. Here is a similar example.
The I Love Math Club has 24 members. In how many different ways could they elect a president, vice president, secretary, and treasurer?
There are 24 ways to choose a president, which leaves 23 ways to choose a vice president, and so on.
21 = 255,024
There are 255,024 different ways to elect a president, vice president, secretary, and treasurer.
A recent city council election had nine candidates. In how many different ways could they finish first through ninth?
There are nine candidates that could finish first, eight candidates that could finish second, seven candidates that could finish third, . . . , two candidates that could finish eighth, and one candidate that could finish ninth.
1 = 362,880
There are 362,880 different ways that the nine candidates can finish first through ninth.
The product in the previous example is a special product called a factorial. The factorial of a nonnegative integer n, denoted by n!, is the product of all of the integers between 1 and n. We define 0! to be 1, and 1! to be 1 as well. If n objects are to be arranged in an order from 1 to n, this can be done in n! different ways. Many calculators provide a way to calculate factorials directly.
A family is planning a trip to six European countries. In how many different ways could they arrange their trip with respect to what order they visit the countries?
Since we are arranging six countries in order from 1 to 6, 6! will do the trick.
6! = 6
There are 720 different ways that they could arrange their trip.
A permutation is an ordered arrangement of a group of objects. The number of permutations of r objects selected from a set of n distinct objects is given by
n! (n r)!
Lets use some concrete numbers for r and n to help explain exactly what this formula does. Suppose there were ten numbered balls in a bag, and we were going to select three of the balls and record the order that they came out in. So, n would be 10 and r would be 3. This is because we are selecting three out of the ten. The formula would tell us how many different ways there are to select three out of the ten where the order of selection is important. The order of selection is important because it tells us that selecting ball 1, ball 2, and then ball 3 is different from selecting ball 3, ball 2, and then ball 1. To use this formula, we must be sure that each step involves a choice of what began as n distinct objects. Also, none of the n objects can be selected more than once. Finally, there must be a distinction between being chosen at each different step. In other words, the order of selection matters. One final note: many calculators are able to calculate permutations for you.
Since no skater can win more than one of these medals, and winning the gold medal is different from winning the silver medal (order matters), we can use permutations. This is a permutation problem with n = 30 and r = 3.
30P3
30! (30 3)! 30! 27! 30
28 27!
= 24,360
There are 24,360 different ways to award the medals.
If you have to work out permutations by hand instead of using a built-in function on your calculator, working them out in the above fashion is a good idea. We rewrite 30! as 30 29 28 27! because 30! is the product of the integers from 30 down to 1, and 27! is the product of the integers from 27 down to 1. 27! is missing only the product of 28, 29, and 30 from 30!. The reason for rewriting it in terms of 27! is to cancel the 27! in the denominator. Note that we obtained the same result that the Multiplication Principle gave us. Then what is the advantage of doing the problem this way? If your calculator has a built-in function to calculate permutations, then that is easier than multiplying several numbers. This advantage increases as n increases and as the difference between n and r decreases.
This is a permutation problem with n = 24 and r = 4.
24! (24 4) ! 24! 20! 24
22 21 20! 22
This is a permutation problem with n = 9 and r = 9.
9! (9 9) ! 9! 0!
9! 1
= 9! = 362,880
There are some problems for which the order of selection does not matter, but in every other way they are like permutation problems. This type of problem involves combinations. Suppose that a statistics instructor selected three of his 36 students at random and gave the first a free A on the final exam, gave the second a free C on the final exam, and gave the third a 0 on the final. In how many different ways could this be done? It is a permutation problem with n = 36 and r = 3. The order of selection matters because getting an A on the final is a different result from getting a C on the final, and both are different from getting a 0 on the final. Now lets change the problem slightly. Suppose that a statistics instructor selected three of his 36 students at random and gave all three a free A on the final exam. Is there any difference between being the first selected, the second selected, or the third selected? No, because the result is the same: a free A on the final. If the instructor picks Ann, Bob and Carl, or if he picks Carl, Bob and Ann, the results are the same. All three get an A. A combination is similar to a permutation, except that the order in which the objects are selected does not matter. The number of combinations of r objects selected from a set of n distinct objects is given by
Lets compare the formulas for permutations and combinations. Permutations
n! r!(n r)!
The formulas are the same except that the formula for combinations has an extra r! in its denominator. Here is the reason why. Suppose we were selecting three objects in a permutation. The following would be six different results. A, B, C A, C, B B, A, C B, C, A C, A, B C, B, A
But if this were a combination problem, all six would only count as one combination (same letters, but different order). So the number of permutations is six times bigger than the number of combinations. We need to divide the number of permutations by 6 to find the number of combinations. That is where the r ! comes in. In this example, 3! = 6.
A statistics instructor with a class of 36 students has taken leave of his senses, and has decided to select three people to be excused from the final exam with an automatic grade of A. In how many different ways can the instructor select three students?
This is a combinations problem because there is no difference between being selected first, second, or third. We will use 36 for n and 3 for r.
= = = =
3! 3!
36! (36 3)!
36! 33!
35 3!
34 33! 34
The instructor could do this in 7140 different ways.
If the I Love Math Club wanted to select four of its 24 members to go to a conference, in how many different ways could this be done?
It does not matter whether a person is selected first, second, third, or fourth because all four people are going to the conference. This is a combinations problem with n = 24 and r = 4.
The I Love Math Club could select four of its members to send to a conference in 10,626 different ways.
A family dinner special at a local restaurant allows the family to order one entre for each person from a list of 12 entres. If no repetition is allowed, how many different ways could a family of five order dinner?
This is a combination problem with n = 12 and r = 5.
12! 5! 7! 12
10 9 5! 7!
A family of five can order dinner in 792 different ways.
A family dinner special at a local restaurant allows the family to order one entre for each person from a list of 12 entres. If no repetition is allowed, how many different ways could a family of seven order dinner?
12C 7
12! 5!
10 9 7! 5! 10 5!
A family of seven can order dinner in 792 different ways.
Are the two previous answers the same by accident, or is there a reason? Think of it this way. A family could order the five entrees that they really like, or they could pick the seven that they really do not like. Either way, the result is the same. The family has decided on five entrees to order. As long as the two numbers add up to n, the number of combinations will be the same. For example, a family of eight has the same number of ways to order dinner as a family of four. For further convincing evidence that this is no accident, just look at the denominators of the two calculations. They are identical. Now here is an example to tie it all together.
If a local pizza parlor has three different crusts and 15 different toppings to choose from, in how many different ways could you order a four-topping pizza? (one crust, four toppings)
There are two steps to this problem. First, we must determine how many different ways there are to select a crust. Then we determine how many different ways there are to select four of the toppings. Finally, we multiply those results. There are three different ways to select a crust, and select four of the 15 toppings.
or 1365 different ways to
There are 4095 different ways to order a four-topping pizza.
Combinations work exactly the same way, except that we select option from the PRB menu.
(nCr)
1. Draw a tree diagram showing the possible outcomes of four tosses of a coin. How many different ways are there to get (a) No heads? (b) Exactly one head? (c) Exactly two heads? (d) Exactly three heads? (e) Four heads?
2. A couple plans to have five children, and they are interested in what the genders will be. To help them out, draw a tree diagram showing all the possible outcomes. How many different ways are there to have (a) No girls? (b) Exactly one girl? (c) Exactly two girls? (d) Exactly three girls? (e) Exactly four girls? (f) Five girls?
3. A pair of dice are rolled and their sum is recorded. How many different ways are there to roll (a) (b) (c) (d) An 8? A 7 or an 11? At least a 6? Doubles or a sum that is lower than 8?
4. A pair of dice are rolled and their sum is recorded. How many different ways are there to roll (a) (b) (c) (d) A 5? A 2, 3, or 12? A 9 or less? Doubles or a sum that is 8?
5. At a restaurant, you can choose from seven main dishes and 15 side dishes. In how many different ways could you select one main dish and one side dish? 6. An ice cream store has 31 flavors of ice cream and 3 types of cones. In how many different ways could you order a single-scoop ice cream cone (one flavor, one type of cone)? 7. There are 31 professional football teams in the NFL and 10 college football teams in the PAC-10. In how many different ways could a football fan list their favorite NFL team and their favorite PAC-10 team? 8. A pizza parlor has 4 types of crust and 12 different toppings. In how many different ways could you order a one-topping pizza (one type of crust, one topping)? 9. A couple planning their wedding must decide on the menu for the reception. There are ten different appetizer packages, seven main courses, and three dessert packages to choose from. In how many different ways could the couple select an appetizer package, a main course, and a dessert package? 10. A college basketball team has five guards, four forwards, and three centers. The coach must select one guard, one forward, and one center to work at a basketball clinic for kids. In how many different ways could she do this?
3.2 Classical Probability
11. A math instructor gives a weekly multiple choice quiz. There are five questions, and each question has four possible answers, of which only one is correct. (a) In how many different ways could a student answer the five questions? (b) In how many different ways could a student answer all five questions correctly? (c) In how many different ways could a student answer all five questions incorrectly? 12. An algebra instructor gives an arithmetic quiz the first day of class. There are eight true/false questions. (a) In how many different ways could a student answer the eight questions? (b) In how many different ways could a student answer all eight questions correctly? (c) In how many different ways could a student answer all eight questions incorrectly? 13. A softball league has ten teams in it. In how many different ways could the teams finish first and second? 14. A third-grade class has 25 students in it. In how many different ways could the teacher pick one student to clean the chalkboard and another student to empty the wastebaskets? 15. Bruce Springsteen has released 16 CDs. In how many different ways could a fan select three of his CDs to bring on a road trip? 16. A faculty senate has 23 members. In how many ways could they send four of the senators to a workshop on shared governance? 17. A math instructor has a class with 40 students in it. He suddenly gets hungry in the middle of class. In how many different ways could he select a student to go get him a donut, a second student to go get him a breakfast burrito, and a third student to go get him some coffee? 18. A math instructor has a class with 40 students in it. He suddenly gets hungry in the middle of class. In how many different ways could he select three students to go get breakfast for him? 19. In how many different ways could the Today Show select three jurors from a 12person jury to interview? 20. Twenty children sign up to test a new medication. In how many different ways could the doctors select eight of the children to receive the medication? 21. In Californias Super Lotto game, 6 numbers are drawn from 51 possible numbers (151). In how many different ways could 6 numbers be selected from 51? 22. Five cards are drawn from a well-shuffled standard 52-card deck of playing cards. In how many different ways could this be done? 23. Five cards are drawn from a well-shuffled standard 52-card deck of playing cards. In how many different ways could five clubs be drawn? 24. A local Beanie Babies club has 20 members. In how many different ways could they elect a president, a vice president, and a newsletter editor? 25. A pizza parlor has 12 different toppings that you can order for your pizza. In how many different ways could you choose 3 different toppings for a pizza? 26. A pizza parlor has four types of crust, 12 different toppings, and three different cheese blends to choose from. In how many different ways could you order a threetopping pizza (one type of crust, three toppings, and one cheese blend)? 27. A local Beanie Babies club has 20 members. In how many different ways could they elect a president, a vice president, and a three-person newsletter committee? 28. A college basketball team has five guards, four forwards, and three centers. In how many different ways could a coach select a starting lineup of two guards, two forwards, and one center?
Classical Probability
In this section we will calculate how likely it is that something happens, in contrast to the previous section, where we figured out how many different ways something could happen. Something is a rather vague word, so here are some definitions to allow us to proceed.
DEFINITION An experiment is a process that allows us to obtain observations.
This definition agrees with our definition of an experiment in a science class. We perform a certain process, and observe the outcome. In probability, an example of an experiment is flipping a coin and observing whether it is heads or tails. Here are some other experiments.
Roll a six-sided die and observe the number that is on the top of the die. Roll a pair of six-sided dice and observe their sum. Draw a card from a 52-card deck and record its suit. Have a child and record its gender. Randomly select a person and ask whether that person has health insurance.
If we flip a coin and it lands heads-up, heads is the outcome of the experiment. If we roll a single six-sided die, what are the possible outcomes? The possible outcomes are the numbers 1, 2, 3, 4, 5, and 6. When we list all of the possible outcomes of an experiment, we are listing the experiments DEFINITION The sample space (S) for an experiment is sample space. the set of all possible outcomes for that experiment. Returning to the experiment of rolling a single six-sided die, we denote the sample space in the following manner: S = {1, 2, 3, 4, 5, 6}. What DEFINITION An event is a subset of the sample space is the sample space for tossing a coin? S = {heads, for an experiment. tails}, since those are the only possible results (assuming a coin cannot land on its side). An event is usually made up of one or more of the possible outcomes of an experiment. We use capital letters to denote an event. If the experiment was rolling a single six-sided die, some possible events are A: the roll is even, and B: the roll is 5 or higher. Event A is then the set A = {2, 4, 6} and event B is the set B = {5, 6}.
DEFINITION An outcome of an experiment is a particular result of an experiment.
If all of the outcomes in a sample space are equally likely to occur, we denote the probability of event A occurring by P(A) and calculate it using the formula
P(A) = n(A) n(S)
In the formula, n(A) represents the number of different ways that event A can occur. We use n(S) to represent the total number of possible outcomes for the experiment.
A single fair six-sided die is rolled. Find the probability that the roll is even.
The word fair implies that each of the possible outcomes is equally likely to occur. Let A denote the event that the roll is even. Then A = {2, 4, 6}, and n(A) = 3. Since S = {1, 2, 3, 4, 5, 6}, n(S) = 6. Therefore,
1 The probability that the roll is even is 2. We could say that there is a 1 in 2 chance that the roll is even, or that there is a 50% chance that the roll is even. We will later 1 see that if this experiment is repeated over and over, approximately 2 of the rolls will be even.
A single fair six-sided die is rolled. Find the probability that the roll is 5 or higher.
Let B denote the event that the roll is 5 or higher. Then B = {5, 6}, and n(B) = 2. Since S = {1, 2, 3, 4, 5, 6}, n(S) = 6. Therefore,
P(B) = = = n(B) n(S) 2 6 1 3
1 The probability that the roll is 5 or higher is 3.
Suppose we rolled a pair of fair six-sided dice and observed the sum of the dice. How large is the sample space? There are 11 possible sums. The possible sums are 2, 3, 4, . . . , 11, and 12. This does not mean that n(S) = 11. Recall the following table.
Sums of a Pair of Dice
We see that n(S) = 36. We will use the following table to calculate the probabilities of events involving the sum of a pair of fair six-sided dice.
Sum Number of ways
A pair of fair six-sided dice are rolled. Find the probability that their sum is 7.
Let event A be the event that the sum is 7. Of the 36 different outcomes, how many have a sum of 7? Six of them do, so
P(A) = = = n(A) n(S) 6 36 1 6
1 The probability that the sum is 7 is 6.
A pair of fair six-sided dice are rolled. Find the probability that their sum is 9 or greater.
Let event B be the event that the sum is 9 or greater. What sums are 9 or greater? The sums 9, 10, 11, and 12. There are 10 of these sums (four 9s, three 10s, two 11s, and one 12), so
P(B) = = = n(B) n(S) 10 36 5 18
The probability that the sum is 9 or greater is
Recall the following information about a standard 52-card deck of playing cards.
There are two colors: red and black. There are 26 cards of each color. There are four suits: diamonds (), hearts (), spades () and clubs (). The diamonds and hearts are red, while the spades and clubs are black. There are 13 cards in each suit. The 13 cards in each suit are A (ace), 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king). There are 4 of each of these 13 cards in a deck. The jack, queen, and king are called face cards, because they have people (with faces) on them. There are 3 face cards in each suit, for a total of 12 face cards in a deck.
A card is drawn from a well-shuffled standard 52-card deck. Find the probability that it is black.
Let A be the event that the card is black.
P(A) = = = n(A) n(S) 26 52 1 2
1 The probability that the card is black is 2.
A card is drawn from a well-shuffled standard 52-card deck. Find the probability that it is a diamond.
Let B be the event that the card is a diamond.
P(B) = = = n(B) n(S) 13 52 1 4
1 The probability that the card is a diamond is 4.
A card is drawn from a well-shuffled standard 52-card deck. Find the probability that it is a face card.
Let C be the event that the card is a face card.
P(C) = = = n(C) n(S) 12 52 3 13
The probability that the card is a face card is
Basic Rules of Probability
Probabilities are real numbers between 0 and 1. If an event is certain to occur, its probability is 1. If an event is certain not to occur, its probability is 0. Special Addition Rule: If events A and B are mutually exclusive, then P(A or B) = P(A) + P(B). Complement Rule: P(A) = 1 P(A does not occur).
From our examples, we should believe that the probability of an event is always between 0 and 1. To convince yourself, change 0 and 1 to equivalent percentages: 0% and 100%. Can there ever be more than a 100% chance that an event occurs? Intuitively, the answer is no. Can there ever be less than a 0% chance that an event occurs? Again, no. Lets convince ourselves in a more mathematical fashion. Can the number of different ways that event A could occur be less than 0 or more than the size of the sample space? The answer is no, and this gives us 0 n(A) n(S). Take this inequality, and divide each piece by n(S), to verify our original claim.
0 n(A) n(S) 0 n(A) n(S) n(S) n(S) n(S) 0 P(A) 1
If an event is guaranteed to happen, then the number of different ways that it could happen must be the same as n(S).
If a single fair six-sided die is rolled, find the probability that it is lower than a 7.
Let A be the event that the roll is less than 7. In this case, S = {1, 2, 3, 4, 5, 6} and A = {1, 2, 3, 4, 5, 6} also, so
P(A) = = n(A) n(S) 6 6
We know that the roll of a die must produce a number less than 7, so the probability that the roll is less than 7 is 1.
If an event is guaranteed not to occur, then the number of different ways that it could occur must be 0. This means that the probability that the event occurs must be 0 as well.
If a single fair six-sided die is rolled, find the probability that it is a 7.
Lets look at the sample space first. In this case, S = {1, 2, 3, 4, 5, 6}. Of those six possibilities, how many of them are a 7? None. If we let B be the event that the roll is 7, then B is an empty set and n(B) = 0. Therefore,
P(B) = = n(B) n(S) 0 6
We know that the roll of a die cannot produce a 7, so the probability that the roll is a 7 is 0.
If two events A and B are mutually exclusive, this means that they cannot both occur at the same time. In such a case, we say that the probability of A and B occurring is 0. In symbols, we write P(A B) = 0. The symbol is the symbol for intersection from set theory. We will use this symbol when we are interested in the probability that both event A occurs and event B occurs. If we know that the two events A and B are mutually exclusive, then P(A or B) = P(A) + P(B). This is known as the Special Addition Rule. We will rewrite P(A or B) as P(A B). The symbol is the symbol for union from set theory. We will use this symbol when we are interested in the probability that either event A occurs or event B occurs, or maybe both event A and event B occur. The key word for union is or. We will be using or in the way that we use the phrase and/or. This is referred to as the inclusive or. For
example, if someone offered you coffee or pie, you are not restricted to only one or the other, but could choose both. To summarize, in symbols we write the Special Addition Rule as
P(A B) = P(A) + P(B)
If the probability that the San Francisco Forty-Niners win the next Super Bowl is 0.4, and the probability that the Green Bay Packers win the next Super Bowl is 0.25, find the probability that the FortyNiners or the Packers win the next Super Bowl.
Let A be the event that the Forty-Niners win, and let B be the event that the Packers win. Are these two events mutually exclusive? Yes, because only one team can win the next Super Bowl.
P(A B) = P(A) + P(B) = 0.4 + 0.25 = 0.65
The probability that the Forty-Niners or the Packers win the next Super Bowl is 0.65.
The probability that Joe Student will get an A on the history exam is 0.25, and the probability that he will get a B is 0.35. Find the probability that Joe gets a B or better on the history exam.
Let A be the event that Joe gets an A, and let B be the event that Joe gets a B. Are these events mutually exclusive? Yes, because Joe cannot get an A and a B on the same exam.
P(B or better) = P(A B)
= P(A) + P(B) = 0.25 + 0.35 = 0.6
The probability that Joe gets a B or better is 0.6.
The Special Addition Rule can be extended to situations where there are three or more mutually exclusive events.
A woman needs to buy a new dress to wear to a wedding. If the probability that she buys the dress at Gottschalks is 0.4, the probability that she buys it at JC Penneys is 0.3, and the probability that she buys it at Mervyns is 0.1, find the probability that she buys the dress at one of these three stores.
Let A be the event that the woman purchases the dress at Gottschalks, B be the event that she buys it at JC Penneys, and C be the event that she buys it at Mervyns.
P(A
C) = P(A) + P(B) + P(C) = 0.4 + 0.3 + 0.1 = 0.8
The probability that she buys the dress at one of these three stores is 0.8.
Why do we use union in the above situations? When we say that the woman buys the dress at one of these three stores, we are really saying that the woman buys the dress at Gottschalks or JC Penneys or Mervyns. Recall that or is the magic word for union. The complement of an event A, denoted A , is made up of the possible outcomes of the experiment that are not included in event A. Since an event and its complement are mutually exclusive by definition, we can use the Special Addition Rule to show that P(A A ) = P(A) + P(A ). Since an event either occurs or does not occur, we also know that P(A A ) = 1. Some easy mathematical manipulation yields the Complement Rule.
P(A) + P(A ) = 1 P(A) = 1 P(A )
Find the probability that it either rains today or does not rain today.
Let R be the event that it rains today, and R be the event that it does not rain today. Since it either has to rain or not rain today, the probability is 1.
P(R R ) = 1
If the probability that it rains today is 0.35, find the probability that it does not rain.
Let R be the event that it rains today, so P(R) = 0.35. Since we are interested in the complement of event R, we can apply the Complement Rule.
P(R ) = 1 P(R) = 1 0.35 = 0.65
The probability of no rain today is 0.65.
If the probability that the San Francisco Forty-Niners win the next Super Bowl is 0.4, and the probability that the Green Bay Packers win the next Super Bowl is 0.25, find the probability that a team other than the Forty-Niners or the Packers win the next Super Bowl.
Let A be the event that the Forty-Niners win the next Super Bowl, and let B be the event that the Packers win the next Super Bowl. What we are interested in is the
complement of A Bthat the winner is not the Forty-Niners or the Packers. Use the Complement Rule.
P(A team other than the Forty-Niners or Packers win) = 1 P(Forty-Niners or Packers win) = 1 P(A B)
= 1 (0.4 + 0.25) = 1 0.65 = 0.35
The probability that a team other than the Forty-Niners or Packers win the next Super Bowl is 0.35.
The probability that Joe Student will get an A on the history exam is 0.25, and the probability that he will get a B is 0.35. Find the probability that Joe gets a C or lower on the history exam.
Let A be the event that Joe gets an A on the exam, and let B be the event that Joe gets a B on the exam. Note that C or lower is the complement of B or better.
P(C or lower) = 1 P(B or better) = 1 P(A B)
= 1 (P(A) + P(B)) = 1 (0.25 + 0.35) = 1 0.6 = 0.4
The probability that Joe gets a C or lower on the exam is 0.4.
The probability that a woman buys a dress to wear at a wedding from Gottschalks, JC Penneys, or Mervyns is 0.8. Find the probability that she buys it somewhere other than these three stores.
Let A be the event that the woman buys the dress at Gottschalks, JC Penneys, or Mervyns. We are interested in the complement of event A.
P(A ) = 1 P(A) = 1 0.8 = 0.2
The probability that she buys the dress somewhere else is 0.2.
A pair of fair six-sided dice are rolled. Find the probability that the sum is not 7.
Let A be the event that the sum is 7. (Although S would be a natural choice for the event that the sum is seven, recall that S is used for the sample space.) We are
interested in P(A ), so we will use the Complement Rule. Since 6 of the 36 possible 6 outcomes have a sum of 7, P(A) = 36 .
P(A ) = 1 P(A) = 1 = = 30 36 5 6 6 36
The probability that the sum is not 7 is 5. 6
A card is drawn from a well-shuffled standard 52-card deck of playing cards. Find the probability that it is not an ace.
Let A be the event that a card is an ace. We are interested in P(A ), so we will use the Complement Rule and the fact that there are four aces in a deck of 52 cards.
P(A ) = 1 P(A) = 1 = = 48 52 12 13
12 13.
The probability that the card is not an ace is
For any two events A and B, we can find the probability that A or B occurs using the General Addition Rule:
P(A B) = P(A) + P(B) P(A B)
Note that the two events do not need to be mutually exclusive in order to use this rule. However, if the two events are mutually exclusive, this formula will work as well. The only difference between the Special Addition Rule (P(A B) = P(A) + P(B)) and the General Addition Rule (P(A B) = P(A) + P(B) P(A B)) is that in the General Addition Rule we subtract P(A B). If events A and B are mutually exclusive, P(A B) = 0, and subtracting 0 does not change the result. Lets use the General Addition Rule.
The probability that a foreign visitor to Californias Central Valley will visit Sequoia Park is 0.15, and the probability that a foreign visitor will visit Yosemite is 0.22. The probability that the foreign visitor visits both Sequoia Park and Yosemite is 0.08. Find the probability that a foreign visitor to Californias Central Valley will visit Sequoia Park or Yosemite.
Let A be the event that the visitor goes to Sequoia Park, and let B be the event that the visitor goes to Yosemite. P(A) = 0.15, P(B) = 0.22, and P(A B) = 0.08. Now use the General Addition Rule.
B) = P(A) + P(B) P(A = 0.15 + 0.22 0.08 = 0.29
The probability that a foreign visitor goes to Sequoia Park or Yosemite is 0.29.
We can use a Venn diagram to help us with problems involving the General Addition Rule. When drawing a Venn diagram for probabilities, we begin with a rectangle representing the sample space S. We then draw two circles inside the rectangle for events A and B in such a way that the two circles overlap. We place P(A B) in the region where the circles overlap. We then subtract P(A B) from P(A) to determine what value to place in the rest of circle A. We repeat this for circle B. Finally, the regions inside the box should add up to 1. Here is the Venn diagram that goes along with the previous example.
We put the probability associated with the intersection in the center region. We then subtract the 0.08 from 0.15 (giving us 0.07) and from 0.22 (giving us 0.14). If we add 0.07 + 0.08 + 0.14, we find that the probability of A or B is 0.29. Where does the 0.71 come from? Subtract 0.29 from 1.
A card is drawn from a well-shuffled standard 52-card deck of playing cards. Find the probability that the card is a diamond or a face card.
Let A be the event that the card is a diamond, and let B be the event that the card 13 12 3 is a face card. P(A) = 52, P(B) = 52, and P(A B) = 52 because there are three cards in the deck that are diamonds and face cards.
P(A B) = P(A) + P(B) P(A = = = 13 12 3 + 52 52 52 22 52 11 26
The probability that a card is a diamond or a face card is
For Exercises 14, a fair six-sided die is rolled. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Find Find Find Find the the the the probability probability probability probability that that that that the the the the roll roll roll roll is is is is a 4. at least 4. less than 4. a prime number.
For Exercises 510, a pair of fair six-sided dice are rolled. Find the probability that the sum is 4. Find the probability that the sum is 4 or higher. Find the probability that the sum is 7 or 11. Find the probability that the sum is 2, 3, or 12. Find the probability that the sum is odd. Find the probability that the sum is even.
For Exercises 1114, a card is drawn from a well-shuffled standard 52-card deck of playing cards. 11. 12. 13. 14. Find Find Find Find the the the the probability probability probability probability that that that that the the the the card card card card is is is is red. a club. a jack. a queen or an ace.
A child has 12 blocks: 4 red, 4 blue, and 4 green. All of the blocks are put inside a bucket. 15. The child reaches into the bucket and randomly picks a block. Find the probability that the block is red. 16. The child reaches into the bucket and randomly picks a block. Find the probability that the block is not red. 17. The child throws two of the green blocks away. The child reaches into the bucket and randomly picks a block. Find the probability that the block is green. 18. The child throws two of the green blocks away. The child reaches into the bucket and randomly picks a block. Find the probability that the block is red. For Exercises 1921, a fair six-sided die is rolled. 19. Find the probability that the roll is at least a 1. 20. Find the probability that the roll is at least a 10. 21. Find the probability that the roll is a 31. 2 For Exercises 2224, a pair of fair six-sided dice are rolled. 22. Find the probability that the sum is 13. 23. Find the probability that the sum is 13 or less. 24. Find the probability that the sum is 31. 2 For Exercises 2528, a card is drawn from a well-shuffled standard 52-card deck of playing cards. 25. 26. 27. 28. Find Find Find Find the the the the probability probability probability probability that that that that the the the the card card card card is is is is a king or a queen. a face card or an ace. a spade or a heart. black or red.
29. The probability that Bob Dylan wins the Grammy award for album of the year is 0.4, and the probability that Paul McCartney wins it is 0.15. Find the probability that Bob Dylan or Paul McCartney wins the award. 30. The probability that Jeff Gordon wins the next Daytona 500 is 0.35, and the probability that Dale Earnhart Jr. wins the race is 0.17. Find the probability that Jeff Gordon or Dale Earnhart Jr. wins the next Daytona 500. 31. Joe Star is a prized baseball free agent. Joes agent states that there is a 75% chance that his client will sign with the Yankees, but there is only a 10% chance that he
3.3 Table Probabilities, Conditional Probability
will sign with the Braves. Find the probability that Joe Star will sign with a team other than the Yankees or Braves. 32. The probability that an algebra student earns a final grade of A is 0.1, the probability of earning a final grade of B is 0.12, and the probability of earning a final grade of C is 0.17. Find the probability that an algebra student passes with a grade of C or higher. 33. Find the probability that it either snows or does not snow today. 34. Find the probability that a customer at a coffee shop either buys a donut or does not buy a donut. 35. A pair of fair six-sided dice are rolled. Find the probability that the sum is even or not even. 36. A card is drawn from a well-shuffled standard 52-card deck of playing cards. Find the probability that the card is red or not red. For Exercises 3740, a card is drawn from a well-shuffled standard 52-card deck of playing cards. 37. 38. 39. 40. Find Find Find Find the the the the probability probability probability probability that that that that the the the the card card card card is is is is not not not not a queen. a face card. red. a diamond.
41. The probability that it snows today is 0.25. Find the probability that it does not snow today. 42. The probability that Bob Dylan wins the Grammy award for album of the year is 0.4. Find the probability that Bob Dylan does not win the award. 43. The probability that Jeff Gordon wins the next Daytona 500 is 0.35. Find the probability that someone else wins the next Daytona 500. 44. The probability that an algebra student earns a final grade of A is 0.1, the probability of earning a final grade of B is 0.12, and the probability of earning a final grade of C is 0.17. Find the probability that a student does not pass with a grade of C or better. 45. A card is drawn from a well-shuffled standard 52-card deck of playing cards. Find the probability that the card is an ace or a diamond. 46. A card is drawn from a well-shuffled standard 52-card deck of playing cards. Find the probability that the card is red or a face card. 47. A pair of fair six-sided dice are rolled. Find the probability that the sum is 5 or that at least one of the dice is a 3. 48. A pair of fair six-sided dice are rolled. Find the probability that the sum is 8 or that doubles are rolled. 49. The probability that a student passes his or her math class is 0.8, the probability that a student makes the Deans List is 0.4, and the probability that a student passes his or her math class and makes the Deans List is 0.25. Find the probability that a student passes his or her math class or makes the Deans List. 50. The probability that a college freshman takes a math class the first semester is 0.75. The probability that a college freshman takes an English class the first semester is 0.7. The probability that a college freshman takes both a math class and an English class the first semester is 0.6. Find the probability that a college freshman takes a math class or an English class the first semester.
Table Probabilities, Conditional Probability
One hundred fifty students at the College of the Sequoias were asked their preference of burgers. Their responses were categorized as Burger King, McDonalds, In-N-Out, and Other. Here are the results.
Burger King Male Female Total 15 35 50
McDonalds 18 12 30
In-N-Out 35 10 45
Other 15 10 25
Total 83 67 150
If a student from this survey is selected at random, find the probability that the student is male.
Let M be the event that the student is male.
P(M ) = = n(M ) n(S) 83 150
The probability that the student is male is
If a student from this survey is selected at random, find the probability that the student is female.
Let F be the event that the student is female.
P(F) = = n(F) n(S ) 67 150
The probability that the student is female is
If a student from this survey is selected at random, find the probability that the student prefers Burger King.
Let B be the event that the student prefers Burger King.
P(B) = = = n(B) n(S) 50 150 1 3
1 The probability that the student prefers Burger King is 3.
If a student from this survey is selected at random, find the probability that the student is male and prefers Burger King.
Let M be the event that the student is male, and let B be the event that the student prefers Burger King. Of the 150 students, 15 are male and prefer Burger King. The 15 students are highlighted in the following table.
Burger King Male Female Total P(M B) = = 15 150 1 10 15 35 50
The probability that the student is male and prefers Burger King is
If a student from this survey is selected at random, find the probability that the student is male or prefers Burger King.
Let M be the event that the student is male, and let B be the event that the student prefers Burger King. We could use the General Addition Rule.
P(M B) = P(M) + P(B) P(M = = 83 50 15 + 150 150 150 59 75
59 75 .
The probability that the student is male or prefers Burger King is
We could have also used the table. We need to include all the students in the top row (male) as well as the students in the left column (Burger King). The 118 students are highlighted in the following table.
Burger King Male Female Total 15 35 50 McDonalds 18 12 30 In-N-Out 35 10 45 Other 15 10 25 Total 83 67 150
Note that 15 + 18 + 35 + 15 + 35 gives us 118, the same total as we had from the General Addition Rule.
P(M B) = = 118 150 59 75
If a male from the survey is selected at random, find the probability that he prefers Burger King.
Without using any notation, lets use our intuition to find the probability. In this problem we are selecting one of the males, and there are 83 to choose from.
How many of those 83 prefer Burger King? Fifteen of the 83 prefer Burger King. 15 It stands to reason that the probability is 83. This problem is an example of conditional probability.
A problem is known as a conditional probability problem if there is a given condition that restricts the sample space. In the previous problem, we knew that the student being selected was male, not just a student. The size of the sample space was reduced from the 150 students to the 83 male students. The probability of event A occurring, given that event B has occurred (B is the condition) is denoted P(A|B). This is read The probability of A given B. The known condition is put after the vertical line, and the event whose probability we are interested in is put before the vertical line.
A child has a set of 12 plastic blocks in a bucket. Four of the blocks are red, 4 of the blocks are blue, and 4 of the blocks are green. If a child draws a green block from the bucket, find the probability that the next block drawn from the bucket will be red.
Let R be the event that a block is red, B be the event that a block is blue, and G be the event that a block is green. The condition that has changed our sample space of 12 blocks is that a green block has already been drawn from the bucket. This leaves us with only 11 blocks, 4 of which are red.
P(R |G) = = n(R ) n(S*) 4 11
The notation S* is used to identify the fact that the sample space has been restricted. The probability that a red block is drawn if a green block has already 4 been drawn is 11.
A child has a set of 12 plastic blocks in a bucket. Four of the blocks are red, 4 of the blocks are blue, and 4 of the blocks are green. If a child draws a green block and a red block from the bucket, find the probability that the third block drawn from the bucket will be red.
Let R be the event that a block is red, B be the event that a block is blue, and G be the event that a block is green. We are interested in the probability of drawing a red block given that a green block and a red block have already been drawn. In symbols, we write this as P(R |GR). Since one red block has already been removed, there are 3 red blocks remaining in the bucket. Also, since 2 blocks have been removed from the bucket, there are 10 blocks remaining in the bucket.
P(R |GR) = = n(R ) n(S*) 3 10
The probability of drawing a red block after drawing a green block and a red block 3 is 10.
If a club is drawn from a well-shuffled standard 52-card deck of playing cards, find the probability that the next card drawn is a diamond.
Let C be the event that a club is drawn, and let D be the event that a diamond is drawn. We are interested in P(D |C). After the club is drawn, how many diamonds remain in the deck? There are still 13 diamonds (none have been removed). After the club is drawn, how many cards remain in the deck? There are 51 cards in the deck, because 1 card has been removed.
P(D |C) = 13 51
The probability that a diamond is drawn after a club has been drawn is
If a diamond is drawn from a well-shuffled standard 52-card deck of playing cards, find the probability that the next card drawn is a diamond.
Let D be the event that a diamond is drawn. We are interested in P(D |D). After the diamond is drawn, how many diamonds remain in the deck? There are now 12 diamonds because 1 has been removed. After the diamond is drawn, how many cards remain in the deck? There are 51 cards in the deck, because 1 card has been removed.
P(D|D) = = 12 51 4 17
The probability that a diamond is drawn after a diamond has been drawn is
Five cards are dealt from a well-shuffled standard 52-card deck of playing cards. Four of the cards are diamonds, and 1 is not a diamond. Find the probability that the sixth card dealt is a diamond.
Let D be the event that a diamond is dealt, and let N be the event that a nondiamond is dealt. We are interested in P(D |DDDDN ). After the first 5 cards are dealt, how many diamonds remain in the deck? There are now 9 diamonds because 4 have been removed. After the first 5 cards are dealt, how many cards remain in the deck? There are 47 cards in the deck, because 5 cards have been removed.
P(D|DDDDN) = 9 47
The probability that the sixth card dealt is a diamond is
Now lets revisit the burger preference example.
Let M be the event that a male is selected, and B be the event that the person prefers Burger King. Since we know that a male is selected, we are looking for P(B |M).
P(B |M) = = n(B ) n(S*) 15 83
If a male from the survey is selected at random, the probability that he prefers 15 Burger King is 83.
If a person from the survey who prefers Burger King is selected at random, find the probability that the person is male.
Let B be the event that the person prefers Burger King, and M be the event that a male is selected. Since we know that a person who prefers Burger King is selected, we are looking for P(M|B).
P(M |B) = = = n(M ) n(S*) 15 50 3 10
If a person from the survey who prefers Burger King is selected at random, the 3 probability that the person is male is 10.
If a female from the survey is selected at random, find the probability that she prefers McDonalds.
Let F be the event that a female is selected, and D be the event that the person prefers McDonalds. Since we know that a female is selected, we are looking for P(D |F).
P(D |F) = = n(D) n(S*) 12 67
Twelve of the 67 females in the survey prefer McDonalds. If a female from the sur12 vey is selected at random, the probability that she prefers McDonalds is 67.
The probability of A given B can be calculated using the conditional probability formula:
P(A|B) = P(A B) P(B)
The best way to remember this formula is to think of the numerator as the probability of both, and the denominator as the probability of the given one. Now we use the formula on the previous example.
P(D | F) = = = P(D P(F) 12/150 67/150 12 67 F)
If a female from the survey is selected at random, the probability that she prefers 12 McDonalds is 67. This agrees with our answer found using the tables.
Reasoning out the probability from the tables may seem easier than applying the formula, but there will be times when we must use the formula (for example, when we do not have a table).
Sixty percent of the households in Visalia have the Visalia Times Delta delivered to their house. Fifteen percent of the households in Visalia have the Visalia Times Delta and the Fresno Bee delivered to their house. If we know that a Visalia household has the Visalia Times Delta delivered to their house, find the probability that they also have the Fresno Bee delivered to their house.
Let D be the event that a household has the Times Delta delivered to their house, and let B be the event that a household has the Fresno Bee delivered to their house. We are looking for P(B |D), the probability that a house gets the Bee if it is known that the house gets the Times Delta.
P(B |D) = = P(B D) P(D) 0.15 0.6
If a Visalia household has the Visalia Times Delta delivered to their house, the probability that they also have the Fresno Bee delivered to their house is 0.25. Twentyfive percent of households that have the Times Delta (not all Visalia households) delivered also have the Fresno Bee delivered.
The probability that a student passes their math class is 0.8, and the probability that a student passes their math class and makes the Deans List is 0.25. If a student passes their math class, find the probability that the student makes the Deans List.
Let M be the event that the student passes their math class, and let D be the event that the student makes the Deans List. We are looking for P(D |M).
P(D |M) = = P(D M) P(M) 0.25 0.8
= 0.3125
If we know that a student has passed their math class, the probability that the student makes the Deans List is 0.3125.
Two hundred basketball fans were asked who they felt was responsible for the 19981999 NBA lockoutthe players or the owners. Here are the responses, divided into casual fans and hardcore fans.
Players Are Responsible Casual Fans Hardcore Fans 90 20 Owners Are Responsible 22 53
Both/Unsure 8 7
For Exercises 112, a fan in this survey is selected at random. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Find the probability that the fan is a casual fan. Find the probability that the fan is a hardcore fan. Find the probability that the fan feels the players are responsible for the lockout. Find the probability that the fan feels the owners are responsible for the lockout. Find the probability that the fan is a casual fan and feels that the players are responsible for the lockout. Find the probability that the fan is a hardcore fan and feels that the players are responsible for the lockout. Find the probability that the fan is a casual fan and feels that the owners are responsible for the lockout. Find the probability that the fan is a hardcore fan and feels that the owners are responsible for the lockout. Find the probability that the fan is a casual fan or feels that the players are responsible for the lockout. Find the probability that the fan is a hardcore fan or feels that the players are responsible for the lockout.
11. Find the probability that the fan is a casual fan or feels that the owners are responsible for the lockout. 12. Find the probability that the fan is a hardcore fan or feels that the owners are responsible for the lockout. 13. If a casual fan in this survey is selected at random, find the probability that the fan feels that the players are responsible for the lockout. 14. If a casual fan in this survey is selected at random, find the probability that the fan feels that the owners are responsible for the lockout. 15. If a hardcore fan in this survey is selected at random, find the probability that the fan feels that the players are responsible for the lockout. 16. If a hardcore fan in this survey is selected at random, find the probability that the fan feels that the owners are responsible for the lockout. 17. If a fan in this survey who feels that the players are responsible for the lockout is selected at random, find the probability that the fan is a casual fan. 18. If a fan in this survey who feels that the players are responsible for the lockout is selected at random, find the probability that the fan is a hardcore fan. 19. If a fan in this survey who feels that the owners are responsible for the lockout is selected at random, find the probability that the fan is a casual fan. 20. If a fan in this survey who feels that the owners are responsible for the lockout is selected at random, find the probability that the fan is a hardcore fan. One hundred voters were surveyed in November 1998 for an exit poll. They were asked the following questions:
On most political matters, do you consider yourself liberal, moderate, or conservative? In todays House of Representatives election, did you vote for the Democratic Party
candidate or the Republican Party candidate? Here are the results.
Liberal Democrat Republican 16 3 Moderate 27 23 Conservative 5 26
For Exercises 2124, a voter from this survey is selected at random. 21. Find the probability that the voter voted for the Democratic Party candidate. 22. Find the probability that the voter considers himself liberal. 23. Find the probability that the voter voted for the Democratic Party candidate or considers himself liberal. 24. Find the probability that the voter voted for the Democratic Party candidate and considers himself liberal. 25. If a voter from this survey who voted for the Democratic Party candidate is selected at random, find the probability that the voter considers himself liberal. 26. If a voter from this survey who considers himself liberal is selected at random, find the probability that the voter voted for the Democratic Party candidate. 27. If a voter from this survey who voted for the Republican Party candidate is selected at random, find the probability that the voter considers himself conservative. 28. If a voter from this survey who considers himself conservative is selected at random, find the probability that the voter voted for the Republican Party candidate. 29. If a voter from this survey who voted for the Republican Party candidate is selected at random, find the probability that the voter considers himself moderate or conservative. 30. If a voter from this survey who considers himself moderate or conservative is selected at random, find the probability that the voter voted for the Republican Party candidate. A child has a set of 12 plastic blocks in a bucket. Four of the blocks are red, 4 of the blocks are blue, and 4 of the blocks are green.
31. If a child draws a red block from the bucket, find the probability that the next block drawn from the bucket will be red. 32. If a child draws a red block from the bucket, find the probability that the next block drawn from the bucket will be blue. 33. If a child draws three red blocks from the bucket, find the probability that the next block drawn from the bucket will be red. 34. If a child draws three red blocks from the bucket, find the probability that the next block drawn from the bucket will be green. 35. If a child draws a block from the bucket and does not reveal its color, find the probability that the next block drawn from the bucket will be red. 36. A face card is drawn from a well-shuffled standard 52-card deck of playing cards. Find the probability that the next card drawn is also a face card. 37. Three cards are drawn from a well-shuffled standard 52-card deck of playing cards. All three cards are hearts. Find the probability that the next card drawn is also a heart. 38. Five cards are drawn from a well-shuffled standard 52-card deck of playing cards. Three cards are kings, and the other two are not. Find the probability that the next card drawn is a king. 39. A card is drawn from a well-shuffled standard 52-card deck of playing cards, its suit is observed, and then the card is put back into the deck. After the deck is reshuffled, another card is drawn. If the first card selected is a spade, find the probability that the second card is a spade. 40. How would your answer to Exercise 39 differ if the first card were not put back into the deck before the second card is drawn? Explain your answer in your own words. 41. On par-three golf holes, the probability that a PGA professional golfer hits his first shot on the green is 0.8. The probability that a PGA professional golfer hits his first shot on the green and requires only one putt to put the ball in the hole is 0.2. If a PGA professional golfer hits his first shot on the green, find the probability that he only requires one putt to put the ball in the hole. 42. Eighty percent of the flights from Fresno to Las Vegas depart on time; in other words, the probability that a flight from Fresno to Las Vegas departs on time is 0.8. Seventy-two percent of all flights from Fresno to Las Vegas depart on time and arrive on time. If a flight from Fresno to Las Vegas departs on time, find the probability that it also arrives on time. 43. At a middle school, 48% of the students ride the bus to and from school. Thirty percent of the students ride the bus to and from school and buy their lunch at school. If a student rides the bus to and from school, find the probability that the student buys his or her lunch at school. 44. In a recent exit poll, 60% of the respondents felt that the country was generally going in the right direction. Twenty-four percent of the respondents were Republicans who felt that the country was generally going in the right direction. If a person in this poll felt that the country was generally going in the right direction, find the probability that the person was Republican. 45. In a recent exit poll, 40% of the respondents claimed to be Democrats. If a person was a Democrat, the probability that he or she voted for the Democratic candidate was 0.9. Find the probability that a person is a Democrat and voted for the Democratic candidate.
Independent Events, Multiplication Rule, Law of Large Numbers Independent Events
Two events A and B are said to be independent if the probability of event A is unaffected by event B occurring or not occurring. In other words, A and B are independent if P(A|B) = P(A).
3.4 Independent Events, Multiplication Rule, Law of Large Numbers
Sixty percent of the households in Visalia have the Visalia Times Delta delivered to their house. Twenty-five percent of the households in Visalia have the Fresno Bee delivered to their house. Fifteen percent of the households in Visalia have the Visalia Times Delta and the Fresno Bee delivered to their house. Are the events A household gets the Times Delta delivered to their house and A household gets the Fresno Bee delivered to their house independent?
Let D be the event that a household has the Times Delta delivered to their house, and let B be the event that a household has the Fresno Bee delivered to their house. We know that P(D) = 0.6, P(B) = 0.25, and P(D B) = 0.15. To show that the two events B and D are independent, we must show that P(B) = P(B |D).
P(B|D) = = P(B D) P(D ) 0.15 0.6
= 0.25 = P(B)
Therefore, P(B) = P(B |D), and events B and D are independent. We could have chosen to show that P(B) = P(D|B) to show that events B and D are independent, and that is left to you as an exercise. The events A household gets the Times Delta delivered to their house and A household gets the Fresno Bee delivered to their house are independent, since the probability of a household getting the Fresno Bee delivered is the same whether we know the house gets the Times Delta or not.
There is another way that we can check for independence. If P(A) P(B) = P(A B), then events A and B are independent. Where does this come from? It comes from the formula for conditional probability as shown next.
P(A) = P(A|B) P(A) = P(A)
P(A B) P(B) B)
P(B) = P(A
Lets revisit the previous example.
Let D be the event that a household has the Times Delta delivered to their house, and let B be the event that a household has the Fresno Bee delivered to their house. We know that P(D) = 0.6, P(B) = 0.25, and P(D B) = 0.15. To check for independence, we can check to see whether P(D) P(B) = P(D B).
P(D)
P(B) = (0.6) = 0.15 = P(D
Thus the events A household gets the Times Delta delivered to their house and A household gets the Fresno Bee delivered to their house are independent.
The probability that a student passes their math class is 0.8, the probability that a student makes the Deans List is 0.4, and the probability that a student passes their math class and makes the Deans List is 0.25. Are the events A student passes their math class and A student makes the Deans List independent events?
Let M be the event that the student passes their math class, and let D be the event that the student makes the Deans List. We know that P(M) = 0.8, P(D) = 0.4, and P(M D) = 0.25. To check for independence, we can check to see whether P(M) P(D) = P(M D).
P(M )
P(D ) = (0.8) = 0.32 P(M
Thus the events A student passes their math class and A student makes the Deans List are not independent events.
Once again, lets return to the burger preference example. One hundred fifty students at the College of the Sequoias were asked their preference of burgers. Their responses were categorized as Burger King, McDonalds, In-N-Out, and Other. Here are the results.
For the students in this survey, are the events A student is male and A student prefers Burger King independent events? Let M be the event that a student is male, and let B be the event that a student prefers Burger King. For these two events to be independent, we must show that P(M) P(B) = P(M B). Since 15 of the 150 students in the survey were male and preferred Burger King,
P(M B) = = 15 150 1 10
We know that P(M) =
and P(B) =
1 or 3. Therefore,
P(B ) = =
Since P(M)
P(B) P(M
B), the two events are not independent.
Later we will use such a survey to try and answer the broader question Is a persons burger preference independent of the persons gender? The events in this example were not independent, but that only relates to this sample. Were they close enough to independent to suggest that for all people burger preference is independent of a persons gender? We will examine this later.
Multiplication Rule
If two events A and B are independent, then the probability of A and B occurring is given by the following formula:
P(A B) = P(A)
P(B)
This is one form of the Multiplication Rule. Recall that we use intersection ( ) in circumstances involving the word and. Also recall from our study of counting that we used multiplication when we were interested in the number of ways that one event and another event could occur. The Multiplication Rule is consistent with that idea.
A fair coin is flipped two times. Find the probability that both tosses result in heads.
1 Let H be the event that a toss results in heads. We know that P(H) = 2. We are interested in the probability that both tosses result in heads, or, in other words, that the first toss is heads and the second toss is heads. The result of the second toss does not depend on the result of the first toss, so the two events are independent.
P(H first
H second) = P(H first) = = 1 2 1 4
P(H second)
1 The probability that both tosses are heads is 4.
The probability that a randomly selected American has an unfavorable opinion of the president is 0.35. If two Americans are selected at random, find the probability that both have an unfavorable opinion of the president.
Let A be the event that a randomly selected American has an unfavorable opinion of the president. We know that P(A) = 0.35. The response of the second American does not depend on the response of the first American.
P(A first
A second) = P(A first) = 0.35
P(A second)
The probability that both Americans have an unfavorable opinion of the president is 0.1225.
We can extend the Multiplication Rule to cover three or more events, as in the following example.
The probability that an elementary school student selects soccer as 1 his or her favorite sport is 4. If four elementary school students are selected at random, find the probability that they all select soccer as their favorite sport.
Let A be the event that a student selects soccer. We are interested in the probability that all four students select soccer as their favorite sport. This is equivalent to the probability that the first student selects soccer and the second student selects soccer and the third student selects soccer and the fourth student selects soccer.
P(All 4 select soccer) = P(A) = = 1 4
P(A)
The probability that all four students select soccer is
Multiplication Rule for Dependent Events
We can adapt the Multiplication Rule to cover events that are not independent. If A and B are any two events, then the probability of A and B occurring is given by
P(B|A)
Instead of multiplying the probability of A occurring by the probability of B occurring, we instead multiply by the probability of B occurring if A has already occurred. The best way to remember this is to try to find the probability of the first event occurring, then multiply by the probability of the next event occurring. Take the events one at a time, and multiply their probabilities.
Two cards are drawn from a well-shuffled standard 52-card deck of playing cards. Find the probability that both cards are diamonds.
Let D be the event that a card is a diamond. We are interested in the probability that both cards are diamondsthat is, that the first and second cards are diamonds. What is the probability that the first card is a diamond? Since there are 13 13 diamonds and 52 cards in the deck, the probability is 52. What is the probability that the second card is a diamond? There are now 12 diamonds and 51 cards left in the
deck, so the probability is 12. To find the probability that the first card is a diamond 51 and the second card is a diamond, we multiply these two probabilities.
P(D first D second) = P(D first) = = 13 52
P(D second)
Two cards are drawn from a well-shuffled standard 52-card deck of playing cards. Find the probability that both cards are face cards.
Let F be the event that a card is a face card. What is the probability that the first 12 card is a face card? It is 52, because there are 12 face cards in a standard 52-card 11 deck. The probability that the second card is a face card is 51, because there is one less face card in the deck on the second card, as well as one less card.
P(F first F second) = P(F first) = = 12 52 11 221
P(F second)
The probability that both cards are face cards is
Two cards are drawn from a well-shuffled standard 52-card deck of playing cards. Find the probability that one card is a king and the other card is a queen.
Let K be the event that a card is a king, and let Q be the event that a card is a queen. There are two ways that we could draw a king and a queen. We could draw a king and then a queen, or we could draw a queen and then a king. We will find the probability of each scenario, and then add. The probability that the first card is a 4 4 king is 52, and the probability that the second card is a queen is 51. The probability 4 that the first card is a queen is 52, and the probability that the second card is a king 4 is 51. This gives us
P(K and a Q) = P(K 1st) = = 4 52
P(Q 2nd) + P(Q 1st)
P(K 2nd)
4 4 + 51 52
4 4 + 663 663
The probability of drawing a king and a queen is
Three cards are drawn from a well-shuffled standard 52-card deck. Find the probability that two are aces and one is not.
Let A be the event that a card is an ace, and let N be the event that a card is not an ace. There are three different ways that two of the three cards could be aces: the first two cards could be aces, the first and third cards could be aces, or the last two cards could be aces. (In symbols: AAN, ANA, and NAA.) We must find the probability of each different way occurring, and then add those respective probabilities. Here is the probability of the first way, keeping in mind that there are four aces and 48 non-aces in a deck of 52 cards.
P(A first A second N third) = P(A first) = = 4 52
P(N third)
Now here is the second probability.
P(A first N second A third) = P(A first) = = 4 52
P(N second)
P(A third)
Note that the probability of the second way is exactly the same as the probability of the first. Here is the third way.
P(N first A second A third) = P(N first) = = 48 52
So the probability of drawing two cards that are aces and one that is not is found by adding the three probabilities.
P(2 aces and 1 non-ace) = = 24 24 24 + + 2252 5525 5525 72 5525
We see that the probability is
72 5525.
The previous examples are examples of sampling without replacement, because the card that is drawn is not replaced in the deck before the next card is drawn. Occasionally the sampling is done with replacement. In such a situation, the card would be reinserted in the deck before the next card is drawn. In sampling with replacement, the events are independent because the sample space is not affected by the previous outcome.
Two cards are drawn, with replacement, from a well-shuffled standard 52-card deck of playing cards. Find the probability that both cards are diamonds.
Let D be the event that a diamond is drawn. We are interested in P(D first) P(D second). What is the probability that the first card is a diamond? Since there are 13 13 1 diamonds in a 52-card deck, the probability is 52 or 4. What is the probability that the second card is a diamond? The sample space has not been changed, so the 13 1 probability is 52 or 4.
P(D first) P(D second) = P(D first) = = 1 4 1 16
The probability of drawing two diamonds with replacement is
Note that the probability of drawing two diamonds is higher when the sampling is done with replacement than without it. Can you explain why/how that happened? Think about ityou will get a chance to answer that question soon.
A fifth-grade class has 15 boys and 21 girls in it. Each week the teacher selects a student to clean the board at the end of the day. Find the probability that a boy will be selected two weeks in a row, if the same person cannot serve in consecutive weeks.
What we have here is essentially sampling without replacement, since the student selected one week is exempt the next week. Let B be the event that a boy is selected. We are interested in P(B first) P(B second). What is the probability that the helper will be a boy in the first week? Since there are 15 boys out of 36 stu15 5 dents, the probability is 36 or 12. What is the probability that the helper will be a boy in the second week? Since that leaves only 14 boys out of 35 students, the proba14 2 bility is 35 or 5.
P(B first) P(B second) = P(B first) = = = 15 36 5 12 1 6
P(B second)
1 The probability that a boy cleans the board for two straight weeks is 6.
A fifth-grade class has 15 boys and 21 girls in it. Each week the teacher selects a student to clean the board at the end of the day. Find the probability that a boy will be selected two weeks in a row, if the same person can serve in consecutive weeks.
What we have here is essentially sampling with replacement, since the student selected one week can be selected the next week. Let B be the event that a boy is selected. We are interested in P(B first) P(B second). What is the probability that
the helper will be a boy in the first week? Since there are 15 boys out of 36 stu15 5 dents, the probability is 36 or 12. What is the probability that the helper will be a boy in the second week? Again, there are 15 boys out of 36 students, so the probability 15 5 is 36 or 12.
P(B first) P(B second) = P(B first) = = = 15 36 5 12
The probability that a boy cleans the board for two straight weeks is
Subjective Probability
When we introduced classical probability, it was necessary that all of the outcomes in the sample space were equally likely to occur. This is not always the case, however. Consider the following example.
The possible types of precipitation are rain, sleet, hail, snow, and none. Find the probability that it snows in Las Vegas on the next Fourth of July.
Lets look at the sample space. S = {Rain, Sleet, Hail, Snow, None}. Is the probabil1 ity of snow 5? It would be, according to classical probability, if each of the five outcomes were equally likely. I have relatives who live in Las Vegas who have told me that it is far too warm to snow in July, and this tells me that the probability should 1 be a lot lower than 5.
How do we find the probability of an event occurring when all of the outcomes are equally likely? This is the field of subjective probability. We use history, our own expertise, or some other method to estimate the probability in question.
Find the probability that a professional basketball player makes his next free throw, if he has made 221 out of his previous 425 attempts this season.
221 The best we can do is come up with an estimate of this probability. Does 425 or 0.52 sound reasonable? If in the past the player has made 52% of his free throws, it stands to reason that there is a 52% chance on any single free throw.
Find the probability that a professional basketball player makes his next free throw, if he has made four out of his five attempts this game.
Does it seem reasonable that there is an 80% chance that the player will make his next free throw attempt? There is a little problem with this estimate. It is based on a small number of attempts (only five), and is not as reliable as the estimate in the previous example where there were 425 attempts.
Law of Large Numbers
The free-throw examples lead us to the Law of LAW OF LARGE NUMBERS As the number of trials of an Large Numbers. experiment increases, the proportion of successes in those trials As far as the free throws go, 5 is not a large approaches the probability of success on one trial. number but 425 is. The question is, how large a number is large? We will be able to calculate how many trials are appropriate in a later chapter, but for now we will have to use our intuition. The following Excel project is the Law of Large Numbers in action. It simulates flipping a coin 1000 times. The probability that a coin lands on heads is 0.5, so the percentage of heads should get closer and closer to 50% as the number of tosses increases.
The Law of Large Numbers
Before we begin, we must turn off the Excel feature that automatically recalculates functions. Click on Tools, and then Options. When the Options dialog box appears, click on the tab labeled Calculation. Click on Manual, and then OK. We also must be sure that the Analysis Tool Pak has been added. Click on Tools. If you see a choice that says Data Analysis, then the Tool Pak is added and you may continue. If you do not see Data Analysis, click on Add-ins under Tools. When the dialog box opens up, click on the box for Analysis Tool Pak, and click on OK. In cell A1, type Toss #. In cell B1, type Heads(1), Tails(0). In cell C1, type % Heads. In cell A2, type 1. In cell B2, type =RANDBETWEEN(0,1). This simulates the toss of a coin. The result will either be a 0, which stands for tails, or a 1, which stands for heads. In cell C2, type =AVERAGE($A$2:A2). Click on cell A13. Click on Window, and then Split. This will divide the screen horizontally. Click on Edit, and then Go To. When the dialog box opens, type A1001. Hold down the Shift key and click on cell A2. This should highlight column A. Click on Edit, then Fill, and then Series. When the dialog box appears, click on OK. This will fill the first column with the number of the toss. Highlight cells B2 and C2. Click on Edit, and then Copy. Click on cell B3. Then hold down the Shift key and click on cell C1001. Press the Enter key. This will copy the information into columns B and C. Press the F9 key to recalculate the worksheet. Here is an example of the results (graphically) of the first 50 tosses. Note how the graph appears to be drawn toward 0.5 as we move to the right.
Percentage of Heads, First 50 Tosses 0.7 Percentage of Heads 0.6 0.5 0.4 0.3
21 25 29 33 Number of Tosses
To produce a graph like this for your example, select Chart from the Insert menu. When the dialog box appears, select XY (Scatter) for the chart type and select the Scatter with data points connected by smooth Lines option for the chart subtype. Click on Next to move to step 2. In the box labeled Data Range, type the following: =Sheet1!$A$2:$A$51,Sheet1!$C$2:$C$51 Click on the Series tab. In the box labeled X Values, type =Sheet1!$A$2:$A$51. In the box labeled Y Values, type =Sheet1!$C$2:$C$51. Click on Next to move to step 3. Click on the Titles tab. To add a label for the x axis, such as Number of Tosses, type it in the box labeled Value (X) axis. You can type % Heads in the box labeled Value (Y) axis to label the y axis. Click on Finish to see your graph. Here are the results (graphically) of the first 200 tosses.
Percentage of Heads, First 200 Tosses 0.7
Percentage of Heads
85 99 113 127 141 155 169 183 197 Number of Tosses
Note that the proportion is jumpy for the lower number of tosses, but as the number of tosses approaches 200, the proportion should be drawn toward 0.5. Now here are the results of the 1000 tosses. Note again that as the number of tosses increases, the proportion of heads is drawn in toward 0.5.
Percentage of Heads, 1000 Tosses 0.6 Percentage of Heads 0.55 0.5 0.45 0.4
73 145 217 289 361 433 505 577 649 721 793 865 937 Number of Tosses
To produce these last two graphs, follow the above directions, extending the range of cells down to row 201 and row 1001, respectively. Print out your three graphs, and in your own words, comment on how they show the Law of Large Numbers at work.
1. The probability that a community college student is female is 0.55. The probability that a community college student is on the Deans List is 0.32. The probability that a community college student is female and on the Deans List is 0.17. Is being female independent of being on the Deans List for community college students? 2. The probability that a resident of Visalia attends the County Fair in June is 0.65. The probability that a resident of Visalia attends the Annual Fog Festival in January is 0.2. The probability that a resident of Visalia attends the County Fair and the Fog Festival is 0.13. Is attending the County Fair independent of attending the Fog Festival for Visalia residents? 3. If a fair six-sided die is rolled and a fair coin is tossed, the probability that the roll 1 of the die is a 5 and the toss of the coin is heads is 12. Are the events A roll of a die is 5 and A toss of a coin is heads independent events? 4. List two events that you intuitively feel are independent events. Explain your reasoning in your own words. List two events that you intuitively feel are not independent events. Explain your reasoning in your own words. A fair coin is flipped three times. 5. Find the probability that all three tosses are heads. 6. Find the probability that all three tosses are tails. 7. Find the probability that the first two tosses are heads and the last is tails. 8. Find the probability that two of the tosses are heads and the other is tails. A fair six-sided die is rolled twice. 9. Find the probability that each roll is a 5. 10. Find the probability that neither roll is a 5. 11. The probability that an elementary school child lists basketball as his or her favorite sport is 0.18. If five elementary school children are selected at random, find the probability that they all name basketball as their favorite sport. 12. The probability that an adult lists blue as his or her favorite color is 0.15. If five adults are selected at random, find the probability that at least one fails to name blue as his or her favorite color. 13. In a certain experiment, the probability of an error is 5%, or 0.05. If the experiment is repeated six times, find the probability that at least one of the trials will result in an error. 14. In a certain experiment, the probability of an error is 1%, or 0.01. If the experiment is repeated four times, find the probability that at least one of the trials will result in an error.
Two cards are drawn from a well-shuffled standard 52-card deck of playing cards. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. Find the probability that both cards are diamonds. Find the probability that both cards are red. Find the probability that both cards are kings. Find the probability that the first card is red and the second card is black. Find the probability that the first card is a spade and the second is a diamond. Find the probability that the first card is a spade and the second card is not a spade. Find the probability that one card is a spade and one card is a diamond. Find the probability that one card is a face card and one card is an ace. Find the probability that all three cards are spades. Find the probability that all three cards are face cards. Find the probability that the first two cards are spades and the third card is a heart. Find the probability that the first card is an ace and the last two cards are kings.
Three cards are drawn from a well-shuffled standard 52-card deck of playing cards.
27. A card is drawn from a well-shuffled standard 52-card deck of playing cards, and then replaced in the deck. A second card is then drawn. Find the probability that both cards are diamonds. Now compare your answer to your answer from Exercise 15. Explain why the answer in this exercise (with replacement) is higher than the probability in Exercise 15 (without replacement). A card is drawn from a well-shuffled standard 52-card deck of playing cards, and then replaced in the deck. A second card is then drawn. 28. Find the probability that both cards are aces. 29. Find the probability that one card is a diamond and one card is a heart. 30. Find the probability that one card is a spade and one card is not a spade. 31. An elementary school class has 16 girls and 11 boys. Each day the teacher selects a student to give a current events report to the class. The teacher posts a schedule for the entire week (Monday through Friday) each Friday. Find the probability that a boy is selected for each day next week, if no student can be selected more than once in a week. 32. A jar of M&M chocolate candies has 12 blue candies and 108 candies that are other colors. A child reaches into the jar, pulls out a candy, and then drops the candy back into the jar. If the child does this three times, find the probability that all three candies are blue.
The purpose of this assignment is to estimate the probability that a randomly selected student on your campus is femalein other words, the percentage of the students at your school that are female. Visually survey 200 students on your campus. This can be done in your classes, in the library, in the center of campus, in the parking lot, or in some other fashion. Keep in mind that we want the sample to be representative of the population. Your best chance at this is to be as random as possible. Draw a graph showing the percentage of your survey that is female after 10 students, 25 students, 50 students, 100 students, 150 students, and 200 students. Here is an example of such a graph. (If you use Excel, type 1 for a female student and 0 for a male student.)
Female COS Students 0.8
0.6 0.4 0.2 0 0 50 100 Students 150 200 250
Now use your results to estimate the probability that a student at your school is female.
Unit 2 began with counting, which involves techniques used to determine the number of different outcomes an experiment may have. We began with tree diagrams and the dice chart, but focused on three techniques: the Multiplication Principle, permutations, and combinations. We use multiplication when a choice consists of two or more steps. We multiply the number of different possibilities for the first step by the number of different possibilities for the second step, and then the third step, and so on. If we are selecting r objects, without repetition, from a set of n distinct objects, and the order of selection matters, then we use permutations. If the order of selection does not matter, we use combinations. Probability is a measure of how likely it is that a particular outcome(s) occurs. To apply classical probability, all of the possible outcomes must be equally likely to occur. The General Addition Rule is used when we are interested in the probability of event A or event B occurring. We apply the Complement Rule when we are looking for the probability of the complement of an event occurring. The key word to the complement rule is often not. The rules of conditional probability are applied in situations where the original sample space has been restricted by some condition. This leads to the idea of independent events. If the probability of event A is unaffected by whether B has occurred or not, and vice versa, then the two events are independent. If we are interested in the probability of event A occurring and event B occurring, we often use the Multiplication Rule.
Review Exercises
1. Draw a tree diagram showing the results of three tosses of a coin. In how many ways is it possible to get no heads? one head? two heads? three heads? 2. Freebirds Pizza has four types of crust, three types of cheese, and 12 toppings. In how many ways can a customer order a three-topping pizza with one type of crust and one type of cheese? 3. Passwords at the computer lab consist of two letters of the alphabet, the first of which cannot be I, O, or Q, and three digits, the first of which cannot be 0. How many different passwords are possible with this scheme? 4. A multiple choice test consists of eight questions, each having four possible answers. (a) In how many ways can a student answer the eight questions? (b) In how many ways can a student answer the eight questions correctly? (c) In how many ways can a student answer the eight questions incorrectly? 5. A statistics instructor who has taken leave of his senses decides to randomly pick 5 of his 35 students and excuse them from the final exam. In how many ways can this be done? 6. A local club has 15 members. They decide to form a four-person committee to oversee the planning of meetings. In how ways can this be done under the following conditions? (a) There is to be one person in charge of notifying members, one person in charge of refreshments, one person to run the meeting, and one person to clean up. (b) There is no distinction between the jobs on the committee. (c) One person is to run the meeting, and there is no distinction between the other jobs on the committee. 7. Forty-eight of 60 dentists in a town recommend sugarless gum for their patients that chew gum. If one of these dentists is chosen at random, what is the probability that he or she recommends sugarless gum for their patients who chew gum? 8. Here is a table listing the probability of Tony Gwynn getting no, one, two, three, four, or five hits in a single game.
Number of hits Probability
(a) Find the probability of Tony getting two or more hits in a game. (b) Find the probability of Tony getting fewer than five hits in a game. (c) Find the probability of Tony getting at least one hit in a game for three consecutive games. 9. The probability that a customer at a fast-food restaurant orders a beverage is 0.60. The probability that a customer orders French fries is 0.50. The probability that a customer orders both a beverage and French fries is 0.42. What is the probability that a customer at a fast-food restaurant orders a beverage or French fries? 10. One hundred students were asked their favorite fast-food restaurant: Taco Bell, McDonalds or neither. Here are the results. McDonalds Male Female 35 10 Taco Bell 15 20 Neither 5 15
(a) Find the probability that a student is male and prefers McDonalds. (b) If a male student is selected at random, find the probability that he prefers McDonalds. (c) If a student who prefers Taco Bell is selected at random, find the probability that the student is female. (d) If a student who prefers neither is selected at random, find the probability that the student is female. (e) If a female student is selected at random, find the probability that she prefers McDonalds or Taco Bell. On a par 3 golf hole, the probability that a professional golfer hits his first shot onto the green and makes a birdie is 0.24. The probability that a professional golfer hits his first shot onto the green on a par 3 golf hole is 0.60. Find the probability that a professional golfer who hits his first shot onto the green will make a birdie. If P(A) = 0.30, P(B) = 0.40 and P(A B) = 0.07, are events A and B independent? Find the probability of drawing two black cards from a deck (a) with replacement (b) without replacement Six female students and six male students are nominated for a scholarship. Three scholarships are awarded. If all 12 students are equally qualified for the scholarship, find the probability that the three scholarships are awarded to females. User IDs at the computer lab use four letters followed by four numbers. How many IDs are possible under the following conditions? (a) The first letter cannot be O. (b) The first letter cannot be O, and no letter can be repeated. A student who is 20 years old and weighs 175 lb. owns four credit cards. He purchases three new shirts, two new pairs of shoes, and four new pairs of pants. In how many different ways can this student choose a new outfit to wear (one shirt, one pair of shoes, and one pair of pants)? The I Love Statistics Club has 25 members. In how many ways can they elect (a) a four-person leadership committee (b) a president, a vice president, a treasurer, and a secretary (c) a president, a vice president, and a two-person leadership committee On a ten-question multiple choice test, if a student guesses on all ten questions, the probabilities of getting none, one, two, . . . , ten correct are listed below. 0 0.107 1 0.268 2 0.302 3 0.201 4 0.088 5 0.026 6 0.006 7 0.002 8 0 9 0 10 0
Number right Probability
(a) Find the probability of getting at most two questions correct by guessing. (b) Find the probability of getting at least five questions correct by guessing.
19. Find the probability of drawing two hearts from a well-shuffled deck of cards if the cards are drawn (a) without replacement (b) with replacement 20. Seven males and five females are to be interviewed for a job as a community college instructor. The top four candidates are sent forward to the president for a second interview. If all the candidates are equally qualified, find the probability that (a) four females get a second interview (b) at least one male gets a second interview 21. In the following table, 76 college students are classified according to their class standing and also according to their favorite pizza topping. Anchovies Freshman Sophomore 4 4 Mushrooms 12 11 Onions 8 6 Pepperoni 14 17
(a) Find the probability that a student is a sophomore or prefers mushrooms. (b) If a freshman is selected at random, find the probability that the student prefers pepperoni. (c) Find the probability that a student prefers onions and is a sophomore. (d) If a student who prefers anchovies is selected at random, find the probability that the student is not a freshman. (e) If a student who prefers anchovies or mushrooms is selected at random, find the probability that the student is a sophomore.
Permutations: nPr = Combinations: nCr = n! (n r)! n! (n r)! n(A) n(S) B) = P(A) + P(B)
Classical probability: P(A) = Special Addition Rule: P(A
Complement Rule: P(A) = 1 P(A ) General Addition Rule: P(A B) = P(A) + P(B) P(A P(A B) P(B)
Conditional probability: P(A|B) =
Condition for independent events: P(A|B) = P(A) or P(A) Multiplication Rule (independent events): P(A Multiplication Rule (dependent events): P(A
B) = P(A) B) = P(A)
DISCRETE PROBABILITY DISTRIBUTIONS
Section 4.1 Probability Distributions Section 4.2 Binomial Probabilities Section 4.3 The Poisson Distribution
NORMAL PROBABILITY DISTRIBUTIONS
Section 5.1 The Normal Distribution Section 5.2 Discrete Random Variables That Are Approximately Normal; Normal Approximation to the Binomial Distribution Section 5.3 Finding Values for Given Probabilities
CHAPTER 4 Discrete Probability Distributions
robability is the tool that allows the study of inferential statistics. In this chapter we begin our study of probability distributions by focusing on two specific discrete probability distributions. In Section 4.2 we will learn about the binomial probability distribution. The binomial distribution is used to make inferences about population proportions, or percentages. In Section 4.3 we will study the Poisson distribution.
In this section we continue our investigation into probability by introducing the concept of a probability distribution. A probability distribution associates a probability with each possible outcome of an experiment. For example, here is the probability distribution for flipping a fair coin.
Outcome Heads Tails Probability 0.5 0.5
Another example is the probability distribution for the store at which a woman buys her New Years Eve gown.
Store Gottschalks JC Penneys Mervyns Other Probability 0.4 0.3 0.1 0.2
Before moving on, we introduce the idea of a random variable. A random variable is a variable that takes on a numerical value for each possible outcome of a probability experiment.
A fair coin is tossed 3 times, and the number of heads is recorded.
What is the random variable? The random variable x represents the number of heads in the 3 tosses. What values can the random variable take on? The number of heads in 3 tosses can be anywhere from zero through 3 (0, 1, 2, or 3).
A couple has 6 children, and the number of boys is recorded.
What is the random variable?
4.1 Probability Distributions
The random variable x represents the number of boys out of the 6 children. What values can the random variable take on? The number of boys out of 6 children can be anywhere from 0 through 6.
An adult male is selected at random, and his height is recorded.
What is the random variable? The random variable x represents the height of the male. What values can the random variable take on? The height of an adult male can take on infinitely many positive values.
In the preceding example, the height of an adult male could take on infinitely many values. Recall from Chapter 1 that height was given as an example of continuous data. The random variable for the previous example is called a continuous random variable. There are two types of random variables: continuous and discrete. A continuous random variable is a variable that can assume infinitely many values from a continuous interval (there are no gaps between possible values). A discrete random variable is a variable that can assume either a finite number of values or a countable number of values. Discrete random variables usually involve some sort of a count (0, 1, 2, 3, . . .). If the possible values of a discrete random variable were placed on a number line, there would be gaps between the values. Continuous random variables are usually associated with physical measurements.
Determine whether the random variables for the following experiments are continuous or discrete.
The random variable x (number of heads) is discrete because it can take on only 4 values: 0, 1, 2, and 3. Any variable that can assume only finitely many values is discrete.
A 6-month-old baby girl is selected at random, and her weight is recorded.
The random variable x (weight of the girl) is continuous. Recall that weight is an example of continuous data. The weight of a 6-month-old girl could take on values such as 15.3 pounds, 15.34 pounds, 15.347 pounds, and infinitely many other values in an interval that is centered at approximately 15.5 pounds.
A basketball player who is a poor free-throw shooter must make a free throw before he is allowed to leave practice. The number of free throws that he attempts until he makes one is recorded.
The random variable x (number of free throws attempted) is discrete. The random variable could take on any of the values 1, 2, 3, 4, . . . . Although there are an infinite number of possible values, this number is countable and the variable is discrete.
The remainder of this chapter is devoted to discrete random variables and their probability distributions. In Chapter 5 we will examine a continuous probability distribution called the normal distribution. A probability distribution for a discrete random variable lists each possible value of the variable along with its probability. There are two basic rules that a probability distribution must follow. First, each probability must be between 0 and 1.
0 P(x) 1
Second, since we list all of the possible outcomes, the total of their probabilities must equal 1. P(x) = 1
Are the following valid probability distributions?
x 0 1 2 3
P(x) 0.15 0.25 0.35 0.45
Although each individual probability is between 0 and 1, this is not a valid probability distribution because their sum is not equal to 1.
x 0 1 2 P(x) 0.2 0.3 1.5
This is not a valid probability distribution because P(2) is not between 0 and 1.
x 0 1 2 3 P(x) 0.343 0.441 0.189 0.027
Since each probability is between 0 and 1, and the sum of all of the probabilities is equal to 1, this represents a valid probability distribution.
The use of a probability distribution makes it easier to calculate probabilities.
Use the given probability distribution to find the following probabilities.
FIVE-QUESTION TRUE/FALSE QUIZ (x = number correct) x 0 1 2 3 4 5 P(x) 0.03125 0.15625 0.3125 0.3125 0.15625 0.03125
Find the probability that a randomly selected student gets 3 questions correct.
P(x = 3) = 0.3125
Find the probability that a randomly selected student gets at least 4 questions correct.
P(At least 4) = P(x = 4) + P(x = 5) = 0.15625 + 0.03125 = 0.1875
Find the probability that a randomly selected student gets fewer than 4 questions correct.
P(Fewer than 4) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) = 0.03125 + 0.15625 + 0.3125 + 0.3125 = 0.8125
We could have used the complement rule on the previous problem.
P(Fewer than 4) = 1 (P(x = 4) + P(x = 5)) = 1 (0.15625 + 0.03125) = 1 0.1875 = 0.8125
Twenty-five percent of all Americans have a cellular phone. The probability distribution for the number of Americans with cell phones out of 10 randomly selected Americans is listed here.
Number with a Cell Phone (x) 0 1 2 3 4 5 6 7 8 9 10 Probability P(x) 0.0563 0.1877 0.2816 0.2503 0.1460 0.0584 0.0162 0.0031 0.0004 0.0000 0.0000
(Based on results of a study by Baskerville Communications.)
(a) If 10 Americans are selected at random, find the probability that 4 of them have a cellular phone. Directly from the probability distribution,
(b) If 10 Americans are selected at random, find the probability that at least 7 of them have a cellular phone.
P(At least 7) = P(x = 7) + P(x = 8) + P(x = 9) + P(x = 10) = 0.0031 + 0.0004 + 0.0000 + 0.0000 = 0.0035
(c) If 10 Americans are selected at random, find the probability that at least 2 of them have a cellular phone.
P(At least 2) = 1 (P(x = 0) + P(x = 1)) = 1 (0.0563 + 0.1877) = 1 0.2440 = 0.7560
If we construct a histogram for a probability distribution, we can use the area in the histogram to obtain a graphical representation of the probability distribution. Here is the histogram for the previous probability distribution. 10 Randomly Selected Americans 0.3000 0.2500 0.2000 0.1500 0.1000 0.0500 0.0000
0 1 2 3 4 5 6 7 8 9 10 Number with Cell Phone
The probability that 2 of the 10 Americans have a cell phone is 0.2816, which is also the area (height) of the bar for x = 2. The association between area and probability will become very important when we investigate continuous probability distributions in Chapter 5. Each probability distribution has a mean and standard deviation associated with it, just as a set of data has a mean and standard deviation. The mean is a typical value for the probability distribution and represents the average outcome that we can expect in the long run. Often the mean of a probability distribution is called its expected value. The mean of a probability distribution can be calculated using the following formula:
P(x)
Find the mean of the following probability distribution.
There are two ways to calculate the mean. We can use the formula directly, or we can take a column approach.
= 0(0.03125) + 1(0.15625) + 2(0.3125) + 3(0.3125) + 4(0.15625) + 5(0.03125) = 0 + 0.15625 + 0.625 + 0.9375 + 0.625 + 0.15625 = 2.5
The mean for this probability distribution is 2.5. If we repeated this experiment over and over, the mean number correct would be 2.5. This is also the number of questions that we could expect any single student to get correct, even though it is impossible for any student to get exactly 2.5 correct. Now we will use the column approach. We multiply each value of the variable by its probability, and write that result in the next column. When we total that column we will have the mean.
x 0 1 2 3 4 5 P(x) 0.03125 0.15625 0.3125 0.3125 0.15625 0.03125 x
0 0.15625 0.625 0.9375 0.625 0.15625 = 2.5
Does it make sense that the mean is 2.5? If there are 5 true/false questions, it stands to reason that we can expect to get half of the questions correct by guessing.
Don is a 65% free-throw shooter. This means that he makes 65% of his free throws, or the probability that he makes any individual free throw is 0.65. Given next is the probability distribution for the number of free throws made in 2 attempts by Don. Find the mean for this probability distribution.
x 0 1 2
P(x) 0.1225 0.455 0.4225
This time we will use the column approach.
0 0.455 0.845 = 1.3
The mean for this probability distribution is 1.3. If Don repeated this experiment over and over, the mean number of free throws he would make is 1.3. Does it make sense that he would average 1.3 free throws made out of 2 attempts? 1.3 is 65% of 2, so if he makes 65% of his free throws he would make 1.3 out of 2 on average.
The standard deviation of a probability distribution measures the variability in the random variable. Here is the formula for standard deviation of a probability distribution.
= ((x )2
P(x))
We begin by finding , the mean of the probability distribution. We then subtract the mean from each value of x, and square those differences. After multiplying each of these squares by the probability of x occurring, we total these products. We finish by taking the square root of this total.
Find the standard deviation of the following probability distribution.
As with the mean, we will use a column approach. We begin with the mean.
Next we will calculate the standard deviation.
x 0 1 2 3 4 5 P(x) 0.03125 0.15625 0.3125 0.3125 0.15625 0.03125 x 2.5 1.5 0.5 0.5 1.5 2.5 (x )2 6.25 2.25 0.25 0.25 2.25 6.25 P(x)
(x )2
0.19531 0.35156 0.07813 0.07813 0.35156 0.19531 1.25
= = ((x )2 1.25
= 1.118
The standard deviation of this distribution is 1.118 correct answers.
Twenty-five percent of all Americans have a cellular phone. The probability distribution for the number of Americans with cell phones out of 10 randomly selected Americans is listed next.
Find the mean and standard deviation for this distribution. First, we find the mean.
x 0 1 2 3 4 5 6 7 8 9 10 P(x) 0.0563 0.1877 0.2816 0.2503 0.1460 0.0584 0.0162 0.0031 0.0004 0.0000 0.0000 x
0.0000 0.1877 0.5632 0.7509 0.5840 0.2920 0.0972 0.0217 0.0032 0.0000 0.0000 2.4999
The mean of this probability distribution is approximately 2.5 Americans. Next we calculate the standard deviation.
x 0 1 2 P(x) 0.0563 0.1877 0.2816 x 2.5 1.5 0.5 (x )2 6.25 2.25 0.25 P(x)
0.3519 0.4223 0.0704 (continues)
(continued) x P(x) 3 4 5 6 7 8 9 10 0.2503 0.1460 0.0584 0.0162 0.0031 0.0004 0.0000 0.0000
x 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5
(x )2 0.25 2.25 6.25 12.25 20.25 30.25 42.25 56.25
0.0626 0.3285 0.3650 0.1985 0.0628 0.0121 0.0000 0.0000 1.8741
= = ((x )2 1.8741
The standard deviation of this distribution is 1.369 Americans.
1. Here is the probability distribution for the number of heads in 4 tosses of a fair coin.
x 0 1 2 3 4 P(x) 0.0625 0.2500 0.3750 0.2500 0.0625
(a) (b) (c) (d)
Find the probability that exactly 2 of the tosses are heads. Find the probability that at least 2 of the tosses are heads. Find the probability that at least 1 of the tosses is heads. How many heads can we expect in the 4 tosses?
2. A fair six-sided die is rolled 8 times. Here is the probability distribution for the number of 3s.
x 0 1 2 3 4 5 6 7 8 P(x) 0.2326 0.3721 0.2605 0.1042 0.0260 0.0042 0.0004 0.0000 0.0000
Find the probability that between 3 and 5 rolls, inclusive, are 3s. Find the probability that more than 3 rolls are 3s. Find the probability of at least two 3s. How many 3s should we expect?
3. Here is the probability distribution for the number of candy bars sold at a house by a child selling the candy for a school fundraiser.
x 0 1 2 3 P(x) 0.4 0.25 0.2 0.15
(a) Find the probability that the child fails to sell a candy bar at a particular house. (b) Find the probability that the child sells at least 2 candy bars at a particular house. (c) How many candy bars should the child expect to sell at a house? 4. In Californias Super Lotto game, the lottery draws 6 numbers from a drum containing 51 numbers (151). Players select 6 numbers for their ticket, and the payoff they receive depends on how many of their numbers match the 6 that were selected by the lottery. Here is a probability distribution for the number of numbers that a ticket matches.
x 0 1 2 3 4 5 6 P(x) 0.4522656426 0.4070390784 0.1240972800 0.0157583848 0.0008245666 0.0000149921 0.0000000555
(a) Players do not receive a payout unless they match at least 3 of the selected numbers. Find the probability that a ticket receives a payout. (b) Find the probability that a ticket does not receive a payout. (c) How many numbers should a player expect to match on a ticket? 5. Construct a probability distribution for x; the sum of 2 fair dice. Use the probability distribution to answer the following. (a) (b) (c) (d) Find the probability of rolling a sum of at least 6. Find the probability of rolling a sum that is higher than 9. Find the probability of rolling a sum that is even. What sum should we expect when rolling 2 fair dice?
6. Complete the following probability distribution, and calculate its mean.
x 0 1 2 3 4 5 6 P(x) 0.28 0.31 0.19 0.07 0.03 0.01
x 0 1 2 3 4 5 6 7 8 P(x) 0.04 0.05 0.07 0.15 0.22 0.14 0.11 0.09
8. The mean for the following probability distribution is 3. Complete the probability distribution.
x 0 1 2 3 4 5 P(x) 0.1 0.1 0.15
For the following probability distributions, calculate their mean and standard deviations. (Note: Probabilities may not add to be 1 due to rounding.) 9. Eight randomly selected children, ages 1217, x = number who smoke cigarettes
(Based on the 1998 Health and Human Services National Household Survey on Drug Abuse.)
10. Six randomly selected Americans, ages 2534, x = number who rent their home
x 0 1 2 3 4 5 6 P(x) 0.0905 0.2673 0.3292 0.2162 0.0799 0.0157 0.0013
(Based on the results of a study by the Census Bureau.)
11. Seven randomly selected residents of Mississippi, x = number who have health insurance
x 0 1 2 3 4 5 6 7 P(x) 0.2097 0.3670 0.2753 0.1147 0.0287 0.0043 0.0004 0.0000
(Source: Associated Press.)
12. Five randomly selected drivers, x = number who admit to running red lights
x 0 1 2 3 4 5 P(x) 0.0165 0.1049 0.2671 0.3400 0.2164 0.0551
(Based on the results of a study by the U.S. Department of Transportation.)
13. x = number of bicycle crashes on a universitys bike paths on a given day
x 0 1 2 3 4 5 6 7 8 9 10 P(x) 0.1225 0.2572 0.2700 0.1890 0.0992 0.0417 0.0146 0.0044 0.0011 0.0003 0.0001
14. x = number of structure fires in a town on a given day
x 0 1 2 3 P(x) 0.7408 0.2222 0.0333 0.0033
15. x = number of cars stolen in a city on a given day
x 0 1 2 3 4 5 6
P(x) 0.4966 0.3476 0.1217 0.0284 0.0050 0.0007 0.0001
16. x = number of interceptions thrown by an NFL quarterback in a given game
Binomial Probabilities
In this section we will examine one particular type of probability problem, which involves the binomial distribution. We could have done this problem in Chapter 3; now we will try to derive a formula to assist us. We begin with a solution that could have appeared in Chapter 3.
A 1996 study of first-time college freshmen by the Higher Education Research Institute shows that 20% of men planned to major in computer science or engineering. If 5 men who are first-time college freshmen are selected at random, find the probability that exactly 2 of them plan to major in computer science or engineering.
We must first decide in how many different ways 2 of the 5 freshmen could plan to major in computer science or engineering, with the other 3 planning to major in other than computer science or engineering. There are 10 different ways. One possible way is that the first 2 freshmen plan to major in computer science or engineering, and the last 3 plan on a different major. Here is the probability of that one way occurring. Keep in mind that the probability that a freshman male plans to major in computer science or engineering is 0.2, and the probability that a freshman male plans on another major is 0.8.
(0.2)(0.2)(0.8)(0.8)(0.8) = (0.2)2(0.8)3 = 0.02048
Here are the 10 different ways, along with their probabilities. We let Y represent a freshman male who plans to major in computer science or engineering, and we let N represent a freshman male who plans to major in something else.
1 and 2 do (YYNNN): 1 and 3 do (YNYNN): 1 and 4 do (YNNYN):
(0.2)(0.2)(0.8)(0.8)(0.8) = (0.2)2(0.8)3 (0.2)(0.8)(0.2)(0.8)(0.8) = (0.2)2(0.8)3 (0.2)(0.8)(0.8)(0.2)(0.8) = (0.2)2(0.8)3
4.2 Binomial Probabilities
1 and 5 do (YNNNY): 2 and 3 do (NYYNN): 2 and 4 do (NYNYN): 2 and 5 do (NYNNY): 3 and 4 do (NNYYN): 3 and 5 do (NNYNY): 4 and 5 do (NNNYY):
(0.2)(0.8)(0.8)(0.8)(0.2) = (0.2)2(0.8)3 (0.8)(0.2)(0.2)(0.8)(0.8) = (0.2)2(0.8)3 (0.8)(0.2)(0.8)(0.2)(0.8) = (0.2)2(0.8)3 (0.8)(0.2)(0.8)(0.8)(0.2) = (0.2)2(0.8)3 (0.8)(0.8)(0.2)(0.2)(0.8) = (0.2)2(0.8)3 (0.8)(0.8)(0.2)(0.8)(0.2) = (0.2)2(0.8)3 (0.8)(0.8)(0.8)(0.2)(0.2) = (0.2)2(0.8)3
Note that the probability is the same for each way, so we may just multiply (0.2)2(0.8)3 by 10. The probability that 2 of the 5 freshman males plan to major in computer science or engineering is
(0.2)2(0.8)3 = 0.2048
This problem is an application of the binomial distribution. For a problem to be a binomial problem there must be a fixed number of trials. We label the number of trials as n. In the above example there are 5 trialseach of 5 freshman males is asked about his major. Each outcome of a binomial trial must be classified as a success or a failure. The success is the outcome that we are interested in the probability of, and the failure is any other outcome. In the above example the success is that a freshman male plans to major in computer science or engineering, and the failure is that a freshman male plans to major in something other than computer science or engineering. Finally, each trial must be independent. This means that the probability of a success on any individual trial remains the same for each trial. In the above example, one freshmans choice of major does not affect another freshmans choice of major. These trials are independent, and the probability that any freshman male plans to major in computer science or engineering is 0.2. As you read a problem, keep an eye out for the idea of x out of n. Binomial problems involve phrases such as 2 out of 5, between 3 and 6 out of 10, and at least 1 out of 7. Also watch for a probability of success. Although most problems will list a probability of success on any individual trial, in some cases the probability of success will be assumed to be known (probability of heads, probability that a child is a girl, probability of getting a multiple choice question correct if there are 4 possible answers, and so on). If a binomial experiment consists of n trials, and the probability of success is p, then the probability of x successes in n trials is given by the following formula:
P(x) = nCx
(1 p)nx
Before we use the formula, lets analyze it. The number of different ways to have x successes in n trials is given by nCx . Since there are x successes, the probability of success p must show up x times, and this is represented in the formula by p X. Similarly, since there are n x failures, the probability of failure (1 p) must show up n x times, and this is represented in the formula by (1 p)nx. Now we revisit the first example, this time applying the binomial formula.
A 1996 study of first-time college freshmen by the Higher Education Research Institute shows that 20% of men planned to major in computer science or engineering. If 5 men that are first-time
college freshmen are selected at random, find the probability that exactly 2 of them plan to major in computer science or engineering. The safest way to attack this problem is by listing the number of trials, the success, the number of successes that we are interested in, the probability of success, the failure, and the probability of failure. Number of Trials: n = 5 Success: A freshman male plans to major in computer science or engineering. Number of successes: x = 2 Probability of success: p = 0.2 Failure: A freshman male plans to major in something other than computer science or engineering. Probability of failure: 1 p = 0.8
P(x) = P(2) =
nCx 5C2
(0.2)2
The probability that 2 of the 5 freshmen males plan to major in computer science or engineering is 0.2048.
Find the probability that 4 of 7 randomly selected 1825-year-olds used alcohol during the past 30 days, if the probability that any 1825-year-old used alcohol during the past 30 days is 0.6. (According to data from the U.S. Department of Health and Human Services, 60% of all 1825-year-olds reported using alcohol within the 30 days prior to the survey.)
Number of trials: n = 7 Success: An 1825-year-old has used alcohol during the past 30 days. Number of successes: x = 4 Probability of success: p = 0.6 Failure: An 1825-year-old has not used alcohol during the past 30 days. Probability of failure: 1 p = 0.4
P(x) =
P(4) = 7C4
When calculating binomial probabilities, we will always round to 4 decimal places. The probability that 4 of the 7 randomly selected 1825-year-olds used alcohol during the past 30 days is 0.2903.
In the appendix, tables are provided with binomial probabilities that have already been calculated. If the number of trials (n) is 20 or less, and the probability of success (p) is 0.05, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, or 0.95, then we can use these tables to find binomial probabilities. Here is the table for n = 7.
n = 7 x 0 1 2 3 4 5 6 7 0.05 0.6983 0.2573 0.0406 0.0036 0.0002 0 0 0 0.1 0.4783 0.3720 0.1240 0.0230 0.0026 0.0002 0 0 0.2 0.2097 0.3670 0.2753 0.1147 0.0287 0.0043 0.0004 0 0.3 0.0824 0.2471 0.3177 0.2269 0.0972 0.0250 0.0036 0.0002 0.4 0.0280 0.1306 0.2613 0.2903 0.1935 0.0774 0.0172 0.0016
P 0.5 0.0078 0.0547 0.1641 0.2734 0.2734 0.1641 0.0547 0.0078 0.6 0.0016 0.0172 0.0774 0.1935 0.2903 0.2613 0.1306 0.0280 0.7 0.0002 0.0036 0.0250 0.0972 0.2269 0.3177 0.2471 0.0824 0.8 0 0.0004 0.0043 0.0287 0.1147 0.2753 0.3670 0.2097 0.9 0 0 0.0002 0.0026 0.0230 0.1240 0.3720 0.4783 0.95 0 0 0 0.0002 0.0036 0.0406 0.2573 0.6983
To use the table to find the probability from the above example, first locate the column for which the probability of success (p) is 0.6. This is the seventh column from the left. Now move down that column until you reach the row where the number of successes (x) is 4. You should see the value 0.2903, which is the same result that we got using the formula. The tables are easier to use than the formula and we gain an advantage when we need to use the formula repeatedly, as in the following example.
Find the probability that between 4 and 6 out of 7 randomly selected 1825-year-olds used alcohol during the past 30 days, if the probability that any 1825-year-old used alcohol during the past 30 days is 0.6.
Number of trials: n = 7 Success: An 1825-year-old has used alcohol during the past 30 days. Number of successes: x = 4, 5, 6 Probability of success: p = 0.6 Failure: An 1825-year-old has not used alcohol during the past 30 days. Probability of failure: 1 p = 0.4 If we were to use the formula for this example, we would need to use the formula 3 times. However, we can use the tables because the number of trials (n) is 20 or less and the probability of success (p) is one that appears in the tables. It is much easier to add three probabilities from the table than to plug into the formula 3 times.
P(4 through 6) = 0.2903 + 0.2613 + 0.1306 = 0.6822
The probability that between 4 and 6 out of 7 randomly selected 1825-year-olds used alcohol during the past 30 days is 0.6822.
When you first determine that a problem is a binomial problem, you should decide whether you can use the tables. Next you should determine whether the complement rule would be helpful. Consider the following.
Find the probability that at least 1 of 7 randomly selected 1825year-olds used alcohol during the past 30 days, if the probability that any 1825-year-old used alcohol during the past 30 days is 0.6.
Number of trials: n = 7 Success: An 1825-year-old has used alcohol during the past 30 days. Number of successes: x = 1 7 Probability of success: p = 0.6 Failure: An 1825-year-old has not used alcohol during the past 30 days. Probability of failure: 1 p = 0.4 To begin with, we can use the tables for this problem. We could begin with x = 1, and add our way down the table. However at least one 1825-year-old used alcohol during the past 30 days is the complement of zero 1825-year-olds used alcohol during the past 30 days. P(At least 1) = 1 P(0)
= 1 0.0016 = 0.9984
If we had totaled the probabilities from x = 1 through x = 7, we would have obtained a probability of 0.9983. The error of 0.0001 is due to rounding.
The probability that a person has O-negative as his or her blood type is 0.06. If 15 people are selected at random, find the probability that at least one of them has O-negative blood.
Number of trials: n = 15 Success: A person has O-negative as his or her blood type. Number of successes: x = 1 through 15 (at least 1) Probability of success: p = 0.06 Failure: A person does not have O-negative as his or her blood type. Probability of failure: 1 p = 0.94 Again, we have a problem that is best handled by using the complement rule. However, we cannot use the tables to assist us this time. We begin by finding the probability that none of the 15 people have O-negative blood.
(0.06)0
(0.94)15
To find the probability that at least 1 of the 15 people have O-negative blood, we subtract the above probability from 1.
P(At least 1) = 1 P(0) = 1 0.3953 = 0.6047
The probability that at least 1 of 15 randomly selected people has O-negative blood is 0.6047.
A basketball player makes 51% of his free-throw attempts (the probability that he makes any single free-throw attempt is 0.51).
If he attempts 12 free throws in tonights game, find the probability that he makes at least 9 of them. Number of trials: n = 12 Success: The player makes a free throw. Number of successes: x = 9 through 12 Probability of success: p = 0.51 Failure: The player misses a free throw. Probability of failure: 1 p = 0.49 We cannot use the tables for this problem, because the probability of success (0.51) does not show up in the tables. Although we see the phrase at least, we do not need to use the complement rule for this problem. A direct calculation uses the formula 4 times, whereas using the complement rule would require us to use the formula 9 different times (0 through 8).
nCx 12C9
(1 p)
P(x) = P(10) =
1 nCx
P(12) =
P(At least 9) = 0.0604 + 0.0189 + 0.0036 + 0.0003 = 0.0832
The probability that the player makes at least 9 of his 12 free throws is 0.0832.
A company makes a product to help reduce male hair loss. The company claims that 80% of the men that use their product will grow new hair. If 10 randomly selected men use the product, find the probability that at least 3 of them do not grow new hair.
In this example, the success is that a man does not grow new hair. However, the probability given in the problem (80% or 0.8) is not the probability of success. It is the probability of failure in this case, so we must use the complement rule to find the probability of success. Number of trials: n = 10 Success: The man does not grow new hair. Number of successes: x = 3 through 10 Probability of success: p = 0.2 Failure: The man does grow new hair. Probability of failure: 1 p = 0.8 We can use the tables for this problem.
P(At least 3) = 1 P(0, 1, 2) = 1 (0.1074 + 0.2684 + 0.3020) = 1 0.6778 = 0.3222
The probability that at least three men do not grow new hair is 0.3222.
Once again, here is the strategy you should employ when tackling binomial problems. 1. Use the tables whenever possible. 2. Determine whether the complement rule would make the problem easier. There is one more topic that must be discussed when dealing with the binomial distribution. As with any probability distribution, the binomial distribution has a mean and a standard deviation. We can calculate these by the methods of Section 4.1, but this distribution has its own formulas for the mean and standard deviation. Mean and Standard Deviation of the Binomial Distribution
= n =
p(1 p)
The probability that a man who is a first-time college freshman plans to major in computer science or engineering is 0.2. Five firsttime college freshman males are selected at random, and asked to name their major. Find the mean and standard deviation for this probability distribution.
This is a binomial distribution with n = 5 and p = 0.2.
= 5(0.2) = 1 = = = n
5(0.2)(0.8) 0.8
This distribution has a mean of 1 success (1 first-time college freshman plans to major in computer science or engineering), with a standard deviation of 0.89 successes.
It should seem reasonable that we should expect 1 of the 5 first-time freshman males to plan on majoring in computer science or engineering, as 20% of all first-time freshman males feel that way. The mean and standard deviation will take on importance in Section 5.2. For now, we will focus only on their calculation.
The probability that a person has O-negative as his or her blood type is 0.06. Fifteen people are selected at random, and asked whether
their blood type is O-negative or not. Find the mean and standard deviation of this probability distribution. This is a binomial distribution with n = 15 and p = 0.06.
= 15(0.06) = 0.9 = = = n
15(0.06)(0.94) 0.846
This distribution has a mean of 0.9 success (0.9 people out of 15 people have this blood type), with a standard deviation of 0.92 successes.
Binomial problems become more tedious as we have to use the formula more times. For instance, consider the following problem: In a certain county, 45% of registered voters are Republican. If 300 registered voters from that county are selected at random, find the probability that at least 155 are Republicans. To calculate this probability directly, we would need to use the formula 146 times. The complement rule is no help here; we would need to use the formula 155 times. There are two tools that we may use to help us in such a situation. The first is Microsoft Excel and its built-in binomial probability functions. The second tool will be covered in Section 5.2. To find the binomial probability of x successes in n trials, with probability of success p, we use Excels built-in binomial function:
=BINOMDIST(x,n,p,FALSE)
There are four arguments in the parentheses. The first three are the number of successes (x), the number of trials (n), and the probability of success (p). The fourth argument allows you to determine whether you want the probability of exactly x successes (FALSE) or the probability of at most x successes (TRUE). Typing TRUE as the fourth argument finds the cumulative probability of 0 through x successes. We will use the cumulative option when we are interested in the probability of a range of successes. We begin by looking back to the first example of the section.
A 1996 study of first-time college freshmen by the Higher Education Research Institute shows that 20% of men planned to major in computer science or engineering. If 5 men that are first-time college freshmen are selected at random, find the probability that exactly 2 of them plan to major in computer science or engineering.
We are interested in exactly 2 successes in 5 trials, with the probability of success being 0.2. Since we are interested in exactly 2 successes, we do not need to use the cumulative option. Type the following in any cell.
=BINOMDIST(2,5,0.2,FALSE)
Excels output is 0.2048. (The number of decimal places that you get depends on the width of the column. Be sure that there are at least four decimal places.)
A basketball player makes 51% of his free-throw attempts (the probability that he makes any single free-throw attempt is 0.51). If he attempts 12 free throws in tonights game, find the probability that he makes at most 8 of them.
This is a binomial problem with 12 trials, and the probability of success on any trial is 0.51. Since we are interested in the probability of at most 8 successes, we will use the cumulative option. Type the following in any cell.
=BINOMDIST(8,12,0.51,TRUE)
Excel tells us that the probability is 0.9168.
In general, to find the probability of x or fewer successes (or at most x successes) in n trials of a binomial experiment with probability of success p, use the following format.
=BINOMDIST(x,n,p,TRUE)
The probability that a student passes a certain math exam is 0.7. If the exam is given to 15 students, find the probability that at least 12 of them pass.
In this problem, we are interested in the probability of at least 12 successes. We will need to use the complement rule to take advantage of Excels cumulative option.
P(At least 12) 1 P(11 or fewer)
Type the following in any cell.
=1-BINOMDIST(11,15,0.7,TRUE)
The result from Excel tells us that the probability of at least 12 students passing is 0.2969.
In general, to find the probability of at least x successes in n trials of a binomial experiment with probability of success p, use the following format.
=1-BINOMDIST(x-1,n,p,TRUE)
The probability that a student passes a certain math exam is 0.7. If the exam is given to 15 students, find the probability that between 9 and 12 of them pass, inclusive.
To use Excel and its cumulative option, we begin by finding the probability of 12 or fewer successes. From that result, we subtract the probability of 8 or fewer successes. This will leave us with the probability of 9 through 12 successes. Type the following in any cell.
=BINOMDIST(12,15,0.7,TRUE)-BINOMDIST(8,15,0.7,TRUE)
From Excel, the probability is 0.7420.
In general, to find the probability of a successes through b successes in n trials of a binomial experiment with probability of success p, use the following format:
= BINOMDIST(b,n,p,TRUE)-BINOMDIST(a-1,n,p,TRUE)
We will finish with the example from the beginning of this Excel section.
In a certain county, 45% of registered voters are Republican. If 300 registered voters from that county are selected at random, find the probability that at least 155 are Republicans.
We are looking for the probability of at least 155 successes in 300 trials of a binomial experiment with probability of success 0.45. Type the following in any cell:
=1-BINOMDIST(154,300,0.45,TRUE)
The probability that at least 155 are Republicans is 0.0120.
Binomial problems become more tedious as we have to use the formula more times. For instance, consider the following problem: In a certain county, 45% of registered voters are Republican. If 300 registered voters from that county are selected at random, find the probability that at least 155 are Republicans. To calculate this probability directly, we would need to use the formula 146 times. The complement rule is no help here; we would need to use the formula 155 times. There are two tools that we may use to help us in such a situation. The first uses technology, the TI-83, and its built-in binomial probability functions. The second tool will be covered in Section 5.2. To find the binomial probability for x successes in n trials, with probability of success p, we must access the DISTR menu of the TI-83. To do this, press 2ND VARS . Select item 0: binompdf(. Inside the parentheses, type the number of trials, the probability of success, and the number of successes, all separated by commas. After closing the parentheses with ) , press ENTER . We begin by looking back to the first example of the section.
A 1996 study of first-time college freshmen by the Higher Education Research Institute shows that 20% of men planned to major in computer science or engineering. If 5 men that are first-time college
freshmen are selected at random, find the probability that exactly 2 of them plan to major in computer science or engineering. We are interested in exactly 2 successes in 5 trials, with the probability of success being 0.2. Here is what the screen should look like after entering the proper values and pressing ENTER :
The TI-83 tells us that the probability is 0.2048.
By the way, if you leave off the number of successes, the TI-83 will calculate the probability of 0 successes, 1 success, . . . , and n successes. We can store these values to a list for convenient viewing. For the previous example, this is done by pressing 2ND VARS , selecting item 0, entering 5 , 0.2 ) ENTER . After the calcula1 ENTER to store the values to list L1. tions are done, press STO , then 2ND Recall that to view these values press STAT and select item 1 under the EDIT menu. If we need to calculate probabilities for a range of successes, we use item A under the DISTR menu, binomcdf(. If we enter the number of trials n, the probability of success p, and a number of successes x, the TI-83 will calculate the probability that there are x or fewer successes.
This is a binomial problem with 12 trials, and the probability of success on any trial is 0.51. Since we are interested in the probability of at most 8 successes, we will use the cumulative option. Press item A. Enter 12
, 2ND , VARS )
to access the DISTR menu, and select
ENTER . Here is what the
and press
screen should look like:
In this problem, we are interested in the probability of at least 12 successes. We will need to use the complement rule to take advantage of the TI-83s binomcdf function.
P(At least 12) = 1 P(11 or fewer)
Press 1 Enter 15 look like:
, then 0.7
2ND ,
VARS )
to access the DISTR menu, and select item A.
ENTER . Here is what the screen should
The result from the TI-83 tells us that the probability of at least 12 students passing is 0.2969.
To use the TI-83 and its binomcdf function, we begin by finding the probability of 12 or fewer successes. From that result, we subtract the probability of 8 or fewer successes. This will leave us with the probability of 9 through 12 successes. Press
2ND , VARS
to access the DISTR menu, and select item A. Enter 15
. Press
for subtraction. Now press
to access the
DISTR menu, and select item A. Enter 15
. Finally, press
to perform the calculation.
We will finish with the example from the beginning of this TI-83 section.
We are looking for the probability of at least 155 successes in 300 trials of a binomial experiment with probability of success 0.45.
P(At least 155) = 1 P(154 or fewer)
, then
Enter 300
ENTER . Here is what the screen
should look like:
The TI-83 tells us that the probability at least 155 are Republicans is 0.0120.
1. Twenty-five percent of all Americans have a cellular phone. If 10 Americans are selected at random, find the probability that 4 of them have a cellular phone. (Based on results of a study by Baskerville Communications.) 2. According to the U.S. Department of Commerce, 62% of Alaskan households have a computer. If 15 Alaskan households are selected at random, find the probability that 9 of them have a computer. 3. Among women ages 1519, 35% use the birth control pill to prevent unintended pregnancies, according to the Alan Guttmacher Institute. If 12 women between the ages of 15 and 19 are selected at random, find the probability that 3 of them use the birth control pill to prevent unintended pregnancies. 4. According to the 1998 Health and Human Services National Household Survey on Drug Abuse, 18% of children between the ages of 12 and 17 are cigarette smokers. If 8 children between the ages of 12 and 17 are selected randomly, find the probability that 2 of them are cigarette smokers. 5. The probability that a college student works to earn extra money is 0.65. If 10 college students are selected at random, find the probability that at least 7 work to earn extra money. (Based on the results of the Social Science Research Center, Department of Psychology, Old Dominion University Campus Concepts.) 6. Eight percent of the students at a community college are at least 30 years old. If 40 students at that community college are selected at random, find the probability that between 2 and 4 students, inclusive, are at least 30 years old. 7. According to Cruise Lines International Corporation, 28% of the North Americans who have taken a cruise in the past five years were at least 60 years old. If 12 North Americans who have taken a cruise in the last five years are selected at random, find the probability that at least one of them was at least 60 years old at the time. 8. The U.S. Bureau of the Census reports that 33% of Americans between the ages of 25 and 34 rent their home. If 8 Americans between the ages of 25 and 34 are randomly selected, find the probability that at least 2 of them rent their home.
9. The probability that a high school student who takes the SAT has a combined score of 1290 or higher is 0.1. If 15 high school students who took the SAT are randomly selected, find the probability that between 2 and 5 students scored 1290 or above, inclusive. 10. The probability that a high school student who takes the SAT has a combined score of 610 or lower is 0.3. If 12 high school students who took the SAT are randomly selected, find the probability that between 3 and 7 students scored 610 or lower, inclusive. 11. A study by the U.S. Department of Health and Human Services revealed that 60% of 1825-year-olds had drunk alcohol in the past 30 days. If twenty 1825-yearolds are selected at random, find the probability that at least ten of them have drunk alcohol in the past 30 days. 12. According to the Associated Press, in 1998 20% of the people in Mississippi did not have health insurance. If 8 people from Mississippi were selected at random, find the probability that 1, 2 or 3 of them did not have health insurance. 13. A fair coin is flipped 9 times. Find the probability that it lands on heads at least twice. 14. A fair coin is flipped 14 times. Find the probability that it lands on heads at least 7 times. 15. A couple plans to have 6 children. Assuming the probability that a child is a girl is 0.5, find the probability that the couple has at least 3 girls. 16. A couple plans to have 5 children. Find the probability that they have at least 1 boy. 17. An instructor gives a five-question True/False quiz. A student must answer at least four questions correctly to pass the quiz. If a student randomly guesses on all five questions, find the probability that he passes the quiz. 18. An instructor gives a ten-question True/False quiz. A student must answer at least seven questions correctly to pass the quiz. If a student randomly guesses on all ten questions, find the probability that she passes the quiz. 19. An instructor gives an eight-question multiple-choice quiz. There are five possible answers for each question, of which only one is correct. A student must answer at least six questions correctly to pass the quiz. If a student randomly guesses on all eight questions, find the probability that he passes the quiz. 20. An instructor gives a six-question multiple-choice quiz. There are five possible answers for each question, of which only one is correct. A student must answer at least five questions correctly to pass the quiz. If a student randomly guesses on all six questions, find the probability that he passes the quiz. 21. According to the U.S. Transportation Department, 56% of all licensed drivers admit to running red lights. If 8 licensed drivers are selected at random, find the probability that at least 3 do not admit to running red lights. 22. A recent study by Market Data Retrieval showed that 89% of K12 teachers feel that they are at least somewhat prepared to integrate technology into instruction. If nine K12 teachers are selected at random, find the probability that at least two of them feel that they are not prepared to integrate technology into instruction. 23. A 1999 survey of teachers by USA Today revealed that 85% of teachers had not encountered a student with a knife, gun, or other deadly weapon in the last three years. If 30 teachers are selected at random, find the probability that at least 2 teachers had encountered a student with such a weapon in the last three years. 24. Data from the U.S. Department of Commerce shows that 19.5% of Americans age 18 and over hold a bachelors or higher degree. If 7 Americans age 18 and over are selected at random, find the probability that at least 1 does not have a bachelors or higher degree. 25. Information from the Census Bureau tells us that 28% of American men have a bachelors degree or higher. If 15 American men are selected at random, find the mean and standard deviation of the resulting binomial probability distribution (x = number of men with a bachelors degree or higher). 26. According to the March 2000 National Vital Statistics report, approximately 21% of all babies are delivered by Caesarian section. If 18 births are selected at random, find the mean and standard deviation of the resulting probability distribution (x = number of babies delivered by Caesarian section). 27. A coin is flipped 50 times, and the number of heads are recorded. Find the mean and standard deviation of this binomial probability distribution (x = number of heads).
28. A coin is flipped 500 times, and the number of heads are recorded. Find the mean and standard deviation of this binomial probability distribution (x = number of heads). 29. According to information from the Bureau of Labor Statistics, approximately 6% of all adult Americans have more than one job. If 225 adult Americans are selected at random, find the mean and standard deviation of the resulting probability distribution (x = number of adult Americans with more than one job). 30. According to the Morbidity and Mortality Weekly Report by the Centers for Disease Control and Prevention, approximately 70% of all high school students have tried smoking cigarettes. If 889 high school students are selected at random, find the mean and standard deviation of the resulting binomial probability distribution (x = number who have tried smoking).
EXTRA The Geometric Distribution
There are discrete probability distributions other than the binomial distribution. One is the geometric distribution, which is closely related to the binomial distribution. The geometric distribution represents the probability that the first success of a binomial trial occurs on trial x. An example is finding the probability that the first time a coin lands on heads is on the fifth toss. Lets develop the formula for the geometric distribution. Suppose we are conducting a binomial experiment with a probability of success p and probability of failure 1 p. The probability that the first trial is a success is p. For the first success to occur on the second trial, the first trial would have to be a failure, and the second trial would be a success. The probability that the second trial produces the first success is (1 p) p. For the first success to occur on the third trial, there would have to be two failures followed by one success. The probability of this occurring is given by (1 p)2 p. We can now see that for the first success to occur on trial x, there must be x 1 failures followed by one success. The probability of this occurring is given by (1 p)x1 p. This is the general formula for the geometric distribution.
P(x) = (1 p)x1
The probability that a basketball player makes a free throw is 0.8. Find the probability that the player makes his first free throw on his fourth attempt.
For this example, p = 0.8 and x = 4. The probability that the first success occurs on the fourth attempt is given by
P(x) = (1 p)x1 P(4) = (0.2)3
The probability that the player makes his first free throw on his fourth attempt is 0.0064.
When rolling a pair of fair six-sided dice, the probability that the sum is 8 or 9 is 0.25, because 9 of the 36 possible rolls have a sum that is 8 or 9. We will construct a geometric probability distribution for the roll containing the first 8 or 9.
Success: A roll is an 8 or a 9. Probability of success: p = 0.25 Failure: A roll is not an 8 or a 9. Probability of failure: 1 p = 0.75 The appropriate formula here is
x1 0.25 = (0.75)
Here is the probability distribution.
x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 P(x) 0.2500 0.1875 0.1406 0.1055 0.0791 0.0593 0.0445 0.0334 0.0250 0.0188 0.0141 0.0106 0.0079 0.0059 0.0046 0.0033 0.0025 0.0019 0.0014 0.0011 0.0008 0.0006 0.0004 0.0003 0.0003 0.0002 0.0001 0.0001 0.0001 0.0001 0.0000
We could go on to answer any question regarding probabilities for this distribution, such as finding the probability that the fourth roll produces the first success.
1. At the beginning of every football game, there is a coin flip to determine which team gets the coin flip. Each team has a 50% chance of winning the coin flip. Find the probability that a team does not win its first coin flip until the sixth game of the season.
2. A game show invites people across the country to dial a toll-free phone number to try to qualify for their show. To become eligible to be on the show, a caller must correctly answer 3 questions. The show estimates that 8% of all callers will correctly answer the 3 questions. Find the probability that it takes a caller 4 attempts to answer the 3 questions correctly. 3. At a racetrack, 32% of the races are won by the favorite. Find the probability that it takes six races for the first favorite to win a race. 4. The probability that the roll of a single die produces a 3 is 1. Find the probability 6 that it takes 8 rolls to produce the first 3.
Binomial Probabilities through Simulation
In this exploration, we will use Excel to simulate the classic binomial experiment flipping a coin. In this example, we will simulate flipping a coin 10 times, counting the number of heads. We will repeat this 100 times, and create a histogram to show our results. We will use the Data Analysis Toolpak to simulate the tossing of the coin. Select the Tools menu. If you do not see Data Analysis as one of the choices, then you must add it. Select Add-Ins, and when the dialog box opens click the box next to Analysis ToolPak and click OK. We will use Excels built-in function RANDBETWEEN to simulate the coin flip. In cell A1, type the following:
=RANDBETWEEN(0,1)
After you press Enter, this will randomly generate either a 0 or a 1. We will treat a result of 1 as heads, and a result of 0 as tails. Click on cell A1, and then select Copy from the Edit menu. Highlight cells A2 through A10, and select Paste from the Edit menu, which will repeat this random generation 10 times. To total the number of heads, type the following in cell A11:
=SUM(A1:A10)
Record the number of heads for later use. Then repeat the experiment by pressing the F9 key on your keyboard. Record this new number of heads. Repeat this cycle a total of 100 times. We will create a frequency distribution showing the number of times that we got 0, 1, 2, . . . , 10 heads. In cells B1 through B11 type the numbers from 0 through 10. In cell C1 type the number of times that you got 0 heads. In cell C2 type the number of times that you got 1 head. Repeat this until you put the number of times that you get 10 heads in cell C11. Now we can create a histogram from this frequency distribution. Click on Insert, and then on Chart. A histogram is a column chart, which is the default selection. Click on Next to move from Step 1 to Step 2. In Step 2, type C1:C11 in the box labeled Data range. These are the heights of the bars in your histogram. Also, be sure that you click on columns instead of rows in the area that says Series in:. Click on the Series tab, and in the box labeled Category (X) axis labels type the following: =Sheet1!$B$1:$B$11 These are the values that will appear on the x axis. Click on Next to move to Step 3.
4.3 The Poisson Distribution
Step 3 contains some ways to make your chart prettier. Click on the Titles tab to add a title to your graph (in the box Chart Title). To show the value for each class on the chart click on the Data Labels tab, then click on Show Value. Click on the Legend tab, and be sure the box labeled Show Legend is not checked. Click on Next to advance to Step 4. Step 4 has to do with the location of the chart. Select the option As object in, and be sure that it says Sheet 1. Now click on Finish, and your histogram will appear. We can also use this experiment to check the validity of the formulas for the mean and standard deviation of the binomial distribution. We have n = 10 and p = 0.5, so we can find the mean as follows:
= 10(0.5) = 5
Find the standard deviation as follows:
= = = n
10(0.5)(0.5) 2.5
Place the number of heads obtained from the 100 times you repeated this experiment in a column, and calculate the mean and standard deviation. Are they close to the values they were supposed to be? After completing this, repeat the process for 50 flips instead of 10, and again for 100 flips.
The Poisson Distribution
The next discrete probability distribution that we will study is the Poisson distribution. This distribution is named after Simeon Denis Poisson (17811840), a French mathematician. If we know the expected number of successes in a given interval, then we can use the Poisson distribution to calculate the probability of x successes occurring in the interval. We use the Greek letter (lambda) to represent the expected number of successes in the given interval. The given interval may be a measure of time (minute, hour, day, week, year) or something else such as a baseball game, a semester, or even a chocolate chip cookie. If is the expected (or mean) number of successes in an interval, then the probability of having x successes in that interval is given by
P(x) = e
x!
where e is the base of the natural logarithm (e 2.71828).
During a typical hour fishing at Lake Lotsafish, a fisher can expect to catch 3 fish. Find the probability that a fisher catches exactly 2 fish in his first hour fishing.
First, we must determine what the expected number of successes is for this situation. A fisher can expect to catch 3 fish during a typical hour, so we know that = 3. (The interval for this problem is one hour.) We are interested in the probability of exactly 2 fish being caught, so x = 2. As with binomial problems, it is wise to create a table of information before proceeding. Success: A fish is caught. Number of successes: x = 2 Expected number of successes: = 3 Interval: One hour Now we may plug into the formula.
P(x) = P(2) = e x x! e3 32 2!
The probability that a fisher catches 2 fish in his first hour fishing at Lake Lotsafish is 0.2240. As with binomial probabilities, we round to four decimal places.
As for the binomial distribution, there are tables in the appendix that give probabilities associated with certain values of . If is 0.1, 0.2, . . . , 0.8, 0.9, 1, 2, . . . , 8, or 9, then you may look up the probabilities in the tables. One of the tables follows. To find the probability of 2 successes when is 3, locate the column where = 3 and look down until you find the row where x is 2. The correct probability has a box around it, and is in bold. x
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 1 0.3679 0.3679 0.1839 0.0613 0.0153 0.0031 0.0005 0.0001 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0.1353 0.2707 0.2707 0.1804 0.0902 0.0361 0.012 0.0034 0.0009 0.0002 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0.0498 0.1494 0.2240 0.2240 0.1680 0.1008 0.0504 0.0216 0.0081 0.0027 0.0008 0.0002 0.0001 0 0 0 0 0 0 0 0 0 0 4 0.0183 0.0733 0.1465 0.1954 0.1954 0.1563 0.1042 0.0595 0.0298 0.0132 0.0053 0.0019 0.0006 0.0002 0.0001 0 0 0 0 0 0 0 0 5 0.0067 0.0337 0.0842 0.1404 0.1755 0.1755 0.1462 0.1044 0.0653 0.0363 0.0181 0.0082 0.0034 0.0013 0.0005 0.0002 0 0 0 0 0 0 0 6 0.0025 0.0149 0.0446 0.0892 0.1339 0.1606 0.1606 0.1377 0.1033 0.0688 0.0413 0.0225 0.0113 0.0052 0.0022 0.0009 0.0003 0.0001 0 0 0 0 0 7 0.0009 0.0064 0.0223 0.0521 0.0912 0.1277 0.1490 0.1490 0.1304 0.1014 0.0710 0.0452 0.0263 0.0142 0.0071 0.0033 0.0014 0.0006 0.0002 0.0001 0 0 0 8 0.0003 0.0027 0.0107 0.0286 0.0573 0.0916 0.1221 0.1396 0.1396 0.1241 0.0993 0.0722 0.0481 0.0296 0.0169 0.009 0.0045 0.0021 0.0009 0.0004 0.0002 0.0001 0 9 0.0001 0.0011 0.0050 0.0150 0.0337 0.0607 0.0911 0.1171 0.1318 0.1318 0.1186 0.0970 0.0728 0.0504 0.0324 0.0194 0.0109 0.0058 0.0029 0.0014 0.0006 0.0003 0.0001
Unlike the binomial distribution, there is no number of trials, so the number of successes in a given interval has no upper limit. Probabilities are calculated until they round to be equal to zero to four decimal places.
During the typical statistics final exam, 2 students leave the room in tears. Find the probability that between 1 and 4 students, inclusive, leave your statistics final exam in tears.
We can expect 2 students to leave the room in tears during a typical final exam, so we know that = 2. (The interval for this problem is a final exam.) We are interested in the probability of between 1 and 4 students, inclusive, leaving in tears, so x = 1, 2, 3, or 4. Success: A student leaves in tears. Number of successes: x = 1, 2, 3, or 4 Expected number of successes: = 2 Interval: A final exam We can use the tables for this problem, since = 2 and that is one of the values included in our table. P(Between 1 and 4 students) = P(1) + P(2) + P(3) + P(4)
= 0.2707 + 0.2707 + 0.1804 + 0.0902 = 0.8120
The probability that between 1 and 4 students leave your final exam in tears is 0.8120.
During the typical statistics final exam, 2 students leave the room in tears. Find the probability that at least 1 student leaves your statistics final exam in tears.
Again, we know that = 2. We can use the tables once again, but instead of adding our way down the table from 1 success through 9 successes (the last nonzero probability listed for = 2), we can use the complement rule.
P(At least 1 student) = 1 P(0) = 1 0.1353 = 0.8647
The probability that at least 1 student leaves your final exam in tears is 0.8647. Does it make sense that this probability is higher than that of the previous example?
During a typical baseball game, the Cleveland Indians score 6.4 runs. Find the probability that the Cleveland Indians do not get shut out in a gamein other words, that they score at least one run in a game.
The Cleveland Indians score 6.4 runs during a typical baseball game, so we know that = 6.4. (The interval for this problem is one baseball game.) We are interested
in the probability of at least one run being scored, so the complement rule will come in handy again. This time we must use the formula, and this time the problem would be quite tedious without the complement rule. Success: A run is scored. Number of successes: x = at least 1 Expected number of successes: = 6.4 Interval: One baseball game Now we may plug into the formula.
P(x) = P(0) = e x x! e6.4
= 0.0017 P(At least 1) = 1 P(0) = 1 0.0017 = 0.9983
The probability that the Cleveland Indians do not get shut out is 0.9983.
Occasionally, the hardest part of figuring out a Poisson probability is determining how many successes we are interested in. Consider the following example.
A juvenile detention center has enough room for 8 juveniles. On an average night, 5 juveniles are brought to the center to be kept overnight. Find the probability that the center will not have enough room on a given night.
In order for the center to not have enough room, more than 8 juveniles must be brought in. So, we are interested in the probability that at least 9 juveniles are brought to the center. On an average night, 5 juveniles are brought to the center to be kept overnight. This tells us that = 5. Since = 5, we may use the tables for this problem. Success: A juvenile is brought to the center to be kept overnight. Number of successes: x = at least 9 Expected number of successes: = 5 Interval: One night
P(At least 9 juveniles) = P(9) + P(10) + P(11) + P(12) + P(13) + P(14) + P(15) = 0.0363 + 0.0181 + 0.0082 + 0.0034 + 0.0013 + 0.0005 + 0.0002 = 0.0680
The probability that the center does not have enough room on a given night is 0.0680.
Occasionally we will be given a typical value for a certain interval and will be asked for a probability involving another interval. For example, we could be told the number of runs that the Cleveland Indians score in a typical game, and be asked to find the probability that they score 10 runs in the next 2 games. The next example illustrates this idea.
On a typical day, a credit union opens 5 new accounts. Find the probability that the credit union opens exactly 6 accounts in the next two days.
If we can expect 5 new accounts on a given day, in two days we can expect twice as many, or 10 new accounts. This will be our value for . Now we proceed in the same fashion as before. Success: A new account is opened. Number of successes: x = 6 Expected number of successes: = 10 Interval: Two days We must use the formula for this problem, since = 10 is too big for our tables.
P(x) = P(6) = e x x! e 10 106 6!
The probability that the credit union opens exactly 6 accounts in the next two days is 0.0631.
Like the binomial distribution, the Poisson distribution has a mean and a standard deviation. Mean and Standard Deviation of the Poisson Distribution
It should come as no surprise that the mean of the Poisson distribution is . After all, that is the expected number of successes.
During a typical hour fishing at Lake Lotsafish, a fisher can expect to catch 3 fish. Find the mean and standard deviation of the probability distribution for the number of fish caught in an hour of fishing at Lake Lotsafish.
In this problem = 3, so
The mean for this distribution is 3 fish, with a standard deviation of 1.73 fish.
During a typical baseball game, the Cleveland Indians score 6.4 runs. Find the mean and standard deviation of the probability distribution for the number of runs scored in a game by the Cleveland Indians.
In this problem = 6.4, so
The mean for this distribution is 6.4 runs, with a standard deviation of 2.53 runs.
Poisson Probabilities
To find the Poisson probability for x successes, with expected number of successes , we use Excels built in Poisson function.
=POISSON(x, ,FALSE)
There are three arguments in the parentheses. The first two are the number of successes (x) and the expected number of successes (). The third argument allows you to determine whether you want the probability of exactly x successes (FALSE) or the probability of at most x successes (TRUE). Typing TRUE as the third argument finds the cumulative probability of 0 through x successes. We will use the cumulative option when we are interested in the probability of a range of successes. We begin by using Excel to rework the first example.
Success: A fish is caught. Number of successes: x = 2 Expected number of successes: = 3 Interval: One hour Type the following in any cell.
=POISSON(2,3,FALSE)
Excels answer, to four decimal places, is 0.2240. This matches the answer that we calculated using the formula, as well as the result that we got from the Poisson table.
Excels cumulative option for finding Poisson probabilities allows us to directly calculate the probability of x or fewer successes. Here is an example.
A small town has 1.6 cars stolen on an average day. Find the probability that there are 2 or fewer cars stolen on a given day.
Success: A car is stolen. Number of successes: x = 2 or fewer Expected number of successes: = 1.6 Interval: One day Type the following in any cell.
=POISSON(2,1.6,TRUE)
Excel gives us 0.7834, rounded to four places.
In general, to find the probability of x or fewer successes (or at most x successes) in a Poisson distribution with expected number of successes , use the following format.
=POISSON(x, ,TRUE)
We can extend the use of the cumulative probability option to calculate probabilities using the complement rule, as in the next example.
Success: A run is scored. Number of successes: x = At least 1 Expected number of successes: = 6.4 Interval: One baseball game The probability that the Cleveland Indians score at least 1 run in a game is calculated by using the complement rule. The complement of at least 1 is zero. Type the following in any cell.
=1-POISSON(0,6.4,TRUE)
Excel tells us that the probability is 0.9983. This result agrees with our calculations earlier in the section.
In general, to find the probability of x or more successes (or at least x successes) in a Poisson distribution with expected number of successes , use the following format.
=1-POISSON(x-1, ,TRUE)
The last type of problem that we can use Excels assistance on is finding the probability of between a and b successes. Here is an example.
Success: A student leaves in tears. Number of successes: x = 1, 2, 3, or 4 Expected number of successes: = 2 Interval: One final exam To use Excel and its cumulative option, we begin by finding the probability of 4 or fewer successes. From that result, we subtract the probability of less than 1 success. This will leave us with the probability of 1 through 4 successes. Type the following in any cell.
=POISSON(4,2,TRUE)-POISSON(0,2,TRUE)
From Excel, the probability is 0.8120. Again, this is the same probability that we found earlier.
In general, to find the probability of a successes through b successes in a Poisson distribution with expected number of successes , use the following format.
= POISSON(b, ,TRUE)-POISSON(a-1, ,TRUE)
The TI-83 has built-in functions to help us find the Poisson probability for x successes, with expected number of successes . Again, we will need to access the DISTR menu, which contains the functions for calculating Poisson probabilities. We access this menu by pressing 2ND VARS . Option B is poissonpdf, which we use for calculating the probability of exactly x successes. Option C is poissoncdf, which we use to find the probability of x or fewer successes. We will begin by using the TI-83 to rework the first example.
Success: A fish is caught. Number of successes: x = 2 Expected number of successes: = 3 Interval: One hour Access the DISTR menu by pressing 2ND VARS . Select option B: poissonpdf because we are interested in exactly 2 successes instead of a range of successes. On the main screen enter 3 , 2 ) , and press ENTER .
The TI-83s answer, to four decimal places, is 0.2240. This matches the answer that we calculated using the formula, as well as the result that we got from the Poisson table.
Here is an example using option C: poissoncdf. This calculates the probability of x or fewer successes. Again, we need to supply the expected number of successes () and the number of successes that we are interested in (x).
Success: A car is stolen. Number of successes: x = 2 or fewer Expected number of successes: = 1.6 Interval: One day Access the DISTR menu by pressing 2ND VARS . Select option C: poissoncdf because we are interested in a range of successes (2 or fewer). On the main screen enter 1.6 , 2 ) , and press ENTER .
The TI-83s answer is 0.7834, rounded to four places.
We can extend the use of the Poisson functions, poissonpdf and poissoncdf, to calculate probabilities using the complement rule, as in the next example.
Success: A run is scored. Number of successes: x = At least 1 Expected number of successes: = 6.4 Interval: One baseball game The probability that the Cleveland Indians score at least 1 run in a game is calculated by using the complement rule. The complement of at least 1 is zero.
P(At least 1) = 1 P(0)
On the main screen, press 1 and
. Access the DISTR menu by pressing
VARS . Select option B: poissonpdf because we are interested in calculating the
probability of exactly 0 successes and not a range of successes. On the main screen enter 6.4
, and press
ENTER .
The TI-83s answer is 0.9983, rounded to four places. This result agrees with our calculations earlier in the section.
The last type of problem that we can use TI-83s assistance on is finding the probability of between a and b successes. Here is an example.
Success: A student leaves in tears. Number of successes: x = 1, 2, 3, or 4 Expected number of successes: = 2 Interval: One final exam To use the TI-83, we begin by finding the probability of 4 or fewer successes. From that result, we subtract the probability of less than 1 success. This will leave us with the probability of 1 through 4 successes. Access the DISTR menu by pressing screen enter 2 pressing
2ND , 2ND VARS . Select option C: poissoncdf
because we are interested in a range of successes (four or fewer). On the main 4
. Now press
and return to the DISTR menu by
VARS . Select option B: poissonpdf because we are interested in
exactly 0 successes. On the main screen enter 2
The TI-83 gives the answer as 0.8120, rounded to four places. Again, this is the same probability that we found earlier.
1. A college baseball player averages 1.07 hits per game. Assuming that the number of hits per game follows a Poisson distribution with = 1.07, complete the following probability distribution.
Number of Hits 0 1 2 3 4 5 6 7 Probability
2. The number of bike crashes on a universitys bike paths follows a Poisson distribution with a mean of 2.1 crashes per day. Complete the following probability distribution.
Number of Crashes 0 1 2 3 4 5 6 7 8 9 10 Probability
3. The number of structure fires per day in a small city follows a Poisson distribution with a mean of 0.3 fires per day. Complete the following probability distribution.
Number of Fires 0 1 2 3 4 Probability
4. The number of cars stolen per day in a city follows a Poisson distribution with 1.5 cars stolen per day. Complete the following probability distribution.
Number of Cars 0 1 2 3 4 5 6 7 8 Probability
5. During a typical day at the Student Health Center, 4.2 students will come in complaining of stomach problems. Find the probability that 5 students will visit the center today to complain about stomach problems. 6. At a local gym, the number of people who are treated for being accidentally kicked in the head each week in an exercise class follows a Poisson distribution with a mean of 1.8 people treated per week. Find the probability that 3 people from exercise classes at the gym will need to be treated for such an injury this week. 7. During a semester at the university, there is an average of 1.8 students from the Lambda Lambda Lambda fraternity on the Deans List. Find the probability that at least 2 students from the fraternity make the Deans List this semester. 8. The number of interceptions thrown by college quarterbacks in a game follows a Poisson distribution with a mean of 1.7 interceptions per game. Find the probability of a college quarterback throwing at least 3 interceptions in a game. 9. The number of students absent from a statistics class follows a Poisson distribution with a mean of 3 students absent per day (not including test days). (a) Find the probability that no students are absent on a given day. (b) Find the probability that 2, 3, or 4 students are absent on a given day. (c) Find the probability that 6 or more students are absent on a given day. 10. The number of accounts opened at a small credit union follows a Poisson distribution with a mean of 2 new accounts per day. (a) Find the probability that at least 1 new account is opened on a given day. (b) Find the probability that 3, 4, or 5 new accounts are opened on a given day. (c) Find the probability that more than 4 accounts are opened on a given day. 11. The number of defects on an assembly line follows a Poisson distribution with a mean of 8 defects per hour. (a) Find the probability that 5 defects come off of the assembly line in the next hour. (b) Find the probability that fewer than 5 defects come off of the assembly line in the next hour. (c) Find the probability that 5 or fewer defects come off of the assembly line in the next hour. (d) Find the probability that the assembly line produces anywhere from 6 to 10 defects in the next hour. 12. The number of cars sold by a new car dealer follows a Poisson distribution with a mean of 4 cars sold per day. (a) (b) (c) (d) Find the probability that at least 6 cars are sold today. Find the probability that there is at least 1 sale today. On what percentage of days are there fewer than 5 cars sold? If we know that at least 2 cars were sold today, find the probability that exactly 4 cars were sold today.
13. On a typical day, 6 cats are brought in to a low-cost clinic to be spayed. If the clinic only has enough supplies to spay 8 cats per day, find the probability that the capacity will be exceeded today. 14. At a typical lunch meeting at a banquet hall, 5 people will request a vegan lunch. If the banquet hall only has 3 vegan lunches available for todays meeting, find the probability that this will not be enough for todays meeting. 15. The number of teachers absent at an elementary school follows a Poisson distribution with a mean of 0.3 absences per day. If the school has 2 substitute teachers available, find the probability that the school will be understaffed on a given day. 16. On an average day, a florist receives orders for 2 blooming plants. If there are 5 blooming plants in stock, find the probability that there will be enough blooming plants for today. 17. For a typical flight from Las Vegas to New York, 3 travelers do not show up. If the airline overbooked a flight by 5 passengers during advance sales, find the probability that it will have enough seats for this flight.
18. A door-to-door salesman averages 4 magazine subscriptions sold per evening. If the salesman needs at least 6 sales tonight to earn the weekly bonus, find the probability that he gets the bonus tonight. 19. The number of runs scored by a college baseball team follows a Poisson distribution with a mean of 4.4 runs per game. Find the probability that the team scores exactly 10 runs in the next two games. 20. A grocery store has an ATM machine inside. The number of customers who use the machine follows a Poisson distribution with a mean of 2.5 customers per hour. Find the probability that exactly 4 customers use the machine in the next three hours. 21. On a typical day at a small town hospital, 3 mothers give birth. Find the probability that there are at least 10 births in the next two days at the hospital. 22. A software company sells software on the Internet. The number of unsolicited sales follows a Poisson distribution with a mean of 5 sales per day. Find the probability that the company makes at least 15 unsolicited sales over the next two days. 23. Here are the number of accidents on a dangerous stretch of road on 40 randomly selected days.
Number of Accidents 0 1 2 3 4 5 6 7 8 Frequency 6 11 12 6 2 2 1 0 0
It is claimed that the number of accidents follows a Poisson distribution with a mean of 1.5 accidents per day. Does it? Complete the table that follows, and use it to help you determine whether the above table is close to what we would expect from a Poisson distribution whose mean is 1.5. To complete the table, first find the probability of 0 accidents, 1 accident, and so on for a Poisson distribution with a mean of 1.5 accidents per day. To find the expected number of days for each number of accidents, multiply the probabilities by 40 (for 40 days).
Number of Accidents 0 1 2 3 4 5 6 7 8 Probability Expected
24. Do the number of tornadoes that strike Utah per year follow a Poisson distribution with a mean of 1.7 tornadoes per year? Here is a table showing how many tornadoes struck Utah each year for the 46 years from 1950 through 1995.
Number of Tornadoes 0 1 2 Frequency 19 8 5 (continues)
(continued) Number of Tornadoes 3 4 5 6 7 8 9
Frequency 5 3 3 3 0 0 0
Complete the following table, and use it to help you determine whether the number of tornadoes follows a Poisson distribution with a mean of 1.7 tornadoes per year.
Number of Tornadoes 0 1 2 3 4 5 6 7 8 9 Probability Expected
25. At a local gym, the number of people who are treated for being accidentally kicked in the head each week in an exercise class follows a Poisson distribution with a mean of 1.8 people treated per week. Find the standard deviation for this distribution. 26. The number of teachers absent at an elementary school follows a Poisson distribution with a mean of 0.3 absences per day. Find the standard deviation of this distribution. 27. The number of penalties for a college football team follows a Poisson distribution with a mean of 7.2 penalties a game. Find the standard deviation for this distribution. 28. The number of errors that a Braille transcriber makes follows a Poisson distribution with a mean of 2.5 mistakes per page. Find the standard deviation for this distribution.
EXTRA Poisson Distribution as a Limit for the Binomial Distribution
The Poisson distribution was initially developed as a limit for the binomial distribution. In other words, as n increased, the binomial distribution got closer and closer to the Poisson distribution. The fit was better for smaller values of p. The mean of the Poisson distribution can be found by using the formula = n p. We will examine how the binomial distribution approaches the Poisson distribution as n becomes large and p becomes small. We will use four binomial distributions that have a mean of 1: n = 10 and p = 0.1; n = 100 and p = 0.01; n = 1000 and p = 0.001; and n = 10,000 and p = 0.0001. You will notice that as n becomes larger and p becomes smaller, the binomial probability distributions become closer to the Poisson probability distribution with = 1.
Complete the following probability distributions.
Binomial n = 10 p = 0.1 Binomial n = 100 p = 0.01 Binomial n = 1,000 p = 0.001 Binomial n = 10,000 p = 0.0001
x 0 1 2 3 4 5 6 7
Poisson = 1 0.3679 0.3679 0.1839 0.0613 0.0153 0.0031 0.0005 0.0001
What can we say as n is increasing?
n this chapter we examine another probability distribution, the normal distribution. The normal distribution differs from the probability distributions in the previous chapter because it is a continuous, rather than discrete, distribution. The random variables used in this chapter may take on infinitely many values, including decimal values. For the most part, we will not be looking at counts of successes as we did in the previous chapter. Another major difference is the fact that the normal distribution is represented by a curve, and we will be using areas to represent probabilities. In Section 5.1, we will be introduced to the family of normal distributions and their properties. We will also use the normal distribution to find probabilities. In Section 5.2, we will learn how to make adjustments in order to handle probabilities for discrete random variables. Here we will learn how to approximate binomial probabilities using the normal distribution, under the correct conditions. The normal approximation to the binomial distribution will give us the tools to make inferences about a population proportion or percentage. Finally, in Section 5.3 we will reverse the whole process. If given a probability, we will learn how to find the value of the random variable associated with that probability.
The Normal Distribution
A normal distribution is a continuous probability distribution. In Chapter 4, we saw that the binomial and Poisson distributions were discrete because the random variables in those distributions could only take on values that were whole numbers. If asked to find the probability that at least 11.9 of 20 people waiting in line would not be able to register for the class that they wanted, we would see that the value 11.9 is not a possible value for the random variable. This is because we are counting the number of people that were not able to register for the class they wanted, which has to be a whole number. On the other hand, if we were asked to find the probability that a can of soda had at least 11.9 ounces of soda in it, we would see that 11.9 is a possible valueas are 11.93, 11.972, and infinitely many other decimal values. In the first sentence of the previous paragraph, we referred to a normal distribution rather that the normal distribution. This is because there are infinitely many normal distributions. A normal distribution is determined by its mean and standard deviation. Here is the graph of a normal distribution with mean and standard deviation .
The graph of a normal distribution is bell-shaped and symmetric. The mean is the value on the horizontal axis where the peak of the graph occurs. The standard deviation, as you might expect, has to do with how spread out the distribution is. The heights of adult males are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. The heights of adult females are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. The graphs of these two distributions are shown next on the same set of axes.
5.1 The Normal Distribution
Note that the curve to the right is for male heights. The mean for male heights is greater than, or to the right of, the mean for female heights because the peak of the curve for male heights is to the right of the peak of the curve for female heights. The standard deviation for male heights is greater than the standard deviation for female heights. You can see that the curve for the male heights is flatter and more spread out.
Which of the following two normal distributions has the greater mean?
The distribution whose graph is to the right (#2) has the larger mean, because its peak occurs farther to the right.
The two normal distributions graphed here have the same mean. Which of the two has the greater standard deviation?
The distribution labeled as #1 has the greater standard deviation, because the curve is more spread out from the center.
CHAPTER 5 Normal Probability Distributions
A normal distributions graph is bell-shaped. There are more values located near the center, or mean, of the distribution. There are fewer and fewer values as we move out from the center. The graph of the distribution continues forever in both directions, without ever touching the horizontal axis. The horizontal axis is an asymptote for the graph of a normal distribution. When drawing the graph of a normal distribution by hand, be sure that the curve never touches the horizontal axis. Normal distributions are symmetric with respect to the mean. The area underneath the curve, but above the horizontal axis, is equal to 1. The fact that the total area is equal to 1 allows us to use the area underneath the curve to represent probabilities. Recall from the last chapter that the total of all probabilities for a probability distribution is equal to 1. Again, since a normal distribution is symmetric, the area underneath the right half of the curve is equal to 0.5, and the same can be said for the left half. Before moving on to determine probabilities involving normal distributions, we must discuss one special normal distribution called the standard normal distribution. The standard normal distribution has a mean of 0 and a standard deviation of 1. Its graph is shown next. The standard normal distribution is sometimes referred to as the z-distribution. This has to do with the fact that we will later use z-scores to help find probabilities.
For any continuous probability distribution, including the standard normal distribution, we find the probability that the random variable is in a certain interval by finding the area underneath the curve in that interval. For example, if we wanted to find the probability that a standard normal variable (z) was between a and b, denoted P(a < z < b), we would need to find the area underneath the curve between a and b. An example of such an area is shown next.
P (a < z < b)
One important fact to take note of is that we can only find the probability that a random variable is in an interval. The probability that a continuous random variable is exactly equal to a certain value is 0. Usually, to find the area underneath a curve, we would need to use calculus. However, there is a table (Table C) that will help us find the area underneath the curve
for any region for the standard normal distribution. Table C gives us the area underneath the curve that lies to the left of any z-value (z*) for the standard normal distribution. For the graph shown here, our table gives us the shaded area.
P (z < z *)
z*
There are three different types of intervals that we will be finding probabilities for. The first is to the left of a certain value, the second is to the right of a value, and the third is in between two values.
Find the following probabilities.
(a) P(z < 1.96) With all standard normal probability problems, begin by drawing the graph and shading the appropriate region. We draw a normal curve, with a mean of 0. We place 1.96 on the right side of the curve, draw a line segment above it, and then shade to the left. We shade to the left because we are looking for the probability that z is less than 1.96. The values that are less than 1.96 are to the left of 1.96. The graph is shown next, with the shaded area representing P(z < 1.96).
P (z < 1.96)
Since we are looking for the shaded region to the left of our value, we can use Table C directly to find our probability. We find the positive values of z on the second page of the table. We look down the column on the left side until we find 1.9, and then we move across until we find the column with a 6 above it (this is the second decimal place of our z-value). The value that you should see there is 0.9750. Therefore, P(z < 1.96) = 0.9750. Would this problem have turned out differently if we were asked for P(z 1.96) instead of P(z < 1.96)? No. Recall that the probability of any particular value occurring is 0. Therefore,
P(z 1.96) = P(z < 1.96) + P(z = 1.96) = P(z < 1.96) + 0 = P(z < 1.96)
(b) P(z < 1.25) Again, we begin with the graph.
Since we are looking for the area of the region to the left of 1.25, we can get this area directly from the table. The area associated with 1.25 is 0.1056, so P(z < 1.25) = 0.1056.
Often, we will be looking to find the area to the right of a value. Our table does not give us this value directly. As a matter of fact, it gives us the complement of the area that we are looking for. We will take the area from the table, and subtract it from 1 to find the area we are looking for. The following picture illustrates this idea.
P (z > z*) = 1 P (z < z*) P (z < z*)
z* 1
(a) P(z > 1.28) Here is the graph for this region.
P (z > 1.28)
Our table gives us the area of the unshaded region for this graph, which is 0.8997. To find the area of the shaded region, we subtract this area from 1, which is the area underneath the entire curve. Therefore,
P(z > 1.28) = 1 0.8997 = 0.1003
(b) P(z > 1.5) Again, we begin with the graph.
P (z > 1.5) = 1 0.0668 = 0.9332
0.0668 2 1.5 1 0 1 2
Again, the table gives us the unshaded area to the left of 1.5, and we subtract that area from 1 to find the area of the shaded region.
P(z > 1.5) = 1 0.0668 = 0.9332
The probability that z is greater than 1.5 is 0.9332.
The third possible type of shaded region occurs when we shade between two z-values. Consider the following graph, which is associated with P(a < z < b), where a is less than b.
We can think of the shaded region as being the region to the left of b with the region to the left of a cut out of it. We can find the area of these two regions separately from the table, and then subtract the area to the left of a from the area to the left of b to give us the area of the region between a and b.
(a) P(1.23 < z < 0.97)
First, sketch the shaded region. The area of the region to the left of 0.97 is 0.8340. The area of the region to the left of 1.23 is 0.1093. We subtract to find the area of the shaded region.
P (1.23 < z < 0.97)
P(1.23 < z < 0.97) = 0.8340 0.1093 = 0.7247
Therefore, P(1.23 < z < 0.97) = 0.7247. (b) P(1.53 < z < 0.75) Here is the graph.
0.2266 0.0630 2 1 1.53 0.75 0 1 2
Using the table, we will subtract the area associated with the z-value on the right minus the area associated with the z-value on the left. P(1.53 < z < 0.75) = 0.2266 0.0630
Therefore, P(1.53 < z < 0.75) = 0.1636. (c) P(0.5 < z < 0.88) Here is the graph.
P (0.5 < z < 0.88) 0.8106 0.6915
P(0.5 < z < 0.88) = 0.8106 0.6915 = 0.1191
Therefore, P(0.5 < z < 0.88) = 0.1191.
A nonstandard normal distribution is a normal distribution that does not have a mean of 0 and a standard deviation of 1. An example of such a distribution is the height of adult females, which are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. In order to be able to find probabilities associated with nonstandard normal distributions, we must first translate the distribution into a standard normal distribution. The formula for z-scores
z = x
will assist us with that. Suppose we want to find the probability that a randomly selected female is 5 feet tall (60 inches) or shorter. The graph for this probability looks like this:
The values in any normal distribution are distributed in the same fashion. If we can translate this graph into one involving the standard normal (z) distribution, then we can find the area of the shaded region. As you may verify in the next example, a height of 60 inches corresponds to a z-value of 1.44, while a height of 63.6 inches corresponds to a z-value of 0. We can use those facts to create the following graph, from which we may find the area of the shaded region.
2 1.441 (60")
0 (63.6")
The heights of adult females are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. Find the probability that a randomly selected female is 5 feet tall (60 inches) or shorter.
We begin by identifying the random variable in this problem. We let x represent the height of an adult female. Now, write the probability in terms of x.
P(z 60)
The next step is to change this probability into a probability involving the standard normal distribution by using the z-score formula z = (x ), as follows.
P(z 60) = P z 60 63.6 2.5
= P(z 1.44)
We finish by drawing the graph, and using our table.
P(z 1.44)
Since we are looking for the area of the region to the left of z = 1.44, we can obtain this area directly from the table. The probability that an adult female is 5 feet tall or shorter is 0.0749.
The important idea to grasp from the previous example is that P(x 60) is exactly the same as P(z 1.44). All we have done is change our original problem into an equivalent problem involving the standard normal distribution. We will take a look at two more examples involving the heights of adult females.
The heights of adult females are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. Find the probability that a randomly selected female is between 5 2 tall (62 inches) and 5 7 tall (67 inches).
We will let x represent the height of an adult female, the random variable in this problem. We will write the probability that we are looking for in terms of x. Then we will convert it to a problem involving z, draw a graph, and use our table.
P(62 x 67) = P 62 63.6 67 63.6 z 2.5 2.5
= P(0.64 z 1.36)
Here is the graph.
P(62 x 67)
0.2611 2 1 0.64 0 1 1.36 2
To find the area of the shaded region, we subtract the area of the region to the left of 0.64 from the area of the region to the left of 1.36.
P(0.64 z 1.36) = 0.9131 0.2611 = 0.6520
The probability that an adult female is between 5 2 and 5 7 tall is 0.6520.
The heights of adult females are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. Find the probability that a randomly selected female is 6 4 tall or taller.
Before beginning here, we should be able to reason that this will be a small probability. As we walk down the street, it is a rare occasion that we see an adult female that is at least 6 4 tall. We will let x represent the height of an adult female.
P(x 76) = P z 76 63.6 2.5
When we go to the table, we notice that the highest z-value is 3.50. For any z-value above 3.50, we will use the area associated with 3.50. (We will use a similar approach for negative z-values below 3.50.) The area of the unshaded region is 0.9998, so the probability that an adult female is at least 6 4 tall is 1 0.9998 or 0.0002. This is an estimate of the probability; to find the actual probability we would need to use 4.96 for a z-value instead of 3.50. Later, we will see that the use of technology will allow us to find this probability, which is 0.000000353.
The lengths of human pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. Find the probability that a pregnancy lasts 280 days or longer.
We will let x represent the length of a human pregnancy.
P(x 280) = P z 280 268 15
= P(z 0.8)
P(z 0.8)
0.7881 1 0.7881
P(z 0.8) = 1 0.7881 = 0.2119
The probability that a pregnancy lasts at least 280 days is 0.2119.
Normal Probabilities
Microsoft Excel provides us with two functions that we can use in this section. The first helps us to calculate probabilities involving the standard normal (z) distribution. The second helps us to calculate probabilities involving nonstandard normal distributions. To find the probability that a standard normal variable is less than z, we use the function
=NORMSDIST(z)
We will look at the second function later. Our first example involves the area of a region to the left of a z-value.
Find P(z < 1.96).
As by hand, we begin with the graph.
P(z < 1.96)
Now type the following in any cell.
=NORMSDIST(1.96)
Excel gives us 0.9750 to four decimal places, which agrees with our previous answer.
In general, any time you need to find P(z < a), type =NORMSDIST(a) in any cell. The next example deals with finding the area to the right of a z-value.
Find P(z > 1.5).
Again, we begin with the graph.
0.0668 2 1.5
Since we have shaded to the right of z = 1.5, we will need to use the complement rule. This is because Excel gives us the area of the unshaded region to the left, not the shaded region to the right. Type the following in any cell.
=1-NORMSDIST(-1.5)
Excel tells us that the probability is 0.9332, which agrees with our previous answer. By the way, if you would like Excel to round your probabilities to four decimal places, type the following in any cell instead.
=ROUND(1-NORMSDIST(-1.5),4)
In general, any time you need to find P(z > a), type =1-NORMSDIST(a) in any cell. The next example deals with finding the area between two z-values.
Find P(1.23 < z < 0.97).
Recall that we want to subtract the area associated with the z-value on the left from the z-value on the right. Type the following in any cell.
=NORMSDIST(0.97)-NORMSDIST(-1.23)
Excel gives us 0.7246, which differs from our previous answer of 0.7247. This is due to the fact that the table rounded each probability before we subtracted, whereas with Excel we wait to round until after the subtraction.
In general, any time you need to find P(a z b), type
=NORMSDIST(b)-NORMSDIST(a)
When working with nonstandard normal distributions, we use the function
=NORMDIST(x,,,TRUE)
The first argument, x, is the x-value associated with the probability. The second and third arguments are the mean and standard deviation, respectively. The last argument must be TRUE if we are to find the probability to the left of x. (Typing FALSE for the last argument gives you the actual height of the normal curve and not a probability.) We use this function in the same exact way that we used the first function. Here is one example of this function in use.
1 1.44 (60")
To find P(x 60), type the following in any cell.
=NORMDIST(60,63.6,2.5,TRUE)
Excel tells us that the probability that an adult female is 5 feet tall or shorter is 0.0749. This agrees with our previous answer. We could have also changed the problem to a standard normal distribution problem first, and then used Excel.
We would then type the following in any cell, using the first function.
=NORMSDIST(-1.44)
This gives the same result.
The TI-83 uses its built-in function normalcdf for calculating probabilities involving a normal distribution. The basic format for using this function is as follows.
normalcdf(lowerbound, upperbound, , )
This will find the area underneath the curve between lowerbound and upperbound for a normal distribution with mean and standard deviation . If there is no lower bound (area to the left of a value), we enter 1E99 as the lower bound by pressing () 1 2ND , (to access EE) 99. If there is no upper bound (area to the right of , a value), we enter 1E99 as the lower bound by pressing 1 2ND (to access EE) 99. If we are using the standard normal distribution ( = 0, = 1), then we may omit these values and enter only the lower bound and upper bound. Our first example involves the area of a region to the left of a z-value.
Press 2ND VARS to access the DISTR menu, and select item 2: normalcdf. , ENTER . 99 , 1.96 ) On the main screen, press () 1 2ND
The TI-83 gives us 0.9750 to four decimal places, which agrees with our previous answer.
The next example deals with finding the area to the right of a z-value.
Since we have shaded to the right of z = 1.5, we have no upper bound. Press
2ND VARS
to access the DISTR menu, and select item 2: normalcdf. On the
main screen, press () 1.5
The TI-83 tells us that the probability is 0.9332, which agrees with our previous answer.
The next example deals with finding the area between two z-values.
P(1.23 < z < 0.97)
For such a case as this, we have both a lower bound and an upper bound. Press
2ND VARS to access the DISTR menu, and select item 2: normalcdf. On the ENTER . main screen, press () 1.23 , 0.97 )
The TI-83 gives us 0.7246, which differs from our previous answer of 0.7247. This is due to the fact that the table rounded each probability before we subtracted, whereas the TI-83 waits to round until after the subtraction.
Here is an example of how to work with nonstandard normal distributions using the TI-83.
VARS to access the DISTR menu, and select item , 99 , 60 , 63.6 2: normalcdf. On the main screen, press () 1 2ND ,
To find P(x 60), press 2.5
The TI-83 tells us that the probability that an adult female is 5 feet tall or shorter is 0.0749. This agrees with our previous answer. We could have also changed the problem to a standard normal distribution problem first, and then omitted the mean and standard deviation.
Find the given probability. 1. 4. 7. 10. 13. 16. 19. P(z < 1.25) P(z 0.85) P(z > 2.33) P(z > 1.96) P(1.29 < z < 2.13) P(1.57 z 0.79) P(z = 1.65) 2. 5. 8. 11. 14. 17. 20. P(z < 0.8) P(z 1.07) P(z 1.64) P(z > 0.09) P(0.17 z 1.4) P(2 < z < 2) P(z = 0) 3. 6. 9. 12. 15. 18. P(z < 2.16) P(z < 3.75) P(z 0.43) P(z > 4.17) P(2.05 < z < 1.13) P(1 < z < 1)
The heights of adult males are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. Find the probability that an adult male is 21. 22. 23. 24. 25. 26. 27. 28. 6 or taller 5 5 or shorter 5 10 or shorter 5 3 or taller Between 5 7 and 5 9 tall Between 5 4 and 6 4 tall Taller than 6 8 Exactly 5 9 tall
A computer software package generates random numbers that are normally distributed with a mean and standard deviation that are entered by the user. Suppose that the user enters a mean of 400 and a standard deviation of 100. Find the probability that a value will be 29. 30. 31. 32. 33. 34. Between 525 and 650 Above 490 Below 535 Below 357 Above 619 Between 340 and 714
At a certain community college, the time that is required by students to complete the math competency exam is approximately a normal distribution with a mean of 57.6 minutes and a standard deviation of 8 minutes. Find the probability that a student takes 35. 36. 37. 38. 39. 40. 41. 42. Longer than 1 hour to complete the test Less than 1 hour and 15 minutes to complete the test Between 55 minutes and 1 hour and 5 minutes to complete the test At least 45 minutes to complete the test No longer than 40 minutes to complete the test Between 35 and 55 minutes to complete the test Exactly 50 minutes to complete the test If the college has a 90-minute time limit for this test, what percentage of all students do not finish the test?
The lengths of human pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. Find the probability that a human pregnancy lasts 43. 44. 45. 46. At least 282 days Less than 275 days Between 254 days and 261 days More than 240 days
5.2 Discrete Random Variables That Are Approximately Normal
47. 268 days or less 48. Between 253 days and 283 days The heights of 12-month-old boys are approximately normally distributed with a mean of 29.8 inches and a standard deviation of 1.2 inches. Find the probability that a 12-month-old boy is 49. 50. 51. 52. 53. 54. Less than 28 inches tall Between 29 and 31 inches tall Taller than 32 inches Shorter than 33 inches Between 25 and 27 inches tall At least 31.5 inches tall
Generating Random Normal Data
To use Excel to generate random data, we will use the Analysis ToolPak, which is an add-in to Microsoft Excel. To see whether it has already been added, select the Tools menu. If you do not see Data Analysis as one of the choices in this menu, then you have to add in the Analysis ToolPak. From the Tools menu, select Add-Ins. When the dialog box appears, click on the box to the left of Analysis ToolPak so that a checkmark appears, then click OK. Select Data Analysis from the Tools menu. When the dialog box opens, scroll down and select Random Number Generation and click OK. This will open the dialog box for random number generation. We will be generating 1000 random normal values, and using them to investigate certain normal probabilities. To generate one column with 1000 values, enter 1 for Number of Variables and 1000 for Number of Random Numbers. For Distribution, select Normal. After selecting the normal distribution, enter 0 for Mean and 1 for Standard Deviation (this is the standard normal distribution). Enter any integer of your choice for Random Seed. Once this has all been done, click OK and Excel will generate your values on a new worksheet. Now that these values have been generated, we will use them to investigate normal probabilities. First, it will help our efforts if the values have been sorted in ascending order. From the Data menu select Sort. A dialog box opens. Be sure that the box labeled Sort by says Column A, and select Ascending for the order. Click on OK, and your values will be sorted for you. We found earlier in this section that P(z < 1.96) = 0.9750. This means that 97.5% of all z-values should be less than 1.96, or approximately 975 of our 1000 values should be less than 1.96. While attempting this simulation twice, I found that 975 and 972 values were less than 1.96. Go on to verify the following probabilities. Write down the percentage of your values that satisfy the given condition. 1. P(z < 1.25) = 0.1056 3. P(z > 1.5) = 0.9332 5. P(1.53 < z < 0.75) = 0.1636 2. P(z > 1.28) = 0.1003 4. P(1.23 < z < 0.97) = 0.7247 6. P(0.5 < z < 0.88) = 0.1191
Discrete Random Variables That Are Approximately Normal; Normal Approximation to the Binomial Distribution
Although normal distributions are continuous, there are some discrete random variables that are approximately normal. In other words, their distribution roughly follows
a normal distribution. If we were to look at a histogram for the distribution, we would see that it was symmetric and followed the same bell-shaped pattern that a normal distribution does. If we are to use a normal distribution to find probabilities associated with a discrete random variable, we must make a minor adjustment first to convert it to a continuous random variable. This minor adjustment is called a continuity correction. Lets begin by taking a look at a discrete random variable that is approximately normally distributed. An IQ test is designed to produce a distribution that is approximately normally distributed with a mean score of 100 points and a standard deviation of 15 points. IQ scores are always whole numbers. If we let x represent the score on this IQ test, then this random variable is discrete because it can only take on whole numbers as values. It is impossible to have an IQ score of 103.7. The problem we are faced with is that a normal distribution is continuous. If we look at a normal distribution with a mean of 100 and a standard deviation of 15, we see that 103.7 is a possible value. All of the values in between 103 and 104 are possible values. However, in a discrete distribution none of those values are possible. If we are to use a normal distribution, we must decide what to do with all of the space between the possible values. We will simply divide the values in the gap in half, assigning the first half to the previous possible value and the second half to the next possible value. An IQ score of 103 is associated with the interval 102.5 x 103.5. The interval 102.5 x 103.5 is the continuous equivalent of the discrete value x = 103. Changing a discrete value into a continuous interval is what we refer to as a continuity correction. In the following example, we will use the continuity correction and a normal distribution to calculate a probability involving a discrete random variable.
An IQ test is designed to produce a distribution that is approximately normally distributed with a mean score of 100 points and a standard deviation of 15 points. Find the probability that a randomly selected person has an IQ of 103 points.
Let x represent a persons IQ score. The random variable x is discrete, so we must make a continuity correction before we can find this probability.
P(x = 103) = P(102.5 x 103.5) = P 102.5 100 103.5 100 z 15 15
In the first step, cc tells us that a continuity correction has been made. The probability P(x = 103) involves a discrete random variable, whereas the probability P(102.5 x 103.5) now involves a continuous random variable. We have changed a discrete problem into a continuous problem. We then finish the problem off with the process developed in the last section: make the transformation to z, draw the graph, and so on.
We can tell from the graph that this probability must be small.
The probability that a randomly selected person has an IQ of 103 is 0.0235.
When we see a normal distribution problem, we must now first determine whether the continuity correction is necessary. Examine your random variable and determine whether only certain values are possible (discrete, make the continuity correction) or infinitely many values are possible (continuous, no correction). The discrete value x = 103 is associated with the interval 102.5 x 103.5, and the discrete value x = 104 is associated with the interval 103.5 x 104.5. There seems to be a conflict, as 103.5 is part of the interval associated with x = 103 and x = 104. We should assign 103.5 to only one of the intervals, but the inclusion of both endpoints in the interval makes our continuity corrections easier to make. Also, recall that the probability of any one specific value of a continuous random variable occurring is 0, so the probabilities will be the same. We will now focus on how to make the continuity correction if we are looking at a range of values, as opposed to a single value. Here are some possible discrete intervals, and their continuous equivalents.
Discrete x 103 x 103 103 x 110 Continuous x 102.5 x 103.5 102.5 x 110.5
For the first interval, x 103, we begin at 103 and move to the right. As a discrete interval, this consists of the values 103, 104, 105, . . . . As a continuous interval, this consists of all values that are 103 and higher. The only values associated with 103 that are not included are the values to the left of 103specifically, from 102.5 up. In general, we make the correction away from the interval that we are dealing with. Sometimes it is easier to look at the interval on the number line. Here is the original interval.
We will make our continuity corrections away from the given interval, like this.
101 102 102.5 103 104 105 106 107
The interval x 103 becomes x 102.5 with a continuity correction. How about the interval x 103? This interval is to the left of 103, so we make our continuity correction of 0.5 to the right. This gives us the interval x 103.5. If we look at an interval between two values, such as 103 x 110, all of the values inside of the interval are included. We make the continuity correction by extending the interval on the outside. We move 0.5 to the left of 103 and 0.5 to the right of 110, to produce the interval 102.5 x 110.5. Now we will look at a few examples, from start to finish.
The scores on a mathematics placement test are approximately normally distributed with a mean of 41.3 points and a standard deviation of 7.8 points. (Scores are whole numbers only.) Find the probability that a randomly selected student scores 55 or higher on the placement test.
Let x represent the score on the test. This random variable is discrete, since all scores are whole numbers. The probability that we are interested in is P(x 55). When we make the continuity correction, we make it to the left, since our interval goes to the right of 55.
P(x 55) = P(x 54.5) = P z 54.5 41.3 7.8
P(z 1.69) = 1 0.9545 = 0.0455
The probability that a randomly selected student gets a score of 55 or higher is 0.0455.
The number of fares for a Las Vegas taxi driver on Fridays is approximately normally distributed with a mean of 21.5 fares and a standard deviation of 6.3 fares. Find the probability that a driver has between 24 and 26 fares on a randomly selected Friday.
Let x represent the number of fares for a Las Vegas taxi driver on a Friday. The random variable in this problem is discrete, because a taxi driver can only have a whole number of fares for a day. The probability that we are interested in here is P(24 x 26). Since our interval is between these two values, we make the continuity correction outside of this interval, changing it to P(23.5 x 26.5).
P(24 x 26) = P(23.5 x 26.5) 23.5 21.5 26.5 21.5 = P z 6.3 6.3 = P(0.32 z 0.79) P(0.32 z 0.79) = 0.7852 0.6255 = 0.1597
P(0.32 z 0.79)
The probability that a Las Vegas taxi driver has between 24 and 26 fares is 0.1597.
Normal Approximation to the Binomial Distribution
If we have a binomial distribution for which n p 5 and n(1 p) 5, then we can use the normal distribution to approximate binomial probabilities. The reason is that, under these conditions, the binomial distribution is approximately normally distributed. The normal distribution has a mean = n p and a standard deviation = n p(1 p). Recall that these formulas were provided in the section on the binomial distribution. Consider a binomial distribution in which n = 100 and p = 0.5. A histogram showing the probability of having x successes is shown here.
Binomial Distribution, n = 100, p = 0.5 0.09 0.08 0.07
Probability of Success
0.06 0.05 0.04 0.03 0.02 0.01 0
0 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79 82 85 88 91 94 97 100
According to the information above, this binomial distribution should be approximately normally distributed, and it appears to be. The mean of this normal distribution should be
= 100(0.5) = 50
The standard deviation of this normal distribution should be
= = = = 5 n
100(0.5)(0.5) 25
Here is the graph of a normal distribution with = 50 and = 5.
Normal Distribution, Mean = 5, Standard Deviation = 5 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0 0 10 20 30 40 50 60 70 80 90 100
Here are the two distributions side by side. Binomial 0.1 0.08 0.06 0.04 0.02 0 32 35 38 41 44 47 50 53 56 59 62 65 68
0.1 0.08 0.06 0.04 0.02 0 30 Normal
We see that the two distributions are essentially the same, except that the binomial distribution is discrete and the normal distribution is continuous. Next is an example showing how to use the normal approximation to binomial probabilities. It will also show us how good an approximation it is.
A fair coin is flipped 12 times. Find the probability that it lands on heads at least 5 times.
This is a binomial problem, and we will do it first using the binomial tables. Then we will use the normal approximation and compare answers. Number of trials: n = 12 Success: A toss is heads. Number of successes: x = at least 5 Probability of success: p = 0.5 Failure: A toss is not heads. Probability of failure: 1 p = 0.5 We can use the binomial tables here, since n = 12 and 0.5 is one of the probabilities given in the tables.
P(At least 5) = 1 P(0 through 4) = 1 (0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208) = 1 0.1937 = 0.8063
From the direct use of the binomial distribution, we see that the probability that at least 5 of the tosses are heads is 0.8063. This problem satisfies the conditions required for the normal approximation to the binomial distribution as both n p and n(1 p) are at least 5.
p = 12(0.5) = 6
n(1 p) = 12(0.5) = 6
We may now approximate this binomial distribution with a normal distribution with the following mean and standard deviation.
12(0.5)(0.5) 3
We are looking for P(x 5), but we must keep in mind that the binomial distribution is discrete, so a continuity correction is necessary.
P(x 5) = P(x 4.5) = Pz 4.5 6 1.73
The probability that the normal approximation gives us for at least 5 heads in 12 tosses is 0.8078. This is off, slightly, from our exact answer of 0.8063. This error decreases as the number of trials increases.
A fair coin is flipped 1000 times. Find the probability that it lands on heads between 480 and 520 times, inclusive.
This is a binomial problem. First we should construct a table of information. Number of trials: n = 1000 Success: A toss is heads. Number of successes: x = 480 through 520 Probability of success: p = 0.5 Failure: A toss is not heads. Probability of failure: 1 p = 0.5 We cannot use the binomial tables here, since n = 1000 is too large for our tables. Using the binomial formula would require us to plug into the formula 41 times. Lets check the conditions for the normal approximation.
p = 1000(0.5) = 500 5
n(1 p) = 1000(0.5) = 500 5
Since both conditions are satisfied, we continue by finding the mean and standard deviation for this distribution.
= 1000(0.5) = 500
1000(0.5)(0.5) 250
Now, we begin with the probability P(480 x 520).
P(480 x 520) = P(479.5 x 520.5) = P 479.5 500 520.5 500 z 15.81 15.81
Heres the graph.
The approximation of the probability is 0.8064. The actual probability, which can be found using Microsoft Excel, is 0.8052.
This is a binomial problem, so we begin with a table of information. Number of trials: n = 300 Success: A registered voter is Republican. Number of successes: x = at least 155 Probability of success: p = 0.45 Failure: A registered voter is not Republican. Probability of failure: 1 p = 0.55 We cannot use the binomial tables for this problem, since the number of trials is too large. To use the binomial formula directly, we would need to plug into the formula 146 times (155 through 300). The complement rule is of no assistance, because it requires us to plug into the binomial formula 155 times (0 through 154). We check the conditions required to use the normal approximation.
p = 300(0.45) = 135 5
n(1 p) = 300(0.55) = 165 5
= 300(0.45) = 135
300(0.45)(0.55) 74.25
The probability that we are interested in is P(x 155).
P(x 155) = P(x 154.5) = P z 154.5 135 8.62
2 1 0 1 2 2.26
The normal approximation to this probability is 0.0119. A direct calculation using the binomial formula shows that the exact probability is 0.0119. Our approximation is quite accurate in this situation.
In the previous example, we see that it is rather unlikely that we draw a sample that has at least 155 Republicans if 45% of the registered voters are Republican. This could lead us to question whether the sample was truly random. Was there a bias in the way that people were selected to make Republicans more likely to be selected? We do not have enough information at this point to make a conclusion about this. Perhaps the 45% figure is not correct, and should actually be higher. In later chapters we will be able to make inferences about a claimed population proportion such as that 45% of registered voters in this county are Republicans.
Scores for high school sophomores on the mathematics portion of the Preliminary SAT/ National Merit Scholarship Qualifying Test (PSAT) are approximately normally distributed with a mean of 46.1 and a standard deviation of 10.5. Scores on the PSAT are whole numbers, with a low of 20 and a high of 80. 1. Find the probability that a sophomores score on the mathematics portion of the PSAT is 60 or higher. 2. Find the probability that a sophomores score on the mathematics portion of the PSAT is 42 or lower. 3. Find the probability that a sophomores score on the mathematics portion of the PSAT is between 40 and 49. 4. Find the probability that a sophomores score on the mathematics portion of the PSAT is between 65 and 75. 5. What percentage of sophomores earn a score of 52 or lower? The College Board reports that this percentage is 70%. How does your answer compare to the College Boards claim? 6. What percentage of sophomores earn a score of 68 or higher? The College Board reports that this percentage is 5%. How does your answer compare to the College Boards claim? IQ scores are approximately normally distributed with a mean of 100 points, and a standard deviation of 15 points. IQ scores are always in whole numbers. Find the probability that a person has an IQ 7. 8. 9. 10. 11. 12. 106 or lower Between 85 and 95 At least 120 No higher than 80 Higher than 98 Lower than 92
A statistics final exam produces scores that are approximately normally distributed with a mean of 225 points and a standard deviation of 20 points. There are 300 points possible on the exam, and scores are always whole numbers. Find the probability that a student scores 13. 14. 15. 16. 17. 270 points or higher on this exam 180 points or lower on this exam Between 210 points and 255 points on this exam, inclusive At least 200 points on this exam No higher than 240 points on this exam
18. Between 180 points and 230 points on this exam, inclusive SAT math scores are normally distributed with a mean of 512 points and a standard deviation of 112 points. SAT scores range from 200 to 800, and are always a multiple of 10. Because possible scores are 10 points apart, a continuity correction of 5 points is necessary instead of 0.5 points. 19. Find the probability that a high school student scores 400 points or lower on the math portion of the SAT. 20. What percentage of high school students score 640 points or lower on the math portion of the SAT? 21. Find the probability that a high school student scores 700 points or higher on the math portion of the SAT. 22. What percentage of high school students score 470 points or higher on the math portion of the SAT? 23. Find the probability that a high school student scores between 500 points and 600 points, inclusive, on the math portion of the SAT. 24. What percentage of high school students score between 350 points and 550 points, inclusive, on the math portion of the SAT? 25. Explain, in your own words, the conditions that require you to make a continuity correction. 26. Explain, in your own words, what the purpose of a continuity correction is. A fair coin is tossed 500 times. Find the probability that 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. At least 270 tosses are heads At most 240 tosses are heads 225 or fewer tosses are heads At least 285 tosses are heads Between 230 tosses and 245 tosses, inclusive, are heads Between 236 tosses and 264 tosses, inclusive, are heads At least 255 tosses are heads Exactly 255 tosses are heads Between 49% and 51% of the tosses, inclusive, are heads At least 55% of the tosses are heads
At a certain community college, 60% of the students are female. Three hundred students are randomly selected to be included in a survey. Find the probability that 37. 38. 39. 40. 41. 42. At least 195 of the students are female Between 150 and 190 students, inclusive, are female 155 or fewer students are female At least 70% of the students are female At least 130 of the students are male 150 or fewer students are male
43. On the night that Wilt Chamberlain scored 100 points in an NBA basketball game, he made 28 of his 32 free-throw attempts. Wilt was a notoriously poor freethrow shooter, making only approximately 51% of his attempts over his career. Assuming that attempting a free throw is an independent event, use the normal approximation to find the probability that a person who has a 51% chance of making any particular free throw makes at least 28 out of 32 attempts. Does your result suggest that Wilt was extremely fortunate to score 100 points in that game? 44. An election poll is printed in the newspaper, showing that the Republican candidate is well in front of his rival from the Democratic Party. Approximately 46% of the voters in the state that was surveyed are registered as Republicans. However, of the 1200 people polled in the survey, 602 were registered as Republicans. If 46% of the people in the state are registered as Republicans, find the probability that at least 602 of 1200 randomly selected voters from the state are registered as Republicans. What does your result suggest about the validity of the poll? 45. A mathematics competency exam is given at a community college as a requirement for graduation. The test is made up of 55 multiple-choice questions, each
with 5 possible answers, of which only 1 is correct. If a student guesses at all 55 questions, find the probability that the student gets 6 or fewer correct. 46. It is claimed that approximately 25% of Americans have a cellular phone. If 440 Americans are surveyed at random, find the probability that 90 or fewer have a cellular phone. (Based on results of a study by Baskerville Communications.) 47. While being interviewed about the status of education, a politician states that at least 20% of all new teachers leave the profession within the first three years. A random sample of 216 teachers hired three years ago revealed that 33 of them were no longer teaching. If 20% of all new teachers actually leave the profession within the first three years, find the probability that 33 or fewer of the 216 teachers were no longer teaching. 48. According to the U.S. Department of Commerce, 30% of Arkansas households have a computer. If 285 Arkansas households are selected at random, find the probability that 70 or fewer have a computer.
Investigating the Normal Approximation
We will now examine the connection between the binomial distribution and the normal distribution, by using Excel to simulate tossing a coin 20 times and counting the number of heads. We will repeat this experiment 1000 times, and construct a histogram for the number of heads recorded. We will use the Data Analysis Toolpak to simulate the tossing of the coins. Select the Tools menu. If you do not see Data Analysis as one of the choices, then you must add it. Select Add-Ins, and when the dialog box opens click the box next to Analysis ToolPak and click OK. From the Tools menu, select Data Analysis. When the dialog box appears, select Random Number Generation from the list and click on OK. In the box labeled Number of Variables type 20. This will represent the 20 tosses of the coin. In the box labeled Number of Random Numbers type 1000. This will represent the 1000 times we repeat this experiment. In the box labeled Distribution select Bernoulli. This will generate a series of 1s (successes) and 0s (failures). Under Parameters enter 0.5 in the box labeled p Value. We enter 0.5 because the probability that any individual toss is heads is 0.5. Type any number in the box labeled Random Seed; this is a number that is used to start the random number generation. Click OK. The random numbers will be generated on another worksheet. (Be patient; this may take some time.) Once the random generation is complete, type the following in cell U1:
=SUM(A1:T1)
This will total the number of heads the first time we tossed the coin 20 times. Copy cell U1 and paste it into cells U2 through U1000. Now we have the number of heads for each of the 1000 times this experiment was repeated. Now we will create a histogram for the number of heads. We will examine the shape of this histogram to determine whether the number of heads is approximately normally distributed. Before we do, we need to type a list of class boundaries for the histogram. Type the numbers from 0 through 20 in cells V1 through V21. From the Tools menu, select Data Analysis. Select Histogram from the list of choices, and click OK. In the box labeled Input Range type U1:U1000. This is the range of cells containing the number of heads for each set of 20 tosses. In the box labeled Bin Range, type V1:V21. Again, this is the range of cells that we created to store the class boundaries. Be sure to select the box labeled Chart Output, and click OK. Excel will create
5.3 Finding Values for Given Probabilities
a histogram on a new worksheet for you. Here is a sample of what your histogram might look like. Heads on 20 Tosses of a Coin (Repeated 1000 Times) 200 150 100 50 0
4 6 8 10 12 14 16 18 20 Number of Heads
Note that it appears to be roughly symmetrical, and that it is almost bell-shaped. When the conditions to use the normal approximation to the binomial distribution are met, the mean is given by = n p and the standard deviation is given by = n p(1 p). For 20 tosses of a coin, n = 20 and p = 0.5. The mean and standard deviation are as follows.
= 20(0.5) = 10
We will now examine the mean and standard deviation of our 1000 repetitions of this experiment. Go back to the worksheet that contains the randomly generated numbers. In cell W1, type =AVERAGE(U1:U1000). Below that, in cell W2, type =STDEV(U1:U1000). The data that generated the previous histogram had a mean of 9.995 and a standard deviation of 2.322. Both of these values are close to what they are supposed to be. Compare your mean and standard deviation to the values that we should get. Now repeat this example for other values of n and p. 1. n = 30 and p = 0.4 3. n = 50 and p = 0.2 2. n = 50 and p = 0.3 4. n = 100 and p = 0.1
For each case, examine the histogram and comment on whether it appears to be symmetrical and bell-shaped. Also, calculate the mean and standard deviation for your simulation, and calculate the mean and standard deviation for this distribution according to the normal approximation to the binomial distribution. Compare these theoretical values to the values that you calculated from your random number generation.
Finding Values for Given Probabilities
In this section we will examine problems that are essentially the opposite of the problems discussed earlier in this chapter. We will be given a certain probability, and we will try to find the value of the random variable associated with it. An example of such a problem is to find the height that separates the tallest 10% of all adult females
from the rest. We would look for a height x* such that P(x > x*) is approximately equal to 0.1 (10%). We will warm up with problems involving the standard normal distribution, and then later we will look at problems involving nonstandard normal distributions.
Find the z-value for which the area under the standard normal curve to the left of it is 0.01. (For this z-value, 1% of the observations fall below it.)
Now we begin with a graph.
1% or 0.01 2 1 0 1 2
We are looking for the z-value that has an area of 0.01 to the left of it. We look on the inside of the chart for an area of 0.0100. The closest we can find is 0.0099, which is associated with a z-value of 2.33. Therefore, the area to the left of the z-value 2.33 is approximately 0.01; that is, 1% of all z-values fall below 2.33.
Find the z-value for which the area under the standard normal curve to the right of it is 0.005. (For this z-value, 0.5% of the observations fall above it.)
We begin once again with the graph. We know that the area to the right of our z-value is 0.005. In order to use our tables, we need to know the area to the left of our z-value, which is 0.995 (1 0.005).
1 0.005 = 0.995 0.005 2 1 0 1 2
When we look for 0.9950 on the inside of the normal distribution table, we notice that it is exactly between 0.9949 and 0.9951. These two areas correspond to the z-values 2.57 and 2.58, respectively. In such a case where the desired area is exactly between two areas, we will use the mean of the two z-values.
2.57 + 258 2
The z-value that has an area of 0.005 to the right of it is 2.575; that is, 0.5% of z-values are higher than 2.575.
Find the z-values that separate the middle 95% of the area under the standard normal curve from the rest of the curve. (For these z-values, 95% of all z-values lie between them.)
We are looking for two z-values in this problem. We will label them z 1 and z2. Lets begin with a graph to get a sense of the big picture.
0.025 2 z1 1
95% or 0.95 0 1 2 z2
The first value, z 1, is the easiest to find, because it is the z-value with an area of 0.025 to the left of it. We look up 0.0250 in our normal table, and find that the z-value associated with it is 1.96. What is the area to the left of z2? It is 0.975, which can be found by adding 0.95 to 0.025. We look at our normal table for 0.9750, and we find that the z-value associated with it is 1.96. This should come as no surprise, because the normal distribution is symmetric. Therefore, the z-values that separate the middle 95% of the area under the standard normal distribution from the rest are 1.96 and 1.96; that is, 95% of all z-values lie between 1.96 and 1.96.
95% or 0.95 2 1.96 0
Now that we have discussed this idea for the standard normal distribution, we can extend it to nonstandard normal distributions. Reconsider the problem mentioned at the beginning of the section: Find the height that separates the tallest 10% of all adult females from the rest. We will begin by finding the z-value for which the area under the curve to the right of that value is 0.1. Then we will convert that z-value into the height of an adult female.
The heights of adult females are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. Find the height that separates the tallest 10% of adult females from the rest.
We begin by finding the z-value for which the area to the right of it is 0.1 (10%). Here is the graph.
0.1 or 10%
The area to the left of this z-value is 0.9. When we look at our normal distribution table for 0.9000, the closest we can find is 0.8997, which is associated with the z-value 1.28. The next step is to convert this z-value to the height of an adult female. This is done using the formula for z-scores. Since we know values for z, , and , we can solve for x, which in this problem represents the height of an adult female.
z = 1.28 = 1.28
x x 63.6 2.5
2.5 = x 63.6
3.2 + 63.6 = x 66.8 = x
The height that separates the tallest 10% of adult females from the rest is 66.8 inches.
IQ tests are designed to produce a mean of 100 points, with a standard deviation of 15 points. Find the IQ score that separates the lowest 5% of IQ scores from the rest.
0.05 or 5% 0
Since we have the area to the left of the z-value, we can go directly to the normal distribution table and look for 0.0500. It falls exactly between 0.4495 (z = 1.64) and 0.4505 (z = 1.65). We use the mean of these two values.
z = (1.64) + (1.65) 2
Now we transform that z-value to an IQ score.
z = 1.645 = 1.645
x x 100 15
15 = x 100
24.675 + 100 = x 75.325 = x
Therefore, 5% of all people have an IQ below 75.325. Since IQ scores are discrete, we could go on to say that 5% of all people have an IQ that is 75 or lower.
Scores on the math portion of the SAT are approximately normally distributed with a mean of 512 points and a standard deviation of 112 points. Between what two scores are the middle 75% of all SAT math scores?
We will start with the graph and look for the two z-values, z 1 and z2, that separate the middle 75% of z-values from the rest. The first z-value, z 1, has an area of 0.1250 to the left of it. Looking at our table, the closest z-value is 1.15. The second z-value, z2, has an area of 0.8750 (0.1250 + 0.7500) to the left of it. Looking at our table, we see that this z-value is 1.15. We now must convert these two z-values into SAT scores.
75% or 0.7500 0.1250 z1 0 0.1250 z2
z = 1.15
z = 1.15 = x x 512 112
383.2 = x
Therefore, 75% of all SAT math scores are between 383.2 and 640.8. Since SAT scores are discrete (multiples of 10), we could go on to say that the middle 75% of all SAT math scores are between 390 and 640.
Microsoft Excel provides us with a built-in function to find a z-value that is associated with a given probability. It is
=NORMSINV(probability)
Inside the parentheses we type a probability, and Excel returns the z-value for which the area to the left of that value is the same as the probability we typed. For example,
=NORMSINV(0.9750)
returns 1.96, because the area to the left of z = 1.96 is equal to 0.9750.
Find the z-value for which the area under the standard normal curve to the left of it is 0.01.
We begin with a graph.
1% or 0.01
We are looking for the z-value that has an area of 0.01 to the left of it. Type the following in any cell.
=NORMSINV(0.01)
Excel gives us 2.32634, which rounds to 2.33 to two decimal places.
Microsoft Excel also provides us with a built-in function to use when the distribution is a nonstandard normal distribution. Here is the general formula.
=NORMINV(probability, mean, standard deviation)
Inside the parentheses we must provide three values. The first is the probability that we are interested in, or the area to the left of the desired value. The last two values are the mean and standard deviation of the distribution in question.
We begin by drawing the graph, so that we know the area to the left of our desired value.
In this case the probability is 0.9. We can then type the following in any cell.
=NORMINV(0.9,63.6,2.5)
The value that Excel gives us is 66.8. Therefore, the tallest 10% of all adult females are taller than 66.8 inches. Note that this value agrees with the value found earlier in this section.
The TI-83 uses the built-in function invNorm to find the value associated with a given probability or area underneath the curve. The basic format for using this function is as follows.
invnorm(area, , )
It calculates the z-value associated with the area to the left of the value. If we are using the standard normal distribution ( = 0, = 1), then we may omit these values and enter only the lower bound and upper bound. This function can be found in the DISTR menu.
We are looking for the z-value that has an area of 0.01 to the left of it. Press
to access the DISTR menu, and select item 3: invNorm. On the main
) ENTER .
screen, press 0.01
The TI-83 gives us 2.326347877, which rounds to 2.33 to two decimal places.
Here is an example involving a nonstandard normal distribution.
In this case the probability is 0.9. Press
) ENTER . 2ND VARS
to access the DISTR menu,
and select item 3: invNorm. On the main screen, press 0.9
The value that the TI-83 gives us is 66.80387892. Therefore, the tallest 10% of all adult females are taller than 66.8 inches. Note that this value agrees with the value found earlier in this section.
Find the z-value that separates 1. The lower 1% of all z-values from the rest 2. The lower 5% of all z-values from the rest
3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.
The lower 10% of all z-values from the rest The lower 31% of all z-values from the rest The upper 3% of all z-values from the rest The upper 2.5% of all z-values from the rest The upper 15% of all z-values from the rest The upper 20% of all z-values from the rest The upper 60% of all z-values from the rest The upper 72% of all z-values from the rest The middle 80% of all z-values from the rest The middle 77% of all z-values from the rest The middle 93% of all z-values from the rest The middle 50% of all z-values from the rest Find the z-value that is associated with the first quartile of a normal distribution. Find the z-value that is associated with the third quartile of a normal distribution.
The heights of adult males are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. 17. 18. 19. 20. 21. 22. 23. What height separates the shortest 8% of adult males from the rest? What height separates the shortest 25% of adult males from the rest? Find the height that separates the tallest 15% of adult males from the rest. Twenty percent of adult males are above what height? What heights separate the middle 70% of adult males from the rest? What heights separate the middle 50% of all adult males from the rest? A builder wants to make the height of his doorways so that only 3% of adult males will have to duck on entering the room. How high should the doorways be? 24. A person of below-average height wants to form a social club for the vertically challenged. He decides that adult males will have to be in the shortest 1% of all adult males to join. What is the maximum height that an adult male can have and still join this club?
The heights of 12-month-old boys are approximately normally distributed with a mean of 29.8 inches and a standard deviation of 1.2 inches. 25. 26. 27. 28. What height separates the tallest 33% of all 12-month-old boys from the rest? Sixty percent of 12-month-old boys are above what height? Twenty-five percent of 12-month-old boys are shorter than what height? What height separates the shortest 50% of all 12-month-old boys from the rest?
Scores for high school sophomores on the mathematics portion of the Preliminary SAT/National Merit Scholarship Qualifying Test (PSAT) are approximately normally distributed with a mean of 46.1 and a standard deviation of 10.5. Scores on the PSAT are whole numbers, with a low of 20 and a high of 80. 29. A universitys engineering department decides to mail out brochures to students whose test scores were in the top 8%. What is the lowest score that will be sent a letter? 30. A high school decides to honor its students who scored in the top 20% of all students nationally. What is the lowest score that will be honored by the high school? 31. A high school decides to set up a program for students who struggled on the PSAT math exam, so that they may improve their performance. The school decides to invite students who scored in the lower 25% nationally. What is the highest score that will be invited into this program? 32. The College Board reports that 1% of all sophomores score below 25 points on this exam. Use the above mean and standard deviation to determine what score separates the lowest 1% of all scores from the rest. Compare your result to the College Boards claim. If there are differences, explain why they may occur. IQ scores are approximately normally distributed with a mean of 100 points and a standard deviation of 15 points. IQ scores are always in whole numbers. 33. Forty percent of all people have an IQ above what score? 34. Thirty percent of all people have an IQ above what score? 35. Mensa, a group founded in England in 1946 by Roland Berrill and Dr. Lance Ware for people of high intelligence, requires that its members be in the top 2% on any intelligence test. What IQ is the lowest possible for admission to Mensa?
36. A new group is being founded for people who are of above-average intelligence, but fail to qualify for Mensa. The group has decided to accept people who have an IQ that is in the top 10%, but not in the top 2%. What are the lowest and highest IQ a person could have and belong to this group? A statistics final exam produces scores that are approximately normally distributed with a mean of 225 points and a standard deviation of 20 points. There are 300 points possible on the exam, and scores are always whole numbers. 37. 38. 39. 40. Seventy-five percent of all statistics students score above what score on this exam? What scores separate the middle 40% of all scores from the rest? What score is required to put a student in the top 1% of all scores? Five percent of all students score below what score?
SAT math scores are normally distributed with a mean of 512 points and a standard deviation of 112 points. SAT scores range from 200 to 800, and are always a multiple of 10. 41. 42. 43. 44. What score separates the top 3% of all scores from the rest? Eighty percent of all students score below what score? What scores separate the middle 60% of all scores from the rest? An elite college refuses to accept students whose SAT math scores are not in the top 5%. What is the lowest score that a student could get and still be admitted to the school?
EXTRA Is a Set of Data Approximately Normally Distributed?
If we have a set of data, how can we determine whether it is normally distributed? This task may be beyond the level of this text, but there are some techniques that we can use to get a good idea about a set of data and whether it is approximately a normal distribution. The first aspect of the data set that we will examine is its mean and median. If the mean and median are close to each other, then the distribution is roughly symmetric. Second, we will take a look at a histogram for the set of data, to see whether it is bell-shaped like the normal distribution. Finally, we will compare the interquartile range and the standard deviation to each other. If a set of data is normally distributed, then the first quartile is associated with a z-value of 0.67 and the third quartile is associated with a z-value of 0.67. Thus, the interquartile range should be approximately equal to 1.34 standard deviations. If we multiply 1.34 by the standard deviation and find this product to be close to the interquartile range, then this is a good sign that the data follow a normal distribution. The set of data that we will be examining follows. The values are the number of heads in 20 tosses of a fair coin. This experiment was carried out 100 times. 5 14 11 8 14 13 8 9 7 10 10 8 12 9 11 9 12 7 10 10 5 6 11 11 9 11 8 10 16 10 10 8 14 9 10 7 7 13 8 9 10 12 12 9 13 13 9 7 6 10 10 10 11 12 12 13 9 13 9 9 9 5 12 8 10 10 9 9 12 10 7 12 11 11 9 9 13 10 11 9 11 12 7 9 9 10 9 11 9 11 14 9 11 11 11 9 9 9 8 12
The first thing to examine is a histogram of the data. Is the histogram symmetric? Is it bell-shaped? Here is a histogram, generated by Microsoft Excel, for this set of data. Histogram 30 20 10 0 Frequency
The set of data seems to be close to symmetric and bell-shaped, and may come from a population that is normally distributed. We move on to compare the mean and median of the set of data. If the mean and median are vastly different, this is a sign that the distribution is not symmetric. The mean for the above set of data is 9.95, and the median is 10. The mean and median are somewhat close, although this suggests that the data may be negatively skewed rather than symmetric. Finally, we compare the interquartile range and the standard deviation. The first quartile is 9, and the third quartile is 11. This means that the interquartile range is 2. The standard deviation is 2.13. If we multiply the standard deviation by 1.34, we get 2.8542, which is not very close to 2. This makes it difficult to conclude whether or not these values are approximately distributed. (Another procedure for testing the normality of a set of data is introduced in Section 9.3.) Determine whether the following sets of data are approximately normal. 1. SAT math scores of 60 students 280 370 530 550 500 520 0 1 6 6 4 2 1 1 3 2 1 610 640 230 750 790 750 8 9 0 10 3 4 3 0 0 0 1 9 5 5 3 2 2 2 1 1 0 0 660 520 500 440 500 690 6 6 4 9 0 4 0 0 0 1 0 530 540 490 660 320 480 6 1 1 7 0 3 3 0 1 1 3 4 5 8 10 1 2 3 1 3 0 1 1 2 1 1 1 2 3 550 390 390 570 580 510 9 2 9 8 3 2 1 0 1 2 2 480 510 600 710 480 750 5 4 6 5 640 410 680 600 650 510 9 0 5 8 7 7 6 7 500 470 540 510 610 530 640 320 510 590 700 400 580 560 530 570 430 720
2. Number of pull-ups performed by 40 fourth-graders.
3. Number of rejects produced per day on an assembly line, for a 56-day period
4. Number of tornadoes that struck Utah for each year from 1950 through 1995
Year 1950 1951 1952 1953 1954 1955 1956 1957 1958 1959 Number 0 0 0 2 1 2 0 1 0 0 Year 1960 1961 1962 1963 1964 1965 1966 1967 1968 1969 Number 0 1 1 1 1 5 2 2 4 3 Year 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 Number 5 1 0 0 0 0 0 0 0 0 Year 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 Number 0 2 3 0 6 0 3 3 1 6 Year 1990 1991 1992 1993 1994 1995 Number 4 5 4 6 0 2
(Utah Disaster Center Web site)
Unit 3 began with discrete probability distributions. You should be familiar with how to calculate the mean and standard deviation of a probability distribution. From there we shifted focus to the binomial distribution. Look for the idea of x out of n. Also, we must have a probability of success. When trying to find a binomial probability, first try to use the binomial tables. If the tables are not applicable, we must use the formula except in the case where we may apply the normal approximation to the binomial distribution. When reading a Poisson distribution problem, look for the expected number of successes. Poisson problems look similar to binomial problems, and that is no accident. Poisson problems lack the number of trials that are present in binomial probability problems. When calculating Poisson probabilities, try to use the tables first. If the tables are not applicable to your problem, then you must use the formula. We next covered the normal distribution. The information required for normal probability problems is a mean and a standard deviation. Watch out for normal variables that are discrete, as they require a continuity correction. You should know when it is appropriate to use the normal approximation to the binomial distribution, and remember the continuity correction here as well. When finding a value that goes with a given probability (reverse normal), you will always be asked for a value rather than a probability.
1. During a typical week, an assembly line must be shut down 3 times due to accidents. Find the probability that the assembly line gets shut down at least once this week due to an accident. 2. Driving speeds on the 198 are normally distributed with a mean of 71 mph and a standard deviation of 5 mph. Find the probability that a car on the 198 is traveling faster than 75 mph. 3. IQ scores are approximately normally distributed with a mean of 100 and a standard deviation of 15. Find the probability that a randomly selected person has an IQ of at least 90. 4. Fifty-five percent of all Visalia households have cable TV. If 10 Visalia households are selected at random, find the probability that 5 of them have cable TV. 5. Fifty-five percent of all Visalia households have cable TV. If 100 Visalia households are selected at random, find the probability that at least 50 of them have cable TV. 6. The height of 12-month-old boys is normally distributed with a mean of 29.8 inches and a standard deviation of 1.2 inches. Ten percent of 12-month-old boys are above what height? 7. During the typical hour at Lake Lotsafish, I can expect to catch 2.5 fish. Find the probability that I catch 4 fish during my first hour fishing at the lake. 8. A quiz has 10 true/false questions. A student must answer at least 7 correctly in order to pass. If a student guesses on all 10 questions, find the probability that the student passes. 9. At the typical Academy Awards ceremony, 3 people cry as they accept their award. Find the probability that fewer than 3 people cry at this years awards. 10. Fifty-eight percent of all Tulare County residents do not have health insurance. If 10 residents of Tulare County are selected at random, find the probability that at least 3 do not have health insurance. 11. The heights of adult women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. Find the probability that an adult woman is 5 feet tall or taller.
12. A statistics final exam is graded out of 300 possible points and each score must be a whole number. The scores are approximately normally distributed with a mean of 225 points and a standard deviation of 35 points. Find the probability that a student scores 209 or lower. 13. The heights of adult males are normally distributed with a mean of 69 inches and a standard deviation of 2.8 inches. What height separates the shortest 20% of adult males from the rest? 14. A restaurant has five rooms for private parties. On a typical Friday night, three of the private rooms are used. Find the probability that there will not be enough private rooms on a Friday night. 15. A 10-question multiple-choice exam is given in a math class. There are 5 possible answers to each question, of which only 1 is correct. Thus, the probability that a student gets an answer correct by guessing is 0.2. If a student guesses on all 10 questions, find the probability that he gets 6 or less correct. 16. A 100-question multiple-choice exam is given in a math class. There are 5 possible answers to each question, of which only 1 is correct. Therefore, the probability that a student gets an answer correct by guessing is 0.2. If a student guesses on all 100 questions, find the probability that he gets 25 or less correct. 17. On a typical Friday night, Joe D. Bouncer has to throw 2.5 people out of his club. Find the probability that he has to throw 5 people out this Friday night. 18. The lengths of human pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. Find the probability that a human pregnancy lasts between 255 and 275 days. 19. IQ scores are approximately normally distributed with a mean of 100 points and a standard deviation of 15 points. Find the probability that a persons IQ is between 110 and 125. 1 20. The probability of a woman giving birth to twins is approximately 40 or 0.025. If 120 pregnant women are selected at random, find the probability that at least 5 of them have twins.
Mean of a probability distribution: = x Binomial probability: P(x) = nCx Poisson probability: P(x) = Standard units: z = x
P(x) ((x )2
Standard deviation of a probability distribution: = p (1 p)
Conditions for the normal approximation to the binomial distribution: n n(1 p) > 5 Mean of binomial distribution: = n
p > 5 and
Standard deviation of binomial distribution: =
One-Sample Confidence Intervals and Hypothesis Tests
THE CENTRAL LIMIT THEOREM AND CONFIDENCE INTERVALS
Section 6.1 The Sampling Distribution of the Sample Mean; The Central Limit Theorem Section 6.2 Confidence Intervals for a Population Mean (Large Samples) Section 6.3 Confidence Intervals for a Population Mean (Small Samples) Section 6.4 Confidence Intervals for a Population Proportion (Large Samples)
ONE-SAMPLE HYPOTHESIS TESTS
Section 7.1 Hypothesis Test for a Single Population Mean (Large Samples) Section 7.2 Hypothesis Test for a Single Population Mean (Small Samples) Section 7.3 Hypothesis Test for a Single Population Proportion (Large Samples)
ecall that it is the goal of inferential statistics to make inferences about a population parameter, such as a mean or proportion, by taking a random sample and calculating the appropriate sample statistic. We can use this statistic to make an inference about the population parameter. For instance, suppose that we wanted to estimate the proportion of all women who smoke cigarettes while pregnant. It would be impossible to contact every pregnant woman to find out whether she smokes. A random sample of 400 pregnant women revealed that 60 (or 15%) are smoking during their pregnancy. Does this mean that exactly 15% of all pregnant women smoke during their pregnancy? Although it is possible, it is highly unlikely that exactly 15% of all women smoke while pregnant. How far off from the actual population proportion is our sample proportion? We do not know. The difference between the sample proportion and the population proportion is called sampling error. In this chapter we will develop estimates for the sampling error, and use it to construct interval estimates for the population parameter. For example, we will conclude that we are 95% sure that the percentage of all women who smoke is somewhere between 11.50% and 18.50%. In other words, we are 95% sure that our sample is off by no more than 3.50% from the population proportion. Section 6.1 examines the sampling distribution of the sample mean and the Central Limit Theorem. This section provides us with the tools to construct interval estimates for population parameters. In Section 6.2 we will construct an interval for a population mean based on a sample of size 30 or greater. We will also determine how large a sample we should take. In Section 6.3 we will construct an interval for a population mean when our sample is smaller than 30. Finally, in Section 6.4 we will construct an interval for a population proportion, like the one above for the proportion of all pregnant women who smoke.
The Sampling Distribution of the Sample Mean; The Central Limit Theorem
Five students take an IQ test, and here are their scores: Alan 102 Bob 84 Cari 108 Darla 112 Erin 98
We will treat these scores as a population. The mean score is 100.8, and the standard deviation is 9.68297475. Keep these measures in mind for later. We could take samples from this population in an attempt to estimate the mean of the population. Suppose we draw Alans score and Bobs score at random. The mean of these two scores is 93. The sampling error is 7.8, since our sample mean is 7.8 below the actual population mean. Following is a table listing all of the different possible samples, drawn with replacement, of size 2.
Sample Alan, Alan Alan, Bob Alan, Cari Alan, Darla Alan, Erin Bob, Alan Bob, Bob Bob, Cari Bob, Darla Bob, Erin Cari, Alan Cari, Bob Cari, Cari Cari, Darla Cari, Erin Darla, Alan Darla, Bob Darla, Cari Darla, Darla Darla, Erin Erin, Alan Erin, Bob Erin, Cari Erin, Darla Erin, Erin Mean 102 93 105 107 100 93 84 96 98 91 105 96 108 110 103 107 98 110 112 105 100 91 103 105 98
6.1 The Sampling Distribution of the Sample Mean; The Central Limit Theorem
The mean of these sample means is 100.8, and the standard deviation is 6.846897107. What we have listed is the sampling distribution of the sample mean. When all samples of a certain size are drawn from a population, and the sample means are calculated, the distribution that we get is the sampling distribution of the sample mean. Note that the mean of this distribution is exactly the same as the mean of the population. Also, the standard deviation is lower than the standard deviation of the population. Lets examine the shape of this distribution. Here is a table of each sample mean, along with its frequency in the above distribution.
Mean 84 91 93 96 98 100 102 103 105 107 108 110 112
Frequency 1 2 2 2 3 2 1 2 4 2 1 2 1
Histogram of Sample Means 6 5 4 3 2 1 0
85 88 91 94 97 100 103 106 109 112 Sample Means
Note that the distribution does not look like a normal distribution. In general, as the sample size increases, the sampling distribution of the sample means does become approximately normal. This idea is summed up in the next theoremthe Central Limit Theorem. CENTRAL LIMIT THEOREM If all samples of size n are As the sample size increases, the distribution drawn from a population with a mean and standard deviaof the sample means gets closer to a normal dis tion , the distribution of the sample means x will be approxitribution. When the sample size is at least 30, a mately a normal distribution. The mean of the sample means normal distribution approximates the sampling (x ) is the same as the mean of the population . The standistribution of the sample means very well. In the dard deviation of the sample means (x ) is / n. x is Excel Exploration at the end of this section, you called the standard error of the mean, or the standard can see how the distribution of the sample means error. gets closer to a normal distribution as the sample size increases. We must note that if the original population was normally distributed, then the sampling distribution of sample means is normally distributed as well. If we look back to our IQ means, we see that the standard deviation (6.846897107) is equal to the standard deviation of the population divided by 2, where 2 is the size of the samples we took. What standard deviation could we expect if our samples were of size 3?
n 9.68297475 3
= 5.590468078
The standard deviation would be 5.590468078. Note that as the sample size increases, the standard deviation decreases. In other words, the sample means have less variation as the sample size increases.
The heights of adult females are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If we selected all possible samples of size 36 and calculated the mean of each sample, what would the mean of those sample means be? What would their standard deviation be?
The mean of the sample means x would be the same as the mean of the population63.6 inches.
The standard deviation of the sample means, or the standard error, is found as follows.
n 2.5 36
Therefore, the standard error is approximately 0.4167.
Probabilities Involving Sample Means
Now that we know that the sample means are approximately normally distributed, we can do probability problems involving sample means. First, lets begin with a problem from the last chapter.
The heights of adult females are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. Find the probability that a randomly selected adult female is 5 5 (65 inches) or taller.
This is a typical normal probability problem with = 63.6 and = 2.5.
= P(z 0.56) = 1 0.7123 = 0.2877
The probability that a randomly selected adult female is 5 5 or taller is 0.2877.
The major difference for probability problems involving a sample mean is in the conversion to z. Here is the formula that we will use:
x / n
It is very similar to the formula we had been using. We subtract the mean and then divide by the standard deviation. The difference is that the standard deviation for sample means, or standard error, is found by dividing the populations standard deviation by the square root of the sample size. Here is an example.
The heights of adult females are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. Find the probability that the mean height of 36 randomly selected adult females is 5 5 (65 inches) or taller.
We are interested in the probability of a sample mean being at least 65 inches.
65 63.6 P(x 65) = P z 2.5/ 36 = P(z 3.36) = 1 0.9996 = 0.0004
The probability that the mean height of 36 randomly selected adult females is 5 5 or taller is 0.0004.
Thus, we see that it is entirely possible that an adult female is at least 5 5 tall (0.2877), but it is rather unlikely that a group of 36 women have a mean height of at least 5 5 (0.0004). In an individual trial anything is possible, but as the sample size increases our sample mean will be drawn toward the population mean.
IQ scores are normally distributed with a mean of 100 points and a standard deviation of 15 points. If 100 people are randomly selected, find the probability that their mean score is below 97.
The mean of the sample means is 100, the same as the mean of the population. The standard error is 1.5.
n 15 100
Now we find the probability that the mean score is below 97.
97 100 P(x < 97) = P z < 15/ 100 = P(z < 2) = 0.0228
The probability of 100 randomly selected people having a mean IQ score below 97 is 0.0228.
You may be wondering why the previous problem did not require a continuity correctionafter all, IQ scores are discrete. The important thing to keep in mind is that we are interested in means of IQ scores, not IQ scores themselves. The means of IQ scores are continuous.
1. Consider the population of possible rolls of a die: 1, 2, 3, 4, 5, 6. (a) Find the population mean . (b) Find the population standard deviation . (c) If all samples of size 2 are drawn from this population, what should the mean of the sample means ( x ) be according to the Central Limit Theorem? (d) If all samples of size 2 are drawn from this population, what should the stan dard error ( x ) be according to the Central Limit Theorem? (e) Construct all 36 possible samples of size 2 from this population, and calculate their means. (Refer to the dice chart in Section 3.1.) (f) Calculate the mean and standard deviation of the sample means found in (e). Do these results agree with your answers in parts (c) and (d)? 2. IQ scores are normally distributed with a mean of 100 points and a standard deviation of 15 points. (a) If all samples of size 64 are drawn, what should the mean of the sample means ( x ) be according to the Central Limit Theorem? (b) If all samples of size 64 are drawn, what should the standard error ( x ) be according to the Central Limit Theorem? 3. The heights of adult males are normally distributed with a mean height of 69.0 inches and a standard deviation of 2.8 inches. (a) If all samples of size 144 are drawn, what should the mean of the sample means ( x ) be according to the Central Limit Theorem? (b) If all samples of size 144 are drawn, what should the standard error ( x ) be according to the Central Limit Theorem? 4. Community college students have a mean age of 23.5 years and a standard deviation of 7.5 years. (a) If all samples of size 30 are drawn, what should the mean of the sample means ( x ) be according to the Central Limit Theorem? (b) If all samples of size 30 are drawn, what should the standard error ( x ) be according to the Central Limit Theorem? 5. SAT math test scores have a mean of 512 points with a standard deviation of 112 points. (a) If all samples of size 40 are drawn, what should the mean of the sample means ( x ) be according to the Central Limit Theorem? (b) If all samples of size 40 are drawn, what should the standard error ( x ) be according to the Central Limit Theorem? IQ scores are normally distributed with a mean of 100 points and a standard deviation of 15 points. 6. A group of 36 people is selected at random. Find the probability that their mean IQ score is at least 105. 7. A group of 49 people is selected at random. Find the probability that their mean IQ score is less than 110. 8. A group of 81 people is selected at random. Find the probability that their mean IQ score is between 95 and 105.
9. A group of 75 people is selected at random. Find the probability that their mean IQ score is 92 or higher. 10. A group of 50 people is selected at random. Find the probability that their mean IQ score is between 105 and 110. The heights of adult males are normally distributed with a mean of 69 inches and a standard deviation of 2.8 inches. 11. A group of 64 adult males is selected at random. Find the probability that their mean height is below 67 inches. 12. A group of 225 adult males is selected at random. Find the probability that their mean height is between 68 inches and 70 inches. 13. A group of 121 adult males is selected at random. Find the probability that their mean height is 72 inches or more. 14. A group of 417 adult males is selected at random. Find the probability that their mean height is between 67.5 inches and 68.5 inches. 15. A group of 600 adult males is selected at random. Find the probability that their mean height is between 68.7 inches and 69.3 inches. 16. The mean age for a womans first marriage is 25.0 years of age, and the standard deviation is 4.3 years. A random sample of 256 females who have been married at least one time was selected. (a) Find the probability that the mean age for this sample is 24.3 years old or younger. (b) If the mean age of the 256 females was 24.3 years old, is it likely that the true population mean age for a womans first marriage is really 25.0 years? (c) If the mean age of the 256 females was 24.3 years old, is it likely that the sample is truly representative of the population? 17. It is widely known that people living in England drink a lot of teaon average, more than 3 cups per day. Suppose that the mean number of cups of tea that people living in England drink per day is 3.4 cups, with a standard deviation of 1.3 cups per day. (a) Find the probability that a sample of 144 people produces a mean of more than 3.65 cups per day. (b) How likely is it that this sample is a representative sample of the English population? (c) What possible biases could have affected this study? 18. The heights of 12-month-old boys are approximately normally distributed with a mean of 29.8 inches and a standard deviation of 1.2 inches. A sample of thirty-six 12-month-old boys was taken. (a) Find the probability that the mean height of this sample is between 29.5 inches and 31.5 inches. (b) Find the probability that the mean height is below 28 inches. (c) Find the probability that the mean height is at least 30.2 inches. (d) Repeat parts (a)(c) for a sample of 225 instead of thirty-six 12-month-old boys. Discuss, in your own words, what the increased sample size has done to these probabilities, and why. 19. The lengths of human pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. Three samples of women who have recently given birth are taken. One has a size of 49, the second has a size of 121, and the third has a size of 400. For each sample, find the probability that the mean length of pregnancy is (a) over 275 days; (b) less than 266 days; and (c) between 265 and 270 days.
Probability n = 49 n = 121 n = 400
P(x > 275) P(x < 266) P(265 < x < 270)
20. The heights of adult males are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. For samples of 36 adult males, 196 adult males, and 2025 adult males, find (a) P(x > 70.2); (b) P(x < 687); (c) P(68.5 < x < 69.5).
Central Limit Theorem
In this computer project, we will explore the Central Limit Theorem using Excel. First we will take 1000 samples, each of size 5. For each sample, we will calculate the sam ple mean, x . We will then look at the distribution of the sample means by constructing a histogram. We will go on to repeat this for samples of size 30 and 100. Before we begin we must be sure that the Analysis Tool Pak has been added. Pull down the Tools menu. If you see a choice that says Data Analysis, then the Tool Pak is added and you may continue. If you do not see Data Analysis, click on Add-ins under Tools. When the dialog box opens up, click on the box for Analysis Tool Pak, and click on OK. We will now select 1000 samples, each of size 5, from a population that has a uniform distribution. Pull down the Tools menu and then click on Data Analysis. When the dialog box appears, click on Random Number Generation and click OK. In the Random Number Generation dialog box, enter 5 in the box labeled Number of Variables. (This is our sample size.) Next, enter 1000 in the box labeled Number of Random Numbers. (This is the number of samples that we are going to take.) In the box labeled Distribution, click the arrow and select Uniform. In the boxes next to Parameters, type 0 and 100. Click OK. (This may take a few minutes for Excel to generate, so be patient.) We need to split the window in order to see the top and bottom of our worksheet. Click on cell A13. Pull down the Window menu and then click on Split. Pull down the Edit menu and click on Go To. When the dialog box appears, type A1001 and click OK. Click on cell A1. We will now calculate the mean for each of our samples. In cell F1 type =AVERAGE(A1:E1). Click on cell F1. Hold down the SHIFT key and click on cell F1000. Pull down the Edit Menu; select Fill and then Down. This will calculate the mean for all 1000 samples. Pull down the Window menu and select Remove Split. We now want to look at the distribution of the sample means. Pull down the Tools menu and select Data Analysis. Click on Histogram and click OK. When the dialog box appears, type F1:F1000 in the box labeled Input Range. Select Chart Output, and then click OK. In new worksheets, repeat this for samples of size 30 and 100. For a sample of size 30, column AD is the thirtieth column, so the sample means need to be calculated in column AE. We will also use column AE for constructing the histogram. For a sample of size 100, column CV is the hundredth column, so the sample means need to be calculated in column CW. We will also use column CW for constructing the histogram. According to the Central Limit Theorem, the sample means should follow a normal distribution. Here is a sample of the histogram that Excel generates for a sample size of 100. Note that the shape appears to be normal.
6.2 Confidence Intervals for a Population Mean (Large Samples)
Compare the shape of your three histograms. As the sample size increases, does the shape of the histogram look more like a normal distribution?
.1 8 42 457 5 .9 45 95 72 62 44 8 .7 06 87 66 46 1 .4 68 02 48 69 .2 4 29 17 72 49 7 .9 90 32 51 76 .7 51 47 53 79 .5 3 12 62 82 55 6 .2 73 77 57 86 .0 34 92 58 89 .7 3 96 07 92 6
Sample Means
Confidence Intervals for a Population Mean (Large Samples)
We begin with a brief definition of a confidence interval for a population mean before we learn how to construct a confidence interval. The level of confidence mentioned in the definition is DEFINITION A confidence interval for a usually 90%, 95%, or 99%. For a 95% confidence interval population mean is an interval that we for a population mean, we are 95% sure that the interval believe with some level of confidence conactually contains the population mean. When we construct tains the unknown population mean. a 95% confidence interval, we are looking for two values for which there is a 95% chance that the population mean is between them. For now, we will call those values L (Low) and H (High). In symbols, we are looking for L and H such that
P(L H) = 0.95
Thinking back to Chapter 5, we see that the z-values associated with the middle 95% are 1.96 and 1.96. We will start with a probability involving z, and work our way back.
0.95 = P(1.96 z 1.96) = P 1.96 = P 1.96 x 1.96 / n x 1.96 n
= P x 1.96 = P x + 1.96 = P x 1.96
x + 1.96 n x 1.96 n x + 1.96 n
We now know that
P x 1.96
x + 1.96 n
= 0.95 n
This tells us that 95% of the intervals constructed in this fashion will contain the population mean . This also tells us that any interval constructed in this fashion has a 95% chance of containing the population mean. The probability that it does not contain the population mean is 0.05. This complement of the level of confidence is denoted by (alpha).
Constructing a Confidence Interval for a Population Mean
From above, we see that if we multiply 1.96 by the standard error, and then move by that amount from the sample mean to the left and to the right, we get the interval that has a 95% chance of containing the sample mean.
_ x
+E
E in this picture stands for the margin of error. For a 95% confidence interval, we are really saying that there is a 95% chance that the sampling error (difference between the sample mean and population mean) is less than the margin of error. When we are trying to construct a confidence interval for a population mean, the first thing we should do is calculate the margin of error.
E = z /2
where z /2 = critical value = population standard deviation n = sample size We will work on the critical value first. Quite simply, it is the positive z-value that along with its opposite separates the middle percentage declared by the level of confidence from the rest of the curve. Suppose we were going to construct a 90% confidence interval. What z-value separates the middle 90% of the graph from the rest?
90% or 0.9
0.05 0 1.645
The critical value for a 90% confidence interval is 1.645. What is the critical value for a 99% confidence interval?
99% or 0.99 0.0050 2.575 0 0.0050 2.575
The critical value for a 99% confidence interval is 2.575. Lets look at the formula for margin of error once again.
where z /2 = critical value = population standard deviation n = sample size Note that the formula contains the population standard deviation. If we are trying to estimate the population mean, we most likely will not know the population standard deviation either. If a sample has size 30 or greater, then we can use the sample standard deviation as a reliable estimate of the population standard deviation. In Section 6.3 we will learn how to handle a sample that has a size less than 30. Thus, if the population standard deviation is unknown and the sample size is at least 30, we calculate the margin of error using the following.
s n
where z /2 = critical value s = sample standard deviation n = sample size Once the margin of error has been calculated, we plug into the following to obtain our confidence interval.
x E x +E
A random sample of 350 male college students were asked for the number of units they were taking. The mean was 12.3 units, with a standard deviation of 2.50 units. Construct a 90% confidence interval for the mean number of units taken by a male college student.
First we calculate the margin of error. We need the critical value for a 90% confidence interval.
90% or 0.9 0 1.645
The critical value for a 90% confidence interval is z/2 = 1.645. Now plug into the formula for margin of error.
= 1.645 = 0.22
When rounding the margin of error, round to one more decimal place than the sample mean. Our margin of error is 0.22. Now we add this to the sample mean, and subtract it from the sample mean, to obtain the confidence interval.
x E x + E 12.3 0.22 12.3 + 0.22 12.08 12.52
We are 90% confident that the mean number of units taken by all male college students is between 12.08 units and 12.52 units.
A random sample of 400 female college students were asked for the number of units they were taking. The mean was 11.5 units, with a standard deviation of 2.56 units. Construct a 90% confidence interval for the mean number of units taken by a female college student.
Our margin of error is 0.21. Now we add this to the sample mean, and subtract it from the sample mean, to obtain the confidence interval.
We are 90% confident that the mean number of units taken by all female college students is between 11.29 units and 11.71 units.
Lets analyze the two confidence intervals that we just constructed. Male college students: Female college students: 12.08 12.52 11.29 11.71
If someone claimed that the mean number of units taken by male college students is at least 12 units, does our confidence interval support such a statement? Yes since we are 90% sure that the mean number of units is between 12.08 and 12.52, our interval is consistent with the claim. If someone claimed that female students take an average of 12 units, what does our interval say about the claim? Our interval does not contain 12 units, so it is unlikely that the mean number of units taken by female students could be 12. There is a 10% chance that the true mean is not contained in our interval, so there is less than a 10% chance that this claim is accurate. If someone claimed that the average number of units taken by male students is greater than the average number of units taken by female students, what do our intervals tell us? All of the values contained in the confidence interval for male students are greater than the values in the confidence interval for female students. This suggests that there is a good chance that the claim is true. It is the goal of Chapters 7 and 8 to test the validity of such claims. We will revisit these claims then. We will develop formal procedures to test such claims, but they are all based upon confidence intervals. If we increase the level of confidence, what happens to the confidence interval? It grows wider. Since we are trying to catch the unknown population mean in our interval, we need to increase the size of our interval if we are to be more likely to have the actual population mean in our interval. Maybe this next example will help. Suppose you are standing in the outfield with a glove, and you are blindfolded. We will hit fly balls to you, and you will try to catch them by just sticking out your glove and hoping for the best. To have a better chance of catching the ball, you need a bigger glove. To catch the unknown population mean, we need a bigger confidence interval. We can make a mathematical argument for this as well. As the level of confidence increases, so does the critical value z/2. This in turn increases the margin of error, which creates a wider confidence interval.
95% or 0.95 0.0250 1.96 0 0.0250 1.96
The critical value for a 95% confidence interval is z/2 = 1.96. Now plug into the formula for margin of error.
E = z/2 = 1.96 = 0.25
s n 2.56 400
Our margin of error is 0.25. (Note that this is larger than the margin of error that we calculated for a 90% confidence interval.) Now we add this to the sample mean, and subtract it from the sample mean, to obtain the confidence interval.
99% or 0.99 0.0050
As we stated, the margin of error increased as the level of confidence increased. This, in turn, created wider confidence intervals.
Level of Confidence 90% 95% 99%
Margin of Error (E) 0.21 0.25 0.33
Lower Limit of Interval 11.29 11.25 11.17
Upper Limit of Interval 11.71 11.75 11.83
We will look at one more example before moving on.
Are movies too long? A random sample of 45 recent releases had a mean length of 111.1 minutes and a standard deviation of 22.6 minutes. Construct a 95% confidence interval for the mean length of all contemporary movies. Then use the interval to comment on the claim, The average length of movies is at least 2 hours.
s n 22.6 45
x E x + E 111.1 6.60 111.1 + 6.60 104.50 117.70
We are 95% confident that the mean length of contemporary movies is between 104.50 minutes and 117.70 minutes. Since 120 minutes (2 hours) is not included in this interval, and all of the values in the interval are less than 120 minutes, it is not likely that the claim The average length of movies is at least 2 hours is valid.
Finding Critical Values, Constructing Confidence Intervals
To find a critical value with Excel, we need to know what /2 is. To find , subtract the level of confidence from 1. For example, if the level of confidence is 95%, = 0.05. Divide by 2 to obtain /2. For a 95% level of confidence, /2 = 0.025. Here are the Excel functions to find the left-tail and right-tail critical values. Left-tail
=NORMSINV 2
Right-tail
=NORMSINV 1 2
For a 95% level of confidence, type the following in any two cells.
=NORMSINV(0.025) =NORMSINV(1-0.025)
(We use the right-tail value for the critical value, so actually only the second formula is needed.) Here is an example using Excel.
The superintendent of a golf course believes that the mean score for professional golfers on his course is above 72. He randomly samples the scores of 40 professional golfers from the last tournament:
Construct a 95% confidence interval for the mean score for all professional golfers on this course.
We will need the sample statistics (mean and standard deviation). Open a new Excel worksheet, and type the above values in column A (cells A1 through A40). In cell B1, type Sample Mean. In cell B2, type Sample Standard Deviation. In cell B3, type Sample Size. In cell B4, type Critical Value. Now that the framework is set up, we can begin the calculations. To calculate the sample mean, type =AVERAGE(A1:A40) in cell C1. To calculate the sample standard deviation, type =STDEV(A1:A40) in cell C2. Type the sample size (40) in cell C3. In cell C4, type
Excel Worksheet 69 Sample Mean 67 Sample Standard Deviation 69 Sample Size 71 Critical Value 75 78 Margin of Error 70 71 Lower Limit 72 Upper Limit 74 72 76 69 76 70 77 78 79 72 71 75 75 75 78 71 64 71 68 72 71 73 72 73 71 71 75 72 72 78 75 72.7 3.352764 40 1.959961 1.039012 71.66099 73.73901
=NORMSINV(1-0.025).
Now we calculate the margin of error. In cell B6, type Margin of Error. To calculate the margin of error in cell C6, type =C4*C2/SQRT(C3). Finally, we can find the endpoints of the confidence interval. In cell B8, type Lower Limit, and type Upper Limit in cell B9. To find the lower limit, type =C1-C6 in cell C8. We find the upper limit in cell C9 by typing =C1+C6. The results of the worksheet appear to the right. The confidence interval that we obtain is 71.66 73.74. Is the superintendents claim valid? Although most of our interval is above 72, we see that it is possible for the mean to be as low as 71.66, which does not support the claim.
To find a critical value with the TI-83, we need to know what /2 is. To find , subtract the level of confidence from 1. For example, if the level of confidence is 95%, = 0.05. Divide by 2 to obtain /2. For a 95% level of confidence, /2 = 0.025. Here is how to find the left-tail and right-tail critical values using the TI-83. To find the left-tail critical value, press 2ND VARS to get to the DISTR menu.
Select option 3:invNorm(. Type the value for /2. For a 95% confidence interval, type 0.025, and ) . Press ENTER , which will produce the following screen.
To two decimal places this is 1.96. Since the normal distribution is symmetric, the right-tail critical value is 1.96. Here is an example using the TI-83.
Construct a 95% confidence interval for the mean score for all professional golfers on this course. Enter these 40 values in L1. Press option 1, and enter L1 and press
STAT , and move to the CALC menu. Select ENTER .
We need the sample mean (72.7) and the sample standard deviation (3.35). To construct the confidence interval press STAT , and move to the TESTS menu.
Select option 7:ZInterval. For Inpt, select Stats. Enter 3.35 for the standard deviation, 72.7 for the mean, 40 for the sample size, and 0.95 for the confidence level.
Select Calculate and press
The confidence interval that we obtain is 71.662 73.738. Is the superintendents claim valid? Although most of our interval is above 72, we see that it is possible for the mean to be as low as 71.662, which does not support the claim.
Determining the Appropriate Sample Size
When a statistical study is started, often the first question is How large a sample should we take? If we know
how confident we want to be in our results (level of confidence) the amount of sampling error that is acceptable to us (margin of error) the standard deviation of the population
then we can calculate how large a sample is necessary to produce the desired results. We start with the formula for margin of error, and solve it for n.
E = z/2 E E E
n = z/2 n = z/2 n =
z/2 E z/2 E z/2 E z/2 E
n )2 = n =
The formula for calculating how large a sample is necessary is
n = z/2 E
As before, the populations standard deviation is rarely known, so we will need an estimate of the populations standard deviation. If no estimate is available, a preliminary sample must be taken so that we can come up with an estimate of the standard deviation.
Suppose we want to estimate the mean home price in Visalia, California. From previous studies, we know that the standard deviation is approximately $60,000. If we want to be 90% sure that the mean home price in Visalia differs from our sample mean by no more than $5,000, how large a sample should we take?
First, lets list the important information. Level of confidence: 90% z /2 = 1.645 Standard deviation: = $60,000 Error: E = $5,000
n = = z/2 E
1.645 60,000 5,000
= 389.7
Since it is not possible to survey 389.7 homes, we must have a sample of at least 390 homes for the desired results.
We will always round our answers that are not whole numbers up to the next whole number. The reason for this is that our formula gives the minimum sample size necessary, and any size below that would not be adequate.
A pediatrician wants to estimate the mean weight of firstborn babies. The standard deviation for all babies is approximately 1.15 pounds. If she wants to be 95% sure that the mean weight of firstborns differs from her sample mean by no more than 0.25 pounds, how large a sample is required?
First, lets list the important information. Level of confidence: 95% z /2 = 1.96 Standard deviation: = 1.15 Error: E = 0.25
n = =
z/2 E
The pediatrician must sample at least 82 babies to produce the desired results.
1. A random sample of 150 college students regarding their work schedules produced a mean of 25.4 hours worked per week with a standard deviation of 12 hours. Construct a 95% confidence interval for the mean number of hours worked per week by college students. 2. A placement exam is given to 36 randomly selected college students. Their mean score is 63.6 points, with a standard deviation of 15.1 points. Construct a 90% confidence interval for the mean score on this placement exam for all students. 3. Thirty randomly selected college students were asked how many cavities they had. These students had a mean of 3.2 cavities with a standard deviation of 1.65 cavities. (a) Construct a 99% confidence interval for the mean number of cavities for all college students. (b) Explain how you would gather the data to construct this confidence interval, and list any biases that may be present using your method. 4. A random sample of 49 pepper seeds was planted. The length of time between planting and germination was measured. The mean time until germination was 10.4 days with a standard deviation of 2.13 days. Construct a 95% confidence interval for the true length of time for all pepper seeds to germinate. 5. How large are math classes in community colleges in California? A random sample of 100 math classes had a mean of 28.3 students, with a standard deviation of 4.5 students. Construct a 92% confidence interval for the mean class size for all community college math classes. 6. A random sample of 50 community college students produced a mean GPA of 3.18 with a standard deviation of 0.30. (a) Construct an 80% confidence interval for the mean GPA of all community college students. (b) Explain how you would gather the data to construct this confidence interval, and list any biases that may be present using your method. 7. A sample of 150 high school students who took the SAT test had a mean math score of 515 with a standard deviation of 125. Construct a 95% confidence interval for the mean SAT math score for all high school students who take the SAT. 8. A sample of 200 smokers were asked at what age they started smoking. The mean age was 18.3 years with a standard deviation of 4.35 years. Construct a 95% confidence interval for the mean age that all smokers begin smoking. 9. The president of a community college is interested in the mean distance driven to school by his students. A random sample of 40 students produced a mean distance of 11.9 miles, with a standard deviation of 12.29 miles. (a) Construct a 95% confidence interval for the mean distance driven to school by all students of the community college. (b) The president believes that the mean distance driven by all students of the community college is 15 miles. Is the presidents belief consistent with your interval? Explain your answer in your own words. 10. The president of a community college claims that the mean age of the students at his school is less than 25. A random sample of 100 of his students had a mean age of 23.1 years old, with a standard deviation of 7.10 years.
(a) Construct a 99% confidence interval for the mean age of all students at his school. (b) Is the presidents claim consistent with your interval? Explain your answer in your own words. 11. A recent Pop-Up Video on VH1 claimed that the mean age of women at their first marriage is 25. A random sample of 104 married females had a mean age of 22.4 years old at their first marriage, with a standard deviation of 4.63 years. (a) Construct a 90% interval for the mean age of all women at their first marriage. (b) Is their claim consistent with your interval? Explain your answer in your own words. 12. A real estate agent claims that the mean sale price of a home in Visalia is $100,000. A random sample of 44 homes sold in Visalia produced a mean sales price of $115,661 with a standard deviation of $58,538. (a) Construct a 95% confidence interval for the mean sale price of all Visalia homes. (b) Is the agents claim consistent with your interval? Explain your answer in your own words. 13. Construct a 95% confidence interval for the mean cost of all funerals. A survey of 60 funerals showed that they had a mean cost of $5235, with a standard deviation of $1739. 14. A random sample of 529 pregnancies lasted 38.8 weeks with a standard deviation of 2.33 weeks. Construct a 90% confidence interval for the mean length of all pregnancies. 15. A sample of 36 medium-sized oranges had a mean vitamin C content of 64.5 mg, with a standard deviation of 3.14 mg. Construct a 99% confidence interval for the mean vitamin C content of medium-sized oranges. 16. (a) Repeat Exercise 15 using a sample size of 360 rather than 36. How does this affect your interval? Explain why this happens, in your own words. (b) Repeat Exercise 15 using a 95% confidence level instead of 99%. How does this affect your interval? Explain why this happens, in your own words. 17. How long is the typical honeymoon? A random sample of 180 couples who were recently married showed that their honeymoons had a mean length of 8.9 days with a standard deviation of 3.7 days. Construct a 95 % confidence interval for the mean length of all honeymoons. (Based on a survey by Brides magazine.) 18. Many health experts stress the importance of drinking at least eight 8-oz. glasses of water a day. A random sample of 330 adults showed that they drank a mean of 4.6 glasses of water per day, with a standard deviation of 2.79 glasses. (Based on the results of a study by Opinion Research Corporation International for the Water Quality Association.) (a) Construct a 95% confidence interval for the mean number of glasses that an adult drinks per day. (b) Does it appear that we are falling short of the recommendations? Explain how your confidence interval answers that question. 19. A random sample of 40 tax returns of individuals who earned less than $40,000 showed a mean charitable contribution of $527 with a standard deviation of $167. Construct a 95% confidence interval for the mean charitable contribution of individuals who earn less than $40,000. (Based on the results of a survey by Gallup.) 20. A sample of 103 homeowners showed that they spend a mean of 4.3 hours per week working on their lawn and garden during spring and summer, with a standard deviation of 3.62 hours. Construct a 95% confidence interval for the mean number of hours worked per week by all homeowners on their lawn and garden during spring and summer. (Based on the results of a study by Bruskin/Goldring Research for Sears, Roebuck and Company.) 21. The public relations officer at a college wants to estimate the mean IQ of all college students. If she wants to be 95% confident that her sample mean is off by no
more than 2 points, how large a sample is necessary? The standard deviation for IQ scores is 15 points. A pediatrician wants to estimate the mean height of boys on their first birthday. If she wants to be 90% sure that her sample mean is off by no more than 0.25 inches, how large a sample should she take? The standard deviation has been shown to be approximately 1.2 inches. How many hours do college students work per week? How large a sample is required to be 99% sure that a sample mean will be off by no more than 2 hours? An initial study suggested that the standard deviation is approximately 8.3 hours. A high school counselor wants to estimate the mean SAT combined score for high school students. He has been told that the standard deviation for all such scores is approximately 230 points. How large a sample is required to be 95% sure that his sample mean is off by no more than 20 points from the true mean SAT combined score? A statistics instructor wants to determine what the mean score on a statistics exam would be. A preliminary sample yielded a standard deviation of 14.2 points. How large should his sample be in order to be 90% sure that the sample mean differs from the population mean by no more than 3 points? How large a sample should a medical researcher take if he wants to estimate the mean pulse rate of healthy adults with 99% confidence within 3 beats per minute? An initial sample yielded a standard deviation of 6.2 beats per minute.
27. Explain, in your own words, how an increase in sample size affects the margin of error in a confidence interval for a population mean. 28. Explain, in your own words, how an increase in level of confidence affects the margin of error in a confidence interval for a population mean.
What is the mean age of students at your school? Randomly sample at least 30 students at your college and find their age. Use these sample data to construct a 95% confidence interval for the mean age of all students at your school. In addition to your confidence interval, include
Your raw data A histogram for the ages A pie chart showing the breakdown of ages. Set this up in any way that
EXCEL EXPLORATION Confidence Intervals
We will use Excel to explore the idea of confidence intervals and how to interpret them. First, we will take 100 samples, each of size 100. For each sample, we will calculate a confidence interval for . Finally, we will determine what percentage of the confidence intervals actually capture the true value of the population mean. Before we begin we must be sure that the Analysis Tool Pak has been added. Pull down the Tools menu. If you see a choice that says Data Analysis, then the Tool Pak is added and you may continue. If you do not see Data Analysis, click on Add-ins under Tools. When the dialog box opens up, click on the box for Analysis Tool Pak, and click on OK.
We are looking for the age for which we are 99% sure that the population mean age is no higher than it. We begin by finding the z-value that separates the lower 99% from the rest.
The value that we are looking for is z = 2.33. Now we proceed to find what we considered to be the right endpoint of our confidence interval. First, the margin of error.
= 2.33 = 1.65
Now we add the margin of error to the sample mean to obtain the confidence interval.
x + E 23.1 + 1.65 24.75
We are 99% confident that the mean age is 24.75 or lower. The presidents claim appears to be valid.
A right-sided interval works in a similar fashion. We begin by finding the critical value, for which we will use a positive value. Next, we find the margin of error. Finally, we subtract the margin of error from our sample mean.
A television commercial claims that the mean cost of funerals is above $5000. A survey of 60 funerals showed that they had a mean cost of $5235, with a standard deviation of $1739. Construct a 95% right-sided confidence interval for the mean cost of all funerals.
First, we find the critical value.
2 1 1.645
We will use z = 1.645.
= 369.31
Now we subtract the margin of error from the sample mean to obtain the confidence interval.
x E 5235 369.31 4865.69
We are 95% confident that the mean cost is $4865.69 or higher. The commercials claim may be false.
Confidence Intervals for a Population Mean (Small Samples)
In the last section, we saw that the sampling distribution of the sample means followed the z-distribution if either the population standard deviation was known or the sample size was at least 30. If neither of those conditions are true, then the distribution follows Students t-distribution. William Gosset, who was a chemist for the Guinness Brewery, developed this distribution that is appropriate for small samples. The Guinness Brewery did not permit its employees to publish research results, so Gosset published under the pen name Student. The t-distribution is similar to the z-distribution in some ways. The t-distribution is symmetric and bell-shaped with a mean of t = 0. There are different t-distributions; they depend on the sample size. Actually, they depend on the degrees of freedom. The degrees of freedom for a t-distribution are equal to 1 less than the sample size (n 1). As the size of the sample (n) increases, the t-distribution approaches the z-distribution. Therefore, if
the sample size is less than 30 the population mean is unknown we have reason to believe that the population follows a normal distribution
then we can construct a confidence interval for the population mean using the t-distribution. Here is the formula for the margin of error.
E = t/2
where t/2 = critical value of t s = sample standard deviation n = sample size
6.3 Confidence Intervals for a Population Mean (Small Samples)
We construct the interval in the same fashion as with large samples.
x E x + E
We begin our work by learning how to find the critical t-value.
Find the critical value of t necessary to construct a 90% confidence interval for a population mean, if there are 11 degrees of freedom.
A 90% confidence interval goes along with = 0.10. On your t chart, look for the column under the heading 0.10 for a two-tailed value. We use the two-tailed value because our confidence interval focuses on the middle 90%, excluding the two tails. Go down that column until you reach 11 degrees of freedom. Here is the correct row from the chart.
Two-Tailed Value Degrees of freedom 11 0.01 3.106 0.02 2.718 0.05 2.201 0.1 1.796 0.2 1.363 0.5 0.697
The critical value is 1.796.
A 99% confidence interval goes along with = 0.01. On your t chart, look for the column under the heading 0.01 for a two-tailed value. Go down that column until you reach 21 degrees of freedom. Here is the correct row from the chart.
Find the critical value of t necessary to construct a 95% confidence interval for a population mean, if there are 6 degrees of freedom.
A 95% confidence interval goes along with = 0.05. On your t chart, look for the column under the heading 0.05 for a two-tailed value. Go down that column until you reach 6 degrees of freedom. Here is the correct row from the chart.
Two-Tailed Value Degrees of freedom 6 0.01 3.707 0.02 3.143 0.05 2.447 0.1 1.943 0.2 1.440 0.5 0.718
We are now ready to construct some confidence intervals. With the exception of finding the critical value of t, everything else is exactly like the confidence intervals for a population mean based on a large sample.
Mark, an avid golfer, claims that his average drive is greater than 250 yards. To prove this, he measured his 14 drives from a round of golf. Here are the distances, in yards.
Construct a 95% confidence interval for the mean length of Marks drives. Based on your confidence interval, does Marks claim appear to be valid? We begin by calculating the mean and standard deviation for this sample. The mean is 257.5 yards and the standard deviation is 11.67 yards. Next, we need to find the critical value. For a 95% confidence interval, = 0.05. We have 13 degrees of freedom.
The critical value is 2.160. Now we can calculate the margin of error.
Finally, we construct the confidence interval.
We are 95% confident that the mean length of Marks drives is between 250.76 yards and 264.24 yards. This interval is consistent with Marks claim that his average drive is greater than 250 yards.
A report on coins states that the mean weight of pennies minted before 1982 is 3 grams. A random sample of 4 such pennies had a mean weight of 3.11 grams with a standard deviation of 0.027 grams. Construct a 99% confidence interval for the mean weight of all pennies minted before 1982. Based on your interval, does the reports claim seem valid?
We need to find the critical value. For a 99% confidence interval, = 0.01. We have 3 degrees of freedom.
E = t/2 s n
x E x + E 3.11 0.0789 3.11 + 0.0789 3.0311 3.1889
We are 99% confident that the mean weight of pre-1982 pennies is between 3.0311 and 3.1889. It appears that the report is in error because 3 is not part of this interval.
It is important that our sample seems to have come from a population that is normally distributed, or we cannot use the t-distribution to construct a confidence interval for the population mean. Look for outliers, because outliers may suggest a distribution that is skewed and not normal. There is a procedure in Section 9.3 to test whether a set of data follows a normal distribution, but until then you can use the material presented in ExtraIs a Set of Data Approximately Normally Distributed? which follows Section 5.3.
Two-Tailed Critical Values for the t-Distribution
As with the standard normal distribution, Excel can help us find critical values for constructing confidence intervals with the t-distribution. We have to supply Excel with and the number of degrees of freedom. Here is the format.
=TINV(, degrees of freedom)
For example, to find the critical value for a 90% confidence interval with 11 degrees of freedom (sample size of 12), type the following in any cell.
=TINV(0.10,11)
Excel returns a value of 1.795884. (To three decimal places this rounds to 1.796.)
Here is an example that uses Excel.
A calculus exam on applications of the derivative was given to 19 randomly selected first-semester calculus students. Here are their results.
Construct a 95% confidence interval for the mean score for all firstsemester calculus students. We will need the sample statistics (mean and standard deviation). Open a new Excel worksheet, and type the above values in column A (cells A1 through A19). In cell B1, type Sample Mean. In cell B2, type Sample Standard Deviation. In cell B3, type Sample Size. In cell B4, type Critical Value. Now that the framework is set up, we can begin the calculations. To calculate the sample mean, type =AVERAGE(A1:A19) in cell C1. To calculate the sample standard deviation, type =STDEV(A1:A19) in cell C2. Type the sample size (19) in cell C3. In cell C4, type =TINV(0.05,18). Now we calculate the margin of error. In cell B6, type Margin of Error. To calculate the margin of error in cell C6, type =C4*C2/SQRT(C3). Finally, we can find the endpoints of the confidence interval. In cell B8, type Lower Limit, and type Upper Limit in cell B9. To find the lower limit, type =C1-C6 in cell C8. We find the upper limit in cell C9 by typing =C1+C6. Your worksheet should look like this:
53 85 93 36 63 74 76 67 72 66 31 50 86 41 53 74 92 88 74 Sample Mean Sample Standard Deviation Sample Size Critical Value Margin of Error Lower Limit Upper Limit 67.05263 18.72453 19 2.100924 9.024944 58.02769 76.07758
By our calculations, the confidence interval is 58.03 76.08.
As with the standard normal distribution, the TI-83 can help us to construct confidence intervals for a population mean using the t-distribution. We will use the following example to walk through the steps.
Construct a 95% confidence interval for the mean score for all firstsemester calculus students. First, we need to put the 19 values in list L1. You may need to clear the values in L1 first. Once the values are in the list, press STAT and move to the TESTS menu. Select option 8:TInterval. The screen should look like this:
For Inpt: select Data. This will automatically calculate the sample mean and standard deviation that we need to construct the confidence interval. Next to List: be sure that it says L1, which is the list that contains our values. Freq: should have a 1 next to it. Next to C-Level: enter 0.95. Once you have Calculate highlighted, press ENTER . This should produce the following screen.
The 95% confidence interval for the mean score for all first-semester calculus students is 58.028 76.078.
1. A major company has a large fleet of cars. A vice president feels that gas costs are too high, which is because the cars in the fleet get poor gas mileage. A random sample of 15 cars produced a mean mileage of 17.9 miles per gallon with a standard deviation of 3.9 miles per gallon. Construct a 95% confidence interval for the mean mileage for all of the cars in the fleet. 2. Do college students eat enough fruits and vegetables? A random sample of 24 college students produced a mean of 1.7 servings of fruits and vegetables per day with a standard deviation of 2.3 servings per day. (a) Construct a 90% confidence interval for the mean number of servings of fruits and vegetables eaten daily by all college students.
(b) Explain how you would gather the data to construct this confidence interval, and list any biases that may be present using your method. A random sample of 10 NFL football games produced a mean of 39.7 total points per game with a standard deviation of 8.3 points. Construct a 99% confidence interval for the mean total points for all NFL games. A certain brand of candy makes a package that they claim contains at least 110 candies. A random sample of 18 packages had a mean of 123.5 candies with a standard deviation of 7.6 candies. Construct a 95% confidence interval for the mean number of candies contained in all packages of this brand. Six artichoke plants at a farm were selected at random. Here are the number of artichokes produced by each plant last year. 32 17 51 40 36 34 Construct a 99% confidence interval for the mean number of artichokes produced by all artichoke plants. Here are 10 randomly selected blood sugar levels from a laboratory. (Levels were measured in mg/DL after a 12-hour fast.) 105 89 96 135 94 91 111 107 141 83 Construct a 90% confidence interval for the mean blood sugar level of all people after a 12-hour fast. The Dean of Registration at a community college has implemented a new telephone registration system. The company that sold the system to the college claims that the mean length of phone calls is less than 10 minutes. To check this, the Dean randomly selected 20 calls, which had a mean length of 7.1 minutes and a standard deviation of 1.89 minutes. (a) Construct a 95% confidence interval for the mean length of all phone calls using this system. (b) Is the companys claim consistent with your interval? Explain your answer in your own words.
8. How old are the students at a local community college? The president of the college claims that the mean age of the students at her school is 25. A random sample of 25 students produced a mean age of 25.8 years with a standard deviation of 9.83 years. (a) Construct a 95% confidence interval for the mean age of all students at her community college. (b) Is the presidents claim consistent with your interval? Explain your answer in your own words. 9. It is claimed that the mean height of adult females is 65 inches (5 5 ). A random sample of 11 adult females had a mean height of 63.7 inches and a standard deviation of 3.00 inches. (a) Construct a 99% confidence interval for the mean height of all adult females. (b) Is the claim consistent with your interval? Explain your answer in your own words. (c) Explain how you would gather the data to construct this confidence interval, and list any biases that may be present using your method. 10. A cotton farmer is considering growing a new type of cotton. Before he switches all of his fields to the new type, he must be sure that the mean yield per acre is above 850 pounds per acre, which is the yield of his current type of cotton. He randomly selects 7 acres in different fields, and the mean yield for these fields is 941.4 pounds with a standard deviation of 195.5 pounds. (a) Construct a 90% confidence interval for the mean yield per acre for this type of cotton. (b) Do you believe that the mean yield per acre is above 850 pounds per acre? Explain your answer in your own words. 11. Construct a 95% confidence interval for the mean number of trades per day on the New York Stock Exchange. Here are the number of trades reported on the New York Stock Exchange for 8 randomly selected days in 1999.
12. Construct a 90% confidence interval for the mean winning mutual payout for horse races at Santa Anita Park. Here are the winning mutual payouts for 20 randomly selected horse races at Santa Anita Park. $3.40 $4.40 $14.80 $5.80 $14.60 $8.00 $23.80 $4.80 $3.40 $13.00 $5.20 $7.20 $5.40 $4.20 $37.20 $7.00 $44.40 $10.60 $23.80 $10.00
13. These are the listed home prices, in thousands of dollars, of 24 homes for sale in Visalia, California. $92.9 $239.9 $69.9 $84.9 $59.5 $49.9 $125 $175 $97.9 $179.9 $229 $89.9 $125.9 $74.9 $79.5 $112.9 $115 $63.9 $121 $185 $97.9 $329 $98.5 $185
Construct a 95% confidence interval for the mean sale price of homes in Visalia, California. 14. Here are the lengths of eight Bruce Springsteen concerts in minutes. 180 175 165 163 150 158 175 165 Construct a 95% confidence interval for the mean length of a Bruce Springsteen concert. 15. Over a one-week period, 18 customers were randomly selected and the length of time that they had to wait at the check-in counter at an airport had a mean of 11.3 minutes with a standard deviation of 3.22 minutes. (a) Construct a 90% confidence interval for the mean waiting time at an airport check-in counter. (b) Explain how you would gather the data to construct this confidence interval, and list any biases that may be present using your method. 16. These are the scores in 18 randomly selected games by NBA teams. 103 79 114 89 98 85 93 90 99 98 104 81 107 105 109 108 81 111
Construct a 95% confidence interval for the mean number of points scored per game by an NBA team. 17. Ten cups of fresh orange juice were purchased at 10 different grocery stores and analyzed for vitamin C content. The mean content of vitamin C was 120.9 mg, with a standard deviation of 5.06 mg. Construct a 95% confidence interval for the mean vitamin C content of fresh orange juice. 18. A random sample of 23 men who first married in 1998 had a mean age of 26.7 years with a standard deviation of 5.03 years. Construct a 95% confidence interval for the mean age at which men get married. (Based on data from the U.S. Census Bureau.) 19. The results of a study on the effect of smoking on blood pressure appeared in the American Journal of Hypertension (1998). Ambulatory 24-hour blood pressure monitoring was conducted in 29 healthy middle-aged smokers to study the influence of smoking on blood pressure and heart rate. Their daytime heart rates had a mean of 74 beats per minute, with a standard deviation of approximately 34.2 beats per minute. Construct a 95% confidence interval for the mean daytime heart rate for all healthy middle-aged smokers. (The study was conducted by G. Bolinder and U. deFaire.) 20. Twenty cows from a college dairy farm were selected at random. Records showed that the cows first lactation had a mean length of 307.2 days with a standard deviation of 46.03 days. Construct a 95% confidence interval for the mean length of first lactations for all cows.
6.4 Confidence Intervals for a Population Proportion (Large Samples)
In this problem, the sample proportion is 60 out of 400, so p = 60/400 or 0.15. The sample size n is 400. Now we find the critical value, which is done in the exact same way as in Section 6.2.
0.0250 1.96
95% or 0.95
0.0250 0 1.96
Thus, z/2 = 1.96. Now we calculate the margin of error.
E = z/2 = 1.96
0.15 0.85 400
When rounding the margin of error for a proportion, round to four decimal places. Now construct the confidence interval.
p E p + E 0.15 0.0350 0.15 + 0.0350 0.1150 0.1850
We are 95% confident that the proportion of pregnant women who smoke while pregnant is between 0.1150 (11.50%) and 0.1850 (18.50%).
A random sample of 65 community college statistics students revealed that 57 of them were under 30 years old. Construct a 95% confidence interval for the proportion of all community college statistics students who are under 30 years old.
In this problem, the sample proportion is 57 out of 65, so p = 57/65. The sample size n is 65. Now we find the critical value.
Thus, z/2 = 1.96. Now calculate the margin of error.
When rounding the margin of error for a proportion, round to four decimal places. Next construct the confidence interval.
p E p + E 57 57 0.0799 + 0.0799 65 65 0.7970 0.9568
We are 95% confident that the proportion of college statistics students who are under 30 is between 0.7970 (79.70%) and 0.9568 (95.68%).
We will now take a look at how the size of the sample affects the confidence interval.
The university is considering adding a sushi restaurant to their food court. They will only add it if they are sure that more than half of the students are in favor of it. A sample of 30 university students revealed that 18 of them were in favor of adding a sushi restaurant. Construct a 99% confidence interval for the true proportion of university students who are in favor of adding a sushi restaurant. Based on the universitys criteria and your confidence interval, will the university add a sushi restaurant?
The sample proportion is 18/30, or 0.6. The level of confidence is 99%.
The critical value is 2.575. Now we calculate the margin of error.
p (1 p) n
Next, the confidence interval.
p E p + E 0.6 0.2303 0.6 + 0.2303 0.3697 0.8303
Based on this interval, we are not 99% confident that the proportion of all university students in favor of adding a sushi restaurant to the food court is more than 0.5.
We will leave the results as they are, but will base the interval on a sample size of 300 instead of 30.
The university is considering adding a sushi restaurant to their food court. They will only add it if they are sure that more than half of the students are in favor of it. A sample of 300 university students revealed that 180 of them were in favor of adding a sushi restaurant. Construct a 99% confidence interval for the true proportion of university students who are in favor of adding a sushi restaurant. Based on the universitys criteria and your confidence interval, will the university add a sushi restaurant?
The sample proportion is 180/300, or 0.6. The level of confidence is 99%.
0.6 0.4 300
Based on this interval, we are 99% confident that the proportion of all university students in favor of adding a sushi restaurant to the food court is more than 0.5.
The major difference in the above two examples is the size of the sample. As the sample size increases, the margin of error grows smaller. Lets look at the same example, based on a sample of size 3000.
The university is considering adding a sushi restaurant to their food court. They will only add it if they are sure that more than half of the students are in favor of it. A sample of 3000 university students revealed that 1800 of them were in favor of adding a sushi restaurant. Construct a 99% confidence interval for the true proportion of university students who are in favor of adding a sushi restaurant. Based on the universitys criteria and your confidence interval, will the university add a sushi restaurant?
The sample proportion is 1800/3000, or 0.6. The level of confidence is 99%.
0.6 0.4 3000
Heres one more example that you may find interesting. It was done by a student of mine as his semester research project.
A student, interested in personal hygiene, was curious about the percentage of men who wash their hands after using the bathroom. At a local mall, the student observed that 9 of 30 men did not wash their hands after using the mens room. Construct a 90% confidence interval for the percentage of men who do not wash their hands after using a public bathroom.
The sample proportion is 9/30, or 0.3. The level of confidence is 90%.
Thus, we are 90% sure that between 16.24% and 43.76% of all men do not wash their hands after using a public bathroom.
Finding Critical Values, Constructing Confidence Intervals for Population Proportions
To find a critical value with Excel for confidence intervals for a population proportion, we use the same procedure that we applied when constructing confidence intervals for a population mean. This is because both require the use of the z-distribution. Again, we need to know what /2 is. To find , subtract the level of confidence from 1. For example, if the level of confidence is 95%, = 0.05. Divide by 2 to obtain /2. For a 95% level of confidence, /2 = 0.025. Here are the Excel functions to find the lefttail and right-tail critical values. Left-tail
(We use the right-tail value for the critical value, so actually only the second formula is needed.) The following example uses Excel to construct a confidence interval.
A company that makes a drug to reverse hair loss in males claims that 60% of the men who try their drug will grow new hair. A test on 40 randomly selected males experiencing hair loss produced the following results (Yes = grew new hair; No = no new hair).
Yes No Yes Yes No
No Yes Yes No Yes
No Yes Yes No No
No No Yes No
No Yes Yes Yes
Yes No Yes Yes
Construct a 95% confidence interval for the proportion of all men experiencing hair loss that will grow new hair by taking this drug.
Yes No No No Yes No No No Yes No Yes Yes Yes Yes No No Yes No Yes Yes Yes Yes Yes Yes No Yes Yes Yes No No No Yes No No Yes Yes No Yes No Yes
1 Number: Yes 0 Sample Size 0 Sample Proportion 0 Critical Value 1 0 Margin of Error 0 0 Lower Limit 1 Upper Limit 0 1 1 1 1 0 0 1 0 1 1 1 1 1 1 0 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1
22 40 0.55 1.959961 0.154172 0.395828 0.704172
Type the results in column A (cells A1 through A40). In this case, the success is a Yes. We need to change all of the successes into 1s, and the failures into 0s. We do this in order to be able to calculate the proportion of successes. In cell B1, type =IF(A1=Yes,1,0). This will change a Yes into a 1, and anything else into a 0. After you press the Enter key, click on cell B1 again. Next, hold down the Shift key and click on cell B40. Pull down the Edit menu, and click on Fill and then Down. This action converts the rest of column A. In cell C1, type Number: Yes. Next to that in cell D1, type =SUM(B1:B40). This will total column B, and tell us how many men grew new hair. In cell C2, type Sample Size. Type the sample size (40) in cell D2. In cell C3, type Sample Proportion. In cell D3 type =D1/D2 to calculate the sample proportion. In cell C4, type Critical Value. In cell D4, type =NORMSINV(1-0.025) to calculate the critical value. The next step is to calculate the margin of error. In cell C6 type Margin of Error. In cell D6, type =D4*SQRT(D3*(1-D3)/D2) to calculate the margin of error. Finally, we can find the endpoints of the confidence interval. In cell C8, type Lower Limit, and type Upper Limit in cell C9. To find the lower limit, type =D3-D6 in cell D8. We find the upper limit in cell D9 by typing =D3+D6. The confidence interval is 0.3958 0.7042.
The Excel worksheet is shown on the left.
We can use the TI-83 to construct a confidence interval for a population proportion with ease. Here is an example for which we will construct a confidence interval.
Construct a 95% confidence interval for the proportion of all men experiencing hair loss that will grow new hair by taking this drug. Begin by pressing STAT , and moving to the TESTS menu. Select option A: 1-PropZInt. Next to x: enter the number of successes (22). Next to n: enter the sample size (40). After entering the level of confidence, highlight Calculate and press ENTER . Here is the screen that you will see.
The confidence interval is 0.39583 0.70417.
Appropriate Sample Size
We can determine how large our sample needs to be for estimating a population proportion just as we did for estimating a population mean. We begin with the formula for margin of error, and solve for n.
E = z/2 E 2 = z/2 E 2 = z/22 E2
p p p p
(1 p) n (1 p) n
n = z/22
E2 E
n = z/22 n = z/22 z/22 z/2 E
p (1 p) E2 p (1 p) E2
n = n =
The formula for the appropriate sample size to estimate a proportion is
If there is no estimate of the sample proportion given, then we replace p by 0.5. This makes the biggest possible result, which will have us sample too many rather than too few. When substituting for the margin of error, be sure to use a decimal value and not a percentage for E.
A campaign manager is interested in finding out what percentage of registered voters will vote for her candidate in November. She knows that her candidate has been getting approximately 40% of the voters in other polls. How large a sample is required to be 95% confident that the percentage of all registered voters in favor of her candidate is within 2% of the sample percentage?
We have an estimate of the sample proportion p in this problem. We will use 40% for p. The level of confidence is 95%.
Thus, the critical value is 1.96. Finally, we use 0.02 for E.
n = = z/2 E 1.96 0.02
(1 p) 0.6
The rule for rounding is the same as when trying to estimate a meanif your result is not a whole number, round your answer up to the next whole number. Thus, she must sample at least 2305 registered voters.
A financial aid officer is curious to find out what percentage of college students who are parents are single parents. There is no known estimate of the sample proportion. How large a sample is required to be 95% confident that the percentage of single parents is within 4% of the sample percentage?
Since there is no known estimate of the sample proportion, we will use 0.5 for p. The level of confidence is 95%.
The critical value is 1.96. We will use 0.04 for E.
The financial officer must sample at least 601 college students.
1. In a random sample of 400 registered voters in New York State during the summer of 1999 who were asked whether they thought that Hillary Clinton should run for the U.S. Senate, 220 of the respondents said that she should run. Construct a 95% confidence interval for the proportion of all registered voters in New York State that felt that Hillary Clinton should run for the U.S. Senate. 2. A community college is considering offering an intensive algebra course to replace a two-semester algebra sequence. Of the 120 students who were randomly placed into the intensive course, 72 passed the course. Construct a 90% confidence interval for the proportion of all algebra students who could pass this course. 3. What percentage of Americans are not exposed to secondhand smoke? A sample of 200 nonsmokers revealed that only 22 of them did not show traces of a chemical that appears in the blood of people exposed to secondhand smoke. Construct a 99% confidence interval for the percentage of all Americans that are not exposed to secondhand smoke. 4. During the 1998 election, an exit poll of 4400 registered voters revealed that 2376 approved of the job that Bill Clinton was doing as president. Construct a 99% confidence interval for the proportion of all registered voters who approved of the job that Bill Clinton was doing as president. (Based on the results of a USA Today/ Gallup Poll.) 5. A random sample of 250 registered voters revealed that 63 of them feel that education is the most important issue when deciding on a candidate. Construct a 90% confidence interval for the proportion of all registered voters who feel that education is the most important issue when deciding on a candidate.
6. A sixth-grade student claims to have a rigged coinit is supposed to land on heads more often than tails. He flips it 30 times, and 18 of the tosses are heads. Construct a 95% confidence interval for the proportion of times that this coin will land on heads. 7. A local buffet restaurant has many patrons that are senior citizens. The manager of the restaurant claims that at least 60% of her customers are senior citizens. A random sample of 50 diners revealed that 33 of them are senior citizens. (a) Construct a 95% confidence interval for the true proportion of customers that are senior citizens. (b) Is the managers claim consistent with your interval? Explain your answer in your own words. 8. A nurse at a community college claims that less than 25% of college students smoke. A random sample of 30 college students revealed that 6 of them smoke. (a) Construct a 90% confidence interval for the true proportion of college students that smoke. (b) Is the nurses claim consistent with your interval? Explain your answer in your own words. (c) Explain how you would gather the data to construct this confidence interval, and list any biases that may be present using your method. 9. A magazine article claimed that half of all people list pepperoni as their favorite pizza topping. A random sample of 150 people revealed that 69 of them listed pepperoni as their favorite pizza topping. (a) Construct a 99% confidence interval for the true proportion of people that list pepperoni as their favorite pizza topping. (b) Is the articles claim consistent with your interval? Explain your answer in your own words. 10. A 911 operator claims that half of all 911 calls can be classified as nonemergency calls. Thirty-four of 64 randomly selected 911 calls were nonemergency calls. (a) Construct a 95% confidence interval for the true proportion of 911 calls that are nonemergency calls. (b) Is the operators claim consistent with your interval? Explain your answer in your own words. 11. A USA Today/CNN/Gallup Poll of 1005 adults nationally conducted October 2124, 1999, asked If Al Gore were the Democratic presidential nominee and George W. Bush were the Republican nominee, whom would you support? Five hundred fifty-three of those polled said they would support George W. Bush. Construct a 95% confidence interval for the proportion of all voters that supported George W. Bush for president versus Al Gore at this time. 12. A poll of 240 registered voters at least 40 years old revealed that 158 of them voted in the last presidential election. (a) Construct a 90% confidence interval for the proportion of registered voters at least 40 years old that voted in the last presidential election. (b) Explain how you would gather the data to construct this confidence interval, and list any biases that may be present using your method. 13. A sample of 75 renters in a city showed that 40 were under the age of 35. Construct a 99% confidence interval for the percentage of all renters that are under the age of 35. 14. A survey of 200 people aboard a cruise ship showed that 76 of them had children under age 18 at home. Construct a 95% confidence interval for the proportion of all cruise-goers that have children under age 18 at home. (Based on the results of a study by Cruise Line International.) 15. An ABC News poll of 452 leaned Democrats (Democrats and independents who lean toward the Democratic party) was conducted during the period of August 1015, 1999. This was right after rumors began to circulate about actor Warren Beatty seeking the Democratic presidential nomination. The poll showed that 9% of those surveyed supported Beatty for the nomination. Construct a 90% confidence interval for the proportion of all leaned Democrats that supported Warren Beatty for the nomination. 16. A nonscientific Internet poll showed that 73% of 2430 drivers feel that they are above average drivers. Construct a 95% confidence interval for the proportion
of all drivers that feel they are above average drivers. (Based on the results of a poll on abcnews.com.) What percentage of young drivers run red lights? A survey of 124 drivers ages 18 to 25 showed that 89 of them run red lights. (Based on the results of a study by the Social Science Research Center, Department of Psychology, Old Dominion University.) (a) Construct a 90% confidence interval for the proportion of all drivers ages 18 to 25 that run red lights. (b) Explain how you would gather the data to construct this confidence interval, and list any biases that may be present using your method. A sample of 412 adults showed that 93 of them are connected to the Internet from home. Construct a 95% confidence interval for the proportion of all adults that have Internet access at home. From a random sample of 139 incoming freshmen, 79 ended up graduating within six years. Construct a 95% confidence interval for the proportion of all college students that graduate within six years. (Based on results published by the NCAA.) A random sample of 47 people showed that 19 of them get more than 10 e-mails per day at work. Construct a 90% confidence interval for the proportion of all people that get more than 10 e-mails per day at work. (Based on a survey by the John J. Heldrich Center for Workforce Development.) In a survey of 320 women, 64 said that Valentines Day was their favorite occasion to receive flowers. Construct a 99% confidence interval for the proportion of all women who feel that Valentines Day is their favorite occasion to receive flowers. (Based on a study by the International Communications Research for the Society of American Florists.) One hundred parents of children age 4 and younger were asked whether they were very concerned about their childs safety and 89 said that they were. Construct a 95% confidence interval for the proportion of all parents of children age 4 and younger that are very concerned about their safety. (Based on a study by Yankelovich Partners for Gerber Products.) A sample of 340 Americans were asked whether they had ever seen a UFO, and 20 said that they had. Construct a 90% confidence interval for the proportion of all Americans who have seen a UFO. (Based on a study by Yankelovich Partners for Life magazine.) When asked whether they drink alcohol or not, 79 of 441 college students said that they did not. Construct a 95% confidence interval for the proportion of all college students that do not drink alcohol. (Based on the Harvard School of Public Health College Alcohol Study.) A study of 619 twelfth-graders showed that 74 of them smoked at least a half-pack of cigarettes daily. Construct a 99% confidence interval for the proportion of all twelfth-graders that smoke a half-pack or more daily. (Based on results of a study by the American Academy of Pediatric Dentistry.) A random sample of 436 workers by the Society of Financial Service showed that 47% of them felt that it was seriously unethical to inspect lockers or work areas, whereas 41% of 121 senior-level bosses felt the same way. (a) Construct a 95% confidence interval for the proportion of all workers who view this as seriously unethical. (b) Construct a 95% confidence interval for the proportion of all bosses who view this as seriously unethical. (c) Does it appear that the proportion of all workers who view this as seriously unethical is the same as the proportion of all bosses who view this as seriously unethical? Using your confidence intervals, explain why or why not. A random sample of 250 mobile phone users showed that 105 had taken an incoming call while dining out. (Based on a study by the National Restaurant Association.) (a) Construct a 95% confidence interval for the proportion of all mobile phone users who have taken an incoming call while dining out. (b) Based on your confidence interval, does it appear that fewer than 50% of all mobile phone users have taken an incoming call while dining out? A random sample of 1477 high school students revealed that 489 of them had ridden in a car at least once in the preceding 30 days with a driver who had been drinking alcohol. Construct a 90% confidence interval for the proportion of all
high school students who have ridden in a car at least once in the past 30 days with a driver who had been drinking alcohol. (Based on the Morbidity and Mortality Weekly Report by the Centers for Disease Control and Prevention.) A recent survey showed that 802 of 4155 high school students had seriously considered attempting suicide in the past 12 months. Construct a 99% confidence interval for the proportion of all high school students that have seriously considered attempting suicide in the past 12 months. (Based on the Morbidity and Mortality Weekly Report by the Centers for Disease Control and Prevention.) A random sample of 250 high school students that smoke cigarettes revealed that 174 of them were not asked to show proof of age the last time they purchased cigarettes. Construct a 95% confidence interval for the proportion of all high school students that were not asked to show proof of age the last time they purchased cigarettes. (Based on the Morbidity and Mortality Weekly Report by the Centers for Disease Control and Prevention.) A researcher wants to determine what proportion of all high school students have Internet access at home. He has no idea what the sample proportion will be. How large a sample is required to be 95% sure that the sample proportion is off by no more than 5%? A news agency is planning a poll. The agency wants to determine what proportion of American citizens supported NATO involvement in Kosovo. If the agency wants to be 99% sure that the sample proportion differs from the true population proportion by no more than 3%, how large a sample is necessary? A student wants to determine what percentage of college students smoke. How large a sample should she take to be 90% confident that her sample proportion is off by no more than 4.5%? An ambitious health researcher wants to determine the proportion of American teenagers that are overweight. The researcher wants to be within 1% of the actual proportion. How large a sample is necessary to be 95% confident that the sample proportion is within the researchers desired margin of error? A parents organization is trying to determine the percentage of high school athletes that make the Deans List. The leader of the organization believes that the percentage is in the neighborhood of 30%. How large a sample will the organization have to take to be 99% confident that their sample proportion is within 4% of the actual proportion? An orange grower has several hundred orchards that were replanted this year. He suspects that many of the new trees have a root disease. The nursery has agreed to reimburse the grower for the trees that are diseased. The disease can only be detected by digging up the tree, which will kill the tree. The highest known percentage of trees to have this disease is 5%. How many trees must be dug up to be 90% confident that the sample proportion of diseased trees is within 2% of the actual proportion of trees that have the root disease?
Randomly sample at least 100 vehicles in your schools parking lots, and record how many are convertibles. Use these sample data to construct a 95% confidence interval for the proportion of all vehicles at your school that are convertibles. You may choose some other classification of vehicles that interests you (for example, vans or foreign-made cars). In addition to your confidence interval, include
Your raw data A pie chart showing the percentage of vehicles at your school that are convertibles and the percentage that are not. Finally, answer the following questions.
We will use Excel to explore the idea of confidence intervals for population proportions and how to interpret them. We will take 100 samples, each of size 100. Each observation will be either a 1 (success) or a 0 (failure). For each sample, we will calculate a confidence interval for . We will then determine what percentage of the confidence intervals actually capture the true value of the population proportion. Before we begin we must be sure that the Analysis Tool Pak has been added. Pull down the Tools menu. If you see a choice that says Data Analysis, then the Tool Pak is added and you may continue. If you do not see Data Analysis, click on Add-ins under Tools. When the dialog box opens up, click on the box for Analysis Tool Pak, and click on OK. We begin by selecting 100 samples, each of size 100. Pull down the Tools menu and click on Data Analysis. When the dialog box appears, click on Random Number Generation and OK. When the next dialog box appears, enter 100 in the box labeled Number of Variables. (This is the number of samples that we are going to take.) Enter 100 in the box labeled Number of Random Numbers. (This is the sample size.) In the box labeled Distribution, click on Bernoulli. For Parameters, enter 0.4 for the p-value. Usually, the population parameter is unknown. Therefore, we try to estimate it using a confidence interval. However, for the purpose of this simulation it is imperative that we know the true population proportion so that we can find out how many of our confidence intervals contain . Our value for the population proportion is 0.4. To create the samples, click OK. (This may take a few minutes, so be patient.) In cell A101, type =SUM(A1:A100)/100. This will calculate the proportion of successes in the first sample. Click on cell A101, then hold the Shift key and click on cell CV101 to highlight the 101st row. Pull down the Edit Menu; select Fill and then Right. This will calculate the proportion of successes for all 100 samples. In row 102 we will calculate the lower limit of each confidence interval. In cell A102, type =A101-1.96*SQRT(A101*(1-A101)/100). Click on cell A102, then hold the Shift key and click on cell CV102 to highlight the 102nd row. Pull down the Edit Menu, select Fill and then Right. In row 103 we will calculate the upper limit of each confidence interval. In cell A103, type =A101+1.96*SQRT(A101*(1-A101)/100). Click on cell A103. Hold down the Shift key and click on cell CV103. Pull down the Edit Menu; click on Fill and then Right. Now we will determine exactly how many intervals contain the true population proportion, 0.4. In cell A104, type =IF((A102<0.4)*AND(A103>0.4),1,0). If the first confidence interval contains 0.4, the value in cell A104 will be 1; otherwise, it will be 0. To determine whether the other intervals contain 0.4, click on cell A104. Hold down the Shift key and click on cell CV104. Pull down the Edit Menu; click on Fill and then Right. To determine how many intervals contain 0.4, type =SUM(A104:CV104) in cell A105. If things go according to plan, approximately 95 of our intervals should contain the population proportion. Repeat this process for sample sizes of 250 and 1000. Does a different sample size dramatically affect the number of intervals that contain the population proportion? Why or why not?
hypothesis test is a procedure that we use to determine whether a claim about a population parameter is valid, invalid, or somewhere in between. In this chapter we will be performing three different tests that involve using a single sample to make an inference about a population parameter. In Section 7.1 we look at the test for a population mean based on a large sample. This test involves comparing the mean of a population to a certain number, such as The mean age of college statistics students is 24. It is in this section that the procedure for testing a claim is established. In Section 7.2 we will cover the small-sample test for a single population mean. This test, using the t-distribution, is used when the sample size is below 30. Finally, in Section 7.3, we cover the large-sample single proportion test. This test is used when a population proportion is compared to a certain proportion. An example of such a claim would be 13.6% of pregnant women smoke while they are pregnant.
Hypothesis Test for a Single Population Mean (Large Samples)
A college statistics instructor claims that the mean age of college statistics students is 24. Do we just assume that the instructor is right, or is there a way we can find out whether the claim is true? We will develop a five-step procedure to determine the validity of such a claim. The procedure is called a hypothesis test. The test is based on the confidence intervals that we developed in Chapter 6, and in many ways mirrors our legal system. Lets take a look at a confidence interval example to develop some intuition for this process.
A college statistics instructor claims that the mean age of college statistics students is 24. A random sample of 116 college statistics students revealed a mean age of 22.7 years with a standard deviation of 5.68 years. Construct a 95% confidence interval for the true mean age of all college statistics students. Use the interval to determine the validity of the instructors claim.
First, we find the critical value z /2 for a 95% confidence interval ( = 0.05).
We see that z /2 = 1.96. Now we determine the margin of error.
Next we construct the confidence interval.
7.1 Hypothesis Test for a Single Population Mean (Large Samples)
The confidence interval tells us that we are 95% sure that the true mean age for all college statistics students is between 21.67 and 23.73 years old. There is a 5% chance that the mean is not in this interval, which tells us that there is less than a 5% chance that the mean is actually 24. The instructors claim appears to be invalid.
Now we develop the five-step procedure for a hypothesis test. We will come back to this example several times to show how a hypothesis test goes hand in hand with a confidence interval. The five steps are as follows.
State the claim and the hypothesis Choose a level of significance Choose a test statistic State a decision rule Do the calculations and draw conclusions
Step 1: State the Claim and the Hypotheses
The claim is the statement that we are interested in investigating. Each hypothesis test should begin by identifying the claim to be tested, and writing it in words. For the previous example, the claim would be The mean age of college statistics students is 24. Once the claim has been identified, we need to write it in symbolic form. In our claim, we are comparing a mean of a population to the number 24. We let represent the mean age of all college statistics students, and our claim becomes = 24. The next thing that we do in Step 1 is to write the complement of the claim in symbolic form. What is the complement of = 24? It is 24. Suppose the claim had been < 24? What is the complement of less than 24? At first, we might be tempted to say greater than 24, but this would not be correct. The mean could be less than 24, greater than 24, or equal to 24. If the claim only uses one of these, the complement combines the other two. Therefore, the complement of < 24 is 24.
Here are the six different claims that could be made comparing the mean age of all college statistics students to 24.
1. The mean age of all college statistics students is 24. Claim: = 24 Claim: 24 Complement: 24
2. The mean age of all college statistics students is not 24. Complement: = 24 Complement: 24
3. The mean age of all college statistics students is less than 24. Claim: < 24 Claim: 24
4. The mean age of all college statistics students is at least 24. Complement: < 24 Complement: 24 Complement: > 24
5. The mean age of all college statistics students is greater than 24. Claim: > 24 Claim: 24
6. The mean age of all college statistics students is at most 24.
CHAPTER 7 One-Sample Hypothesis Tests
Once the claim and its complement have been established, we must decide which of those two is the null hypothesis (H0) and which is the alternate hypothesis (HA). Of the claim and its complement, the null hypothesis is the statement containing equality. In other words, we are looking for the symbols =, , or . We assume that the null hypothesis is true, and look for sample evidence to suggest that it is false. In our example, we are looking for a sample mean that is significantly lower than 24 or significantly higher than 24 to suggest that the true mean is not 24. The assumption that the null hypothesis is true is equivalent to assuming that a person on trial is innocent (until proven guilty). If there is sufficient evidence to prove that the assumption of the persons innocence is false beyond a reasonable doubt, then we reject our assumption that the person is innocent and declare him guilty. The evidence in a hypothesis test comes from a sample. If there is enough sample evidence to suggest that the null hypothesis is false, we reject the null hypothesis and accept the alternate hypothesis. When we assume that the null hypothesis is true, we are actually assuming that the population mean is equal to the value in the claim. We assume that = 24 whether the null hypothesis is = 24, 24, or 24. We say that the claimed value of the mean is 24. After the null hypothesis has been determined, the alternate hypothesis is the statement that does not contain equality. If we reject the null hypothesis as false, then we accept the alternate hypothesis as true. If we fail to reject the null hypothesis, we are not saying that the null hypothesis is true. We can only say that there is not enough sample evidence to convince ourselves that it is false. In this case, we fail to support the alternate hypothesis as true. When there is not enough evidence to find a defendant guilty, is the defendant then found innocent? No, the defendant is found not guilty. This means that it was not proven that the person was innocent. There was simply not enough evidence to conclude that the person was guilty. Reject null hypothesis: support alternate hypothesis Fail to reject null hypothesis: fail to support alternate hypothesis
Lets take a look at three claims, and try to determine what the null and alternate hypotheses are.
1. The mean number of hours spent studying per week by college students is less than 10 hours. Claim: < 10 Complement: 10 H0: 10 HA: < 10
2. The mean price for a home sold in Visalia, California is $100,000. Claim: = 100,000 Complement: 100,000 Claim: 30,000 Complement: > 30,000 H0: = 100,000 HA: 100,000 H0: 30,000 HA: > 30,000
3. The mean starting salary for a college graduate is at most $30,000.
There are two types of errors that we can make in a hypothesis test. A Type I error is rejecting a null hypothesis that is actually true. This is equivalent to convicting an innocent man. A Type II error is failing to reject a null hypothesis that is actually false.
This is equivalent to letting a guilty man go free. We try to minimize Type I errors in hypothesis testing. We control the chances of making a Type I error by establishing a level of significance.
Step 2: Choose a Level of Significance ()
We want to reject the null hypothesis only if the probability that it is true is very small. By very small we usually use 5% or 1%. In the case of 5%, we reject the null hypothesis if there is less than a 5% chance that the null hypothesis is true. In other words, if the chances that the null hypothesis is true are low enough, then we will declare it false. This 5%, or 1%, is referred to as the level of significance. We use the Greek letter alpha () to stand for the level of significance. This is equivalent to the phrase reasonable doubt in a trial. We will discuss the use of in confidence intervals and hypothesis tests when we get to Step 4. The level of significance that is used most frequently is 0.05. However, if the consequences of rejecting a null hypothesis that is true are serious, it is common to use 0.01 instead. Suppose a company is testing a new drug that they believe lowers high blood pressure. A lower level of significance is appropriate here, because of the serious health implications.
Step 3: Choose a Test Statistic
The test statistic is a formula that will be used to determine the probability that the null hypothesis is true based on the sample evidence. For the large-sample, singlemean test the formula is Test Statistic for a Large-Sample, Single-Mean Test (Population Standard Deviation Known)
z = x / n
where x = = = n =
sample mean claimed population mean population standard deviation sample size
As mentioned earlier in the text, the population standard deviation is often unknown. If the sample size is 30 or greater, the sample standard deviation is an acceptable estimate of the population standard deviation. Test Statistic for a Large-Sample, Single-Mean Test (Population Standard Deviation Unknown)
z = x s/ n
where x = = s = n =
This test statistic is the same formula we used in the previous chapter when trying to find probabilities involving a sample mean. If we are assuming that the value of is correct, this test statistic will allow us to calculate the probability of obtaining a sample mean that is as extreme as the mean of our sample. The test statistic will change for different types of hypothesis tests. The above formula applies only to a test for a single population mean when the sample size is at least 30. In this chapter there will be a second test statistic for a small-sample, singlemean test, and a third for the large-sample, single-proportion test.
Step 4: State the Decision Rule
When we state the decision rule, we are establishing the guidelines for rejecting the null hypothesis. We will be finding the critical value(s) for our test statistic. If the value of the test statistic is beyond the critical value for the test, we reject the null hypothesis. The decision rule for this test depends on two things: the alternate hypothesis and the level of significance. We return to our example to explore the establishment of a decision rule.
A college statistics instructor claims that the mean age of college statistics students is 24. Test his claim at the 0.05 level of significance. A random sample of 116 college statistics students revealed a mean age of 22.7 years with a standard deviation of 5.68 years.
Step 1 Claim in words: The mean age of all college statistics students is 24. Claim: = 24 Complement: 24 H0: = 24 HA: 24 Step 2 Level of significance: = 0.05 Step 3 Test statistic: z = x s/ n
To establish our decision rule, we must first decide what types of sample means suggest that the null hypothesis is false and, more important, that the alternate hypothesis is true. There are two ways that the population mean could be not equal to 24: it could be less than 24, and it could be greater than 24. This is an example of a two-tailed test. It is called a two-tailed test because the sample means that suggest the null hypothesis is false are found in the right tail of the z-distribution ( > 24) and the left tail of the z-distribution ( < 24). To set up our decision rule, we look for the critical values that separate the outer 5% of the z-distribution from the middle 95%. We do this by dividing the level of significance in half ( /2) and placing that amount in each of the tails. We then look for the z-values that go with such a graph. We need to look for the z-value that is associated with an area of 0.0250 (z = 1.96), and the z-value that is associated with an area of 0.9750 (z = 1.96).
0.9500 0.0250 1.96 1 0 1 0.0250 1.96
The shaded regions on the outside are called the regions of rejection. If the value of the test statistic falls into one of these regions, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis. All that remains to be done is to formally state our decision rule.
Decision rule: Reject H0 if z < 1.96 or if z > 1.96.
Two-Tailed Tests A test is two-tailed if our alternate hypothesis involves the symbol . To find our decision rule for a two-tailed test, divide the level of significance and place that amount in the left and right tails. Find the critical values of z for the graph that we have, and reject the null hypothesis if the value of the test statistic is outside of these critical values. We used the symbol z /2 for the critical value in a confidence interval in the previous chapter. Here is why we did that. Suppose we were going to construct a 95% confidence interval for a population mean. The middle 95% leaves 5% of the area under the curve in the two outer tails.
That 5% we call alpha (). If we divided in half, that would put /2 in each tail. Thus, our critical values go along with the critical values that we would obtain by putting /2 in each tail to begin with.
Find the decision rule for a two-tailed test at the 0.01 level of significance.
First we divide by 2, which gives us 0.005. We put 0.005 in the left tail and right tail, which produces the graph shown here.
0.9900 0.005 2 2.575 1 0 1 2 2.575 0.005
We look for the z-values associated with areas of 0.0050 (left tail) and 0.9950 (right tail). From the graph, the decision rule is
Reject H0 if z < 2.575 or if z > 2.575
Right-Tailed Tests Suppose we had the following hypotheses.
H0: 30,000 HA: > 30,000 What type of sample mean would be required to suggest that the null hypothesis was false and that the alternate hypothesis was true? To conclude that > 30,000, we would have to have a sample mean that was significantly higher than 30,000, which is the claimed value of . We find the sample means that are significantly higher than 30,000 in the right tail. The decision rule for a right-tailed test differs from a two-tailed test because the entire level of significance is put in the right tail. It is not divided in half, because there are not two tails to put it in.
Once is placed in the right tail, we find the critical value in the same fashion as we did with a two-tail test.
Find the decision rule for a right-tailed test at the 0.01 level of significance.
We put 0.01 in the right tail, which produces the graph shown here.
0.9900 0.01 2 1 0 1 2 2.33
We look for the z-value associated with an area of 0.9900. From the graph, the decision rule is
Reject H0 if z > 2.33
Left-Tailed Tests Suppose we had the following hypotheses.
H0: 10 HA: < 10 What type of sample mean would be required to suggest that the null hypothesis was false and that the alternate hypothesis was true? To conclude that < 10, we would have to have a sample mean that was significantly lower than 10, which is the claimed value of . We find the sample means that are significantly lower than 10 in the left tail. The decision rule for a left-tailed test differs from the previous tests because the entire level of significance is put in the left tail. It is not divided in half, because there are not two tails to put it in.
Once is placed in the left tail, we find the critical value in the same fashion as before.
Find the decision rule for a left-tailed test at the 0.05 level of significance.
We put 0.05 in the left tail, which produces the graph shown here.
0.9500 0.05 2 1 1.645 0 1 2
Reject H0 if z < 1.645
Step 5: Do the Calculations and Draw Conclusions
The first task in Step 5 is to calculate the test statistic using the sample information. Once that has been accomplished, we check our decision rule to make our decision
about the null hypothesis H0. Then we must draw our final conclusion about the claim being tested. Lets begin with the calculation. Recall our example: A college statistics instructor claims that the mean age of college statistics students is 24. Test his claim at the 0.05 level of significance. A random sample of 116 college statistics students revealed a mean age of 22.7 years with a standard deviation of 5.68 years. Lets calculate the test statistic for our example. There are four values that need to be plugged into this statistic: the sample mean (x ), the claimed population mean (), the sample standard deviation (s), and the sample size (n). Here is what we get when we plug the correct values into the test statistic.
z = = x s/ n 22.7 24 5.68/ 116
If the value of our test statistic is 2.47, what is our decision about the null hypothesis H0? Recall the decision rule:
Reject H0 if z < 1.96 or if z > 1.96
Since 2.47 falls in a rejection region (2.47 < 1.96), our decision is to reject H0.
0.9500 0.0250 1.96 2.47 1 0 1 0.0250 1.96
We still must say something about our claim; after all, we were testing the instructors claim, not the null hypothesis. Our conclusion about the claim depends on two things: whether our claim was the null or alternate hypothesis, and whether the null hypothesis was rejected or not. If the claim is the null hypothesis, then our conclusion is easy to state. In such a case, whatever our decision is about the null hypothesis, our conclusion about the claim is the same. If the claim is the null hypothesis and the null hypothesis is rejected, then the claim is rejected as well. We have proved the claim to be false. If the
claim is the null hypothesis and we fail to reject the null hypothesis, then we fail to reject the claim. In this case we have failed to prove the claim to be false. Please note that this is not the same as proving the claim to be true. When the claim is the null hypothesis, either the claim is proved to be false or we fail to prove that it is false. If the claim is the alternate hypothesis, then we either support it as being true or we fail to support it. If the claim is the alternate hypothesis and the null hypothesis is rejected, then the claim is supported. In other words, if we throw the null hypothesis out as being false, then we accept the alternate hypothesis (the claim) as being true. We have proved the claim to be true. If the claim is the alternate hypothesis and we fail to reject the null hypothesis, then we fail to support the claim. In this case we have failed to prove the claim to be true. The possible conclusions are illustrated in the following chart. You may find it handy to use as a guide as you work through the homework exercises, but you must understand why each conclusion is the way it is. If you cannot understand and be able to draw your own conclusion, then you will never fully understand the process of conducting a hypothesis test.
Reject H0 Claim is H0 Claim is HA Reject the claim Support the claim Fail to Reject H0 Fail to reject the claim Fail to support the claim
In our example, the claim was the null hypothesis. Since our decision was to reject the null hypothesis, then our conclusion is that there is sufficient sample evidence to reject the claim that the mean age of all college statistics students is 24.
One more step that can be added to a hypothesis test is the calculation of a p-value. The p-value for a hypothesis test is the probability of obtaining a sample as extreme as our sample if the null hypothesis is actually true. We will find the probability of obtaining a value as extreme as the test statistic, or more extreme, if the null hypothesis is indeed true. When calculating a p-value, we need to know two things: what tailed test we have and the value of the test statistic.
Two-Tailed P-Values We find a p-value for a two-tailed hypothesis test by putting
the value of the test statistic on a normal distribution graph, finding the area in the tail that it creates, and multiplying by 2. Multiplying by 2 takes into consideration the tail on the opposite side. For our example, we found a test statistic of 2.47. Put it on a normal distribution graph, and find the area in the left tail.
0.0068 2 2.47 1 0 1 2
p-value = 2 P(Z < 2.47) = 2 (0.0068) = 0.0136
This tells us that the probability of obtaining a sample mean as extreme as ours if the null hypothesis is actually true is only 0.0136. This is a much stronger statement than saying that there is less than a 0.05 probability that it is true. The decision rule can be rewritten in terms of the p-value and the level of significance. Recall that we want to reject the null hypothesis only if the probability that it is true is less than the level of significance. This translates to the following:
Reject H0 if p-value <
In our example, since the p-value was less than 0.05, the correct decision is to reject the null hypothesis. What if the level of significance had been 0.01? We would have failed to reject the null hypothesis, because our p-value is greater than 0.01.
Left-Tailed P-Values We find a p-value for a left-tailed hypothesis test by putting
the value of the test statistic on a normal distribution graph, and finding the area to the left of that value. If we had a left-tailed hypothesis test, and the test statistic was calculated to be 1.57, here is how the p-value would be calculated.
p-value = P(Z < 1.57) = 0.0582
0.0582 2 1 0 1 2
Right-Tailed P-Values We find a p-value for a right-tailed hypothesis test by putting the value of the test statistic on a normal distribution graph, and finding the area to the right of that value. If we had a right-tailed hypothesis test, and the test statistic was calculated to be 2.29, here is how the p-value would be calculated.
0.9890 0.0110 2 1 0 1 2 2.29
p-value = P(Z > 2.29) = 1 0.9890 = 0.0110
Our example has been spread out over the entire section. Before moving on, here it is from start to finish, all in one example.
Step 1 Claim in words: The mean age of all college statistics students is 24. Claim: = 24 Complement: 24 H0: = 24 HA: 24 Step 2 Level of significance: = 0.05 Step 3 Test statistic: z = Step 4 Two-tailed test, = 0.05 x s/ n
Decision rule: Reject H0 if z < 1.96 or if z > 1.96. Step 5
= x s/ n 22.7 24 5.68/ 116
Decision: Reject H0. Conclusion: There is sufficient sample evidence to reject the claim that the mean age of all college statistics students is 24. p-value
p-value = 2 = 2
(0.0068) = 0.0136
Now we will look at three more examples in their entirety. Please use them as a reference while you do the homework exercises. Although this process may seem foreign and difficult, it will be used throughout most of the remainder of the book, so you
need to get comfortable with it now. There will be small changes from section to section (different test statistics, different ways of finding the decision rule, . . .) but the general process will remain the same.
Test the claim that the mean price for a dozen roses at a California florist is higher than $45 at the 0.01 level of significance. A random sample of 30 California florists had a mean price of $48.33 per dozen with a standard deviation of $6.07.
Step 1 Claim in words: The mean price for a dozen roses at a California florist is higher than $45. Claim: > 45 Complement: 45 H0: 45 HA: > 45 Note that the claim is the alternate hypothesis, so our conclusion will be either to support the claim or fail to support the claim. Also note that the alternate hypothesis involves the symbol >. This tells us that our test is a right-tailed test. Step 2 Level of significance: = 0.01 Step 3 Test statistic: z = x s/ n
Since we are comparing the mean of a population to a certain number ($45) and the sample size is at least 30, the above test statistic is the appropriate one. Step 4 Right-tailed test, = 0.01
Decision rule: Reject H0 if z > 2.33. Step 5
= x s/ n 48.33 45 6.07/ 30
Decision: Reject H0. Conclusion: There is sufficient sample evidence to support the claim that the mean price for a dozen roses at a California florist is higher than $45. p-value
0.9987 0.0013 0 3.00
p-value = P (Z > 3.00) = 1 0.9987 = 0.0013
Test the claim that the mean amount spent by a customer at Dylans Place, a local fast-food restaurant, is less than $5 at the 0.05 level of significance. A random sample of 34 customers had spent a mean of $4.43 and a standard deviation of $3.23.
Step 1 Claim in words: The mean amount spent by a customer at Dylans Place is less than $5. Claim: < 5 Complement: 5 H0: 5 HA: < 5 Note that the claim is the alternate hypothesis, so our conclusion will be either to support the claim or fail to support the claim. Also note that the alternate hypothesis involves the symbol <. This tells us that our test is a left-tailed test. Step 2 Level of significance: = 0.05 Step 3 Test statistic: z = x s/ n
Since we are comparing the mean of a population to a certain number ($5) and the sample size is at least 30, the above test statistic is the appropriate one. Step 4 Left-tailed test, = 0.05
Decision rule: Reject H0 if z < 1.645. Step 5
z = = x s/ n 4.43 5 3.23/ 34
Decision: Fail to reject H0. Conclusion: There is not sufficient sample evidence to reject the claim that the mean amount spent by a customer at Dylans Place is less than $5. p-value
p-value = P (Z < 1.03) = 0.1515
Test the claim that the mean time Americans spend watching TV per day is 4.4 hours at the 0.05 level of significance. A random sample of 80 people had a mean time of 2.225 hours viewing per day and a standard deviation of 1.088 hours.
Step 1 Claim in words: The mean time spent watching TV by Americans per day is 4.4 hours. Claim: = 4.4 Complement: 4.4 H0: = 4.4 HA: 4.4 Note that the claim is the null hypothesis, so our conclusion will be either to reject the claim or fail to reject the claim. Also note that the alternate hypothesis involves the symbol . This tells us that our test is a two-tailed test. Step 2 Level of significance: = 0.05 Step 3 Test statistic: z = x s/ n
Since we are comparing the mean of a population to a certain number (4.4 hours) and the sample size is at least 30, the above test statistic is the appropriate one. Step 4 Two-tailed test, = 0.05
z = = x s/ n 2.225 4.4 1.088/ 80
Decision: Reject H0. Conclusion: There is sufficient sample evidence to reject the claim that the mean time Americans spend watching TV per day is 4.4 hours. p-value Since 17.88 does not come close to being on our z-table, we will use the biggest value we havenamely, 3.50.
p-value = 2 P (Z < 17.88) < 2 P (Z < 3.50) = 2 (0.0002)
0.0002 3.50 0
The p-value for this test is (much) less than 0.0004.
TWO-TAILED CRITICAL VALUES
Hypothesis Tests for a Single Population Mean
To use Excel to find two-tailed critical values, we first must find /2. If the level of significance () is 0.05, then /2 = 0.025. To find the critical value for the left tail, type the following in any cell.
=NORMSINV(0.025)
This Excel function returns the z-value for which the area to the left of it is equal to 0.025. Round Excels result to two decimal places. To find the critical value for the right tail, type the following in any cell.
=NORMSINV(1-0.025)
We must subtract 0.025 from 1 to obtain the area to the left of the right tail. Again, round the result to two decimal places. For values of /2 other than 0.025, replace 0.025 by the appropriate value. RIGHT-TAILED CRITICAL VALUES To use Excel to find a right-tailed critical value, we need . Suppose the level of significance was 0.01; in other words, = 0.01. To find the critical value for the right tail, type the following in any cell.
=NORMSINV(1-0.01)
Round Excels result to two decimal places. LEFT-TAILED CRITICAL VALUES To use Excel to find a left-tailed critical value, we need . Suppose that = 0.05. To find the critical value for the left tail, type the following in any cell.
Round Excels result to two decimal places. TWO-TAILED p-VALUES If the calculated value of the test statistic is negative, we need to multiply the area to the left of the test statistic by 2. In any cell, we type the following.
=2*NORMSDIST(z-value)
If the test statistic was calculated to be z = 2.47, then we would type
=2*NORMSDIST(-2.47)
Round all p-values to four decimal places. We can also use Excel to calculate two-tailed p-values if the calculated value of the test statistic is positive. In such a case, we type the following in any cell.
=2*(1-NORMSDIST(z-value))
For example, if we calculated the test statistic to be z = 1.23, then we would type the following in any cell.
=2*(1-NORMSDIST(1.23))
We must subtract the result from 1 because Excels NORMDIST function gives us the area to the left of the z-value, but we need the area in the right tail. LEFT-TAILED p-VALUES Since we are looking for the area to the left of a particular z-value, NORMSDIST will work without any adjustments. In any cell, we type the following.
=NORMSDIST(z-value)
Round that answer to four decimal places.
RIGHT-TAILED p-VALUES Because we are looking for the area to the right of a particular z-value, we must subtract the value that NORMSDIST gives us from 1. In any cell, we type the following.
=1-NORMSDIST(z-value)
=1-NORMSDIST(2.29)
Round that answer to four decimal places. Here is an example that uses these tools in performing a hypothesis test.
An algebra instructor has designed a final exam that he believes will produce a mean score of 70. Mr. Thomas, one of his colleagues, disagrees, claiming that the mean score for all algebra students on this exam will be below 70. Mr. Thomas randomly selects 38 algebra students, and gives them the exam. Here are their scores.
Use these sample data to test Mr. Thomas claim at the 0.01 level of significance. We will need the sample statistics (mean and standard deviation).
Open a new Excel worksheet, and type the sample values in column A (cells A1 through A38). In cell B1, type Sample Mean. In cell B2, type Sample Standard Deviation.
In cell B3, type Sample Size. Now that the framework is set up, we can begin the calculations.
To calculate the sample mean, type =AVERAGE(A1:A38) in cell C1. To calculate the sample standard deviation, type =STDEV(A1:A38) in cell C2. Type the sample size (38) in cell C3.
We begin with the hypothesis test by hand, and then will turn to Excel when needed. Step 1 Claim in words: The mean score for all algebra students on this exam is below 70. Claim: < 70 Complement: 70 H0: 70 HA: < 70 Note that the claim is the alternate hypothesis, so our conclusion will be either to support the claim or fail to support the claim. Also note that the alternate hypothesis involves the symbol <. This tells us that our test is a left-tailed test.
In cell B4, type Claimed Mean. Type the claimed value of the mean (70) in
cell C4. Step 2 Level of significance: = 0.01 Step 3 Test statistic: z = x s/ n
Since we are comparing the mean of a population to a certain number (70) and the sample size is at least 30, the above test statistic is the appropriate one. Step 4 Left-tailed test, = 0.01
Type Critical Value in cell B6. We will calculate the critical value for this test
in cell C6. Since this is a left-tailed test, type =NORMSINV(0.01). The value Excel gives us is -2.32634. Our decision rule is then
Reject H0 if z < 2.32634
Here is the critical value using our usual method.
0.01 2 2.33 1 0 1 2
Decision rule: Reject H0 if z < 2.33. Step 5
In cell B7, type Test Statistic. Next we calculate the test statistic in cell C7. Type =(C1-C4)/(C2/SQRT(C3)).
The result Excel gives us is 2.7704. Here is the formula worked out.
z = = x s/ n 62.8 70 16.04/ 38
Decision: Reject H0. Conclusion: There is sufficient sample evidence to support the claim that the mean score for all algebra students on this exam is below 70. p-value
Type p-value in cell B9. We will calculate the p-value for this test in cell C9.
Since we are looking for a left-tailed p-value, type =NORMSDIST(C7). The result that Excel gives us is 0.002799.
p-value = P (Z < 2.77) = 0.5 0.4972 = 0.0028
Here is what your Excel worksheet should look like.
32 Sample Mean 41 Sample Standard Deviation 43 Sample Size 44 Claimed Mean 46 46 Critical Value 47 Test Statistic 48 49 p-value 50 52 54 54 56 57 58 58 60 60 61 65 66 66 67 67 68 68 68 68 71 72 76 83 87 92 92 94 100 62.78947 16.04412 38 70 2.32634 2.7704 0.002799
The TI-83 can help us to perform a hypothesis test, as shown in the following example.
Use these sample data to test Mr. Thomas claim at the 0.01 level of significance.
Put these 38 scores in list L1. We will need sample statistics before continuing on to the hypothesis test. Press list L1, and press
STAT , then use the
key to move to the CALC
2nd 1
menu. Select item 1: 1-Var Stats. On the main screen, press
ENTER . Here is the screen of sample information.
We can use 62.8 for the sample mean, and 16.04 for the sample standard deviation. Since the sample size of 38 is at least 30, we may use the sample standard deviation as an estimate of the population standard deviation. To begin the hypothesis test, Press STAT , then use the key to move to the TESTS menu. Select item 1: Z-Test. Fill in the screen in the following manner.
Inpt: Highlight Stats. We use this option when we type in the standard devi
ation, mean, and sample size manually. We use the Data option to take these statistics from a list. 0: Enter the claimed value of the mean, which is 70. : Enter the value that will be used for the population standard deviation. Again, in this example, we will use the sample standard deviation, which is 16.04. x: Enter the value of the sample mean. You may find that it has already been entered for you. If it has not, use 62.8. n: Enter the sample size. Again, you may find that it has already been entered for you. If it has not, use 38. : Select the form of the alternate hypothesis. The claim for this test is that the mean score for all algebra students on this exam is below 70, so we are testing H0: 70 against HA: < 70. Enter the second option, <0. Highlight Calculate and press ENTER .
You will see the following screen, which lists the alternate hypothesis, calculated value of the test statistic, p-value, sample mean, and sample size.
Explain the following, in your own words. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. How to write the claim in symbols How to write the complement of the claim in symbols How to identify the null hypothesis How to identify the alternate hypothesis The meaning of level of significance The purpose of a test statistic How to determine whether a test is left-tailed, right-tailed, or two-tailed How to formulate the decision rule for a right-tailed test How to formulate the decision rule for a left-tailed test How to formulate the decision rule for a two-tailed test The difference between the decision about the null hypothesis and the conclusion about the claim 12. The meaning of a p-value 13. A skeptical student believes that the mean height of adult males is actually greater than 69 inches. He randomly samples 45 adult males. The sample has a mean height of 70.4 inches with a standard deviation of 2.44 inches. Test the students claim at the 0.05 level of significance. Step 1: (a) State (b) State (c) State (d) State (e) State the the the the the claim in words. claim in symbols. complement of the claim in symbols. null hypothesis (H0). alternate hypothesis (HA).
Step 2: (f) State the level of significance (). Step 3: (g) State the appropriate test statistic. Step 4: (h) What tailed test is this test? (i) State the decision rule.
Step 5: (j) Calculate the test statistic. (k) State your decision about the null hypothesis (H0). (l) State your conclusion about the claim. For the remaining hypothesis tests, use the five-step procedure developed in this section. Refer to Exercise 13 for help. 14. A counselor at a community college claims that the mean GPA for students who have transferred to State University is higher than 3.0. A random sample of 45 such students had a mean GPA of 3.12 with a standard deviation of 0.31. (a) Test the counselors claim at the 0.05 level of significance. (b) Explain how you would gather the data to test this claim, and list any biases that may be present using your method. 15. A student is interested in the mean life span of males in his town. He has heard claims that the mean age at death for males in his town is 70 years. He randomly selects 88 male deaths from the past year. The mean age at death for these 88 men was 68.9 years, with a standard deviation of 16.71 years. Test the claim that the mean age at death for males in his town is 70 years, at the 0.05 level of significance. 16. A student is concerned with the lack of effort among students at her school. She claims that students spend 5 or fewer hours a week studying. A random sample of 100 students produces a mean study time of 5.73 hours per week, with a standard deviation of 3.464 hours. (a) Test the claim that the mean study time per week for these students is 5 or fewer hours at the 0.05 level. (b) Explain how you would gather the data to test this claim, and list any biases that may be present using your method. 17. I recently heard someone say Movies are getting too long! It seems like there are a lot of movies that are over two hours long! I claim that the mean length of movies is less than 2 hours (120 minutes). A random sample of 90 recent releases produced a mean length of 108.44 minutes with a standard deviation of 17.952 minutes. Test my claim at the 0.01 level of significance. 18. At the 0.05 level of significance, test the claim that the mean mileage for cars is less than 20 miles per gallon. A sample of 48 cars had a sample mean of 16.9 miles per gallon, with a standard deviation of 6.83 miles per gallon. 19. A TV commercial claims that the mean cost of a funeral today is above $5000. A survey of 60 funerals showed that they had a mean cost of $5235, with a standard deviation of $1739. (a) Test the commercials claim at the 0.05 level of significance. (b) Explain how you would gather the data to test this claim, and list any biases that may be present using your method. 20. A random sample of 100 women who smoked while pregnant showed that they smoked a mean of 7.4 cigarettes per day with a standard deviation of 4.95 cigarettes. (Based on data contained in Centers for Disease Control and Prevention Monthly Vital Statistics Report.) (a) Test the claim that women who smoke while pregnant smoke a mean of less than 10 cigarettes per day, at the 0.01 level of significance. (b) Explain how you would gather the data to test this claim, and list any biases that may be present using your method. 21. A sample of 212 women who gave birth showed that they gained a mean of 26.2 pounds, with a standard deviation of 10.48 pounds. (Based on data contained in Centers for Disease Control and Prevention Monthly Vital Statistics Report.) (a) Test the claim that the mean weight gain while pregnant is higher than 25 pounds, at the 0.05 level of significance. (b) Explain how you would gather the data to test this claim, and list any biases that may be present using your method. 22. Test the claim that the mean number of prenatal visits for pregnant women is higher than 10 visits at the 0.05 level of significance. A random sample of 441
women who gave birth showed that they had a mean of 11.2 prenatal doctor visits, with a standard deviation of 3.76. (Based on data contained in Centers for Disease Control and Prevention Monthly Vital Statistics Report.) 23. A racing publication reports that the average winning time for a six-furlong race is 1 minute and 10.8 seconds (70.8 seconds). These are the winning times, in seconds, for 63 six-furlong races at Santa Anita Park. (Six furlongs are equal to three-quarters of a mile.) 71.4 71.0 69.2 69.8 70.6 70.6 70.8 68.8 71.0 72.6 70.8 70.2 70.0 72.2 71.6 69.4 69.8 71.0 71.0 69.4 72.2 70.2 71.2 69.6 69.6 70.6 70.8 70.0 70.0 71.0 69.4 71.4 71.2 69.4 71.0 71.0 71.8 71.0 71.6 69.2 70.2 70.4 70.2 70.8 71.0 72.0 68.8 69.8 70.6 72.2 69.0 69.4 69.2 69.2 70.0 71.0 70.4 69.0 72.2 68.6 70.0 70.8 71.6
Test the publications claim at the 0.01 level of significance. 24. A random sample of 120 fillets from farm-raised Atlantic salmon had a mean fat content of 10.5 grams and a standard deviation of 1.12 grams. Use this sample to test the claim that the mean fat content of farm-raised Atlantic salmon fillets is higher than 8.5 grams at the 0.05 level of significance. 25. A random sample of 67 fillets from wild Atlantic salmon had a mean fat content of 8.1 grams and a standard deviation of 1.48 grams. Use this sample to test the claim that the mean fat content of wild Atlantic salmon fillets is lower than 8.5 grams at the 0.05 level of significance. 26. A survey of 65 beer drinkers revealed a mean consumption of 25.2 beers per month, with a standard deviation of 4.96 beers. (Based on the results of a survey by Maritz AmeriPoll.) (a) Test the claim that the mean number of beers consumed per month by beer drinkers is 24 beers (1 case) at the 0.05 level of significance. (b) Explain how you would gather the data to test this claim, and list any biases that may be present using your method. 27. A random sample of 300 incoming flights at an airport showed that travelers had to wait in the baggage claim area for a mean of 18.9 minutes, with a standard deviation of 5.73 minutes. Test the claim that the mean waiting time in the baggage claim area is less than 20 minutes at the 0.01 level of significance. 28. Test the claim that the mean vitamin C content of medium-sized oranges is 66 mg at the 0.05 level of significance. A sample of 36 medium-sized oranges had a mean vitamin C content of 64.5 mg, with a standard deviation of 4.14 mg. 29. A politician claims that the mean number of hours Americans worked per week in a typical week is greater than 40 hours. A sample of 324 American workers produced a mean of 41.3 hours per week with a standard deviation of 10.63 hours per week. Test the politicians claim at the 0.05 level of significance. (Based on a study by the Center for Survey Research and Analysis at University of Connecticut.) 30. A random sample of 74 recently engaged men showed that they had spent a mean of $2982 on the engagement ring, with a standard deviation of $1691. At the 0.05 level of significance, test the claim that the mean amount spent on engagement rings by all men is above $2500. (Based on the results of a survey by Brides magazine.) 31. How many e-mails does a person get at work per day? A random sample of 47 people showed that they receive a mean of 10.7 e-mails at work per day, with a standard deviation of 10.53 e-mails. At the 0.05 level of significance, test the claim that the mean number of e-mails received per day is higher than 5. (Based on a survey by the John J. Heldrich Center for Workforce Development.) 32. Test the claim that the mean cost of a honeymoon is over $3000 at the 0.05 level of significance. A random sample of 180 couples spent a mean of $3657 on their
honeymoons, with a standard deviation of $2109. (Based on a survey by Brides magazine.) A sample of 225 parents were asked how much time they spent per week on school work or school-related activities. The sample produced a mean of 5.6 hours per week, with a standard deviation of 4.4 hours. At the 0.01 level of significance, test the claim that the mean number of hours spent by parents on school work or school-related activities is 5 hours per week. (Based on a study by Yankelovich Partners for LearningPays.com.) One hundred one cruise ship passengers were selected at random, and they had a mean age of 50.1 years old, with a standard deviation of 16.41 years. At the 0.05 level of significance, test the claim that the mean age of a passenger on a cruise ship is below 55 years old. (Based on a study by Polk.) A study of 42 restaurants showed that the mean cost for a sit-down dinner for one person at these restaurants, not including tip, was $27.05 with a standard deviation of $7.20. At the 0.05 level of significance, test the claim that the mean cost for a sit-down dinner for one person at all restaurants is $25. (Based on a study by the National Restaurant Association.) Television networks and movie studios are always interested in the age of the audience they are reaching. A random sample of 104 people at a showing of Toy Story 2 had a mean age of 26.4 years with a standard deviation of 14.82 years. At the 0.05 level of significance, test the claim that the mean age of people going to see Toy Story 2 is less than 30 years. (Based on the results of an informal survey by Disney in six cities.) A television golf analyst claims that the mean drive for professional golfers is 270 yards. Here are the lengths of 63 randomly selected drives from the 100th U.S. Open at Pebble Beach. 305 288 268 280 283 275 262 252 269 283 282 273 270 276 258 259 289 300 290 293 277 279 274 284 287 229 270 268 293 263 261 257 293 262 281 284 262 273 264 270 289 312 296 266 270 261 260 281 289 274 306 254 273 281 266 250 256 286 256 242 261 241 234
At the 0.05 level of significance, test the television golf analysts claim. (Source: On-line scoring for U.S. Open at lycos.com.) What is the appropriate conclusion about the claim under the following conditions, if the level of significance is 0.05? 38. Claim to be tested: Mean is less than 25. p-value: 0.0885 39. Claim to be tested: Mean is greater than 17.1. p-value: 0.0322 40. Claim to be tested: Mean is equal to 129.7. p-value: 0.0039 41. Claim to be tested: Mean is not 25. p-value: 0.0668 42. A colleague tells you that he is trying to verify the claim that households that subscribe to a particular magazine have a mean income that is above $45,000. He is elated that the mean income of 225 randomly selected households that subscribe to the magazine is $49,245. Explain to him why this may not be enough to support his claim, and what other information may be necessary.
The following statistical project contains a major error. Potential errors could be choosing the wrong hypothesis test or not meeting the assumptions required for this test. Find the error, and explain what could be done to correct it.
A student randomly selects 6 elementary schools from his county. He visits these schools and obtains the count of students in 12 different third-grade classes. He uses these data to test the claim that the mean number of students in all third-grade classes in his county is above 20 students, using the hypothesis test introduced in this section. What is wrong with this picture?
I claim that the mean height for adult females is 5 5 (65 inches). You are to test my claim at the 0.05 level of significance. Randomly sample 50 adult females, and find out their height in inches. Use these sample data to test the claim. In addition to your complete hypothesis test, include
Your raw data A histogram for the heights A pie chart showing the percentages that are 5
5 3 and 5 7 , and 5 8 or taller
2 or shorter, between
Explain your technique of sampling. Which type of sampling did you use? Was it truly random? What potential biases may show up in your sample?
The Effect of Sample Size
In this exploration, we will use Microsoft Excel to examine how the sample size affects a hypothesis test. We will be testing the null hypothesis that a population mean is equal to 50 by randomly generating samples from a population whose mean is 48 and whose standard deviation is 15. We know that the mean is not equal to 50 (it is equal to 48), but will our hypothesis-testing procedure be able to reach that conclusion? We will see that one factor that may play a part is the sample size. If we are testing the null hypothesis that = 50 at the 0.05 level of significance, the two-tailed decision rule is to reject the null hypothesis if z < 1.96 or z > 1.96. We will use Excel to generate a sample, and then calculate the test statistic for that sample. Before we begin we must be sure that the Analysis Tool Pak has been added. Pull down the Tools menu. If you see a choice that says Data Analysis, then the Tool Pak is added and you may continue. If you do not see Data Analysis, click on Add-ins under Tools. When the dialog box opens up, click on the box for Analysis Tool Pak, and click on OK. We begin by generating 100 samples of size 30. Pull down the Tools menu and click on Data Analysis. When the dialog box appears, click on Random Number Generation and OK. When the next dialog box appears, enter 100 in the box labeled Number of Variables. (This is the number of samples that we are going to take.) Enter
30 in the box labeled Number of Random Numbers. (This is the sample size.) In the box labeled Distribution click on Normal. For Parameters, enter 48 in the box labeled Mean, and 15 in the box labeled Standard Deviation. Now we will calculate the test statistic. In cell A31, type =AVERAGE(A1:A30). This is the mean of the sample. To calculate the standard deviation of the sample, type =STDEV(A1:A30) in cell A32. Finally, to calculate the test statistic, type the following in cell A33.
=(A31-50)/(A32/SQRT(30))
In the above formula, A31 represents the sample mean, 50 is the claimed population mean, A32 is the sample standard deviation, and 30 is the sample size. Highlight cells A31 through A33, then select Copy from the Edit menu. Highlight the cells in rows 31, 32, and 33 through column CV, then select Paste from the Edit menu. This will calculate the test statistic for each of the 100 samples. In a trial run, using 100 samples, the null hypothesis was not rejected a single time. The results of your 100 samples will most likely be the same. Repeat this procedure for samples of size 50, 100, 250, and 1000. Summarize your results, stating how the increase in sample size affects the test statistic.
Hypothesis Test for a Single Population Mean (Small Samples)
Recall that the sampling distribution of sample means follows the t-distribution for samples of size less than 30. When we are trying to test a claim about a population mean using a small sample (n < 30) and the population standard deviation is unknown, it is appropriate to use the t-distribution as long as the sample comes from a population that is normally distributed. With small samples, outliers suggest that the population is skewed and not normally distributed. There is a hypothesis test in Section 9.3 to test whether a set of data follows a normal distribution, but until then you can use the material presented in ExtraIs a Set of Data Approximately Normally Distributed? which follows Section 5.3. If it appears that the population is not normally distributed, we may instead use a hypothesis test from Section 11.1. Here is the test statistic for the small-sample test for a single population mean.
t = x s/ n degrees of freedom = n 1
The test statistic is identical to the one that we use for a large sample test when the population standard deviation is unknown, but the decision rule will be different. Before we take a look at a few hypothesis tests, we will review how to obtain the decision rule.
Find the decision rule for a right-tailed test at the 0.01 level of significance with 22 degrees of freedom.
Since this is a one-tailed test, we look for 0.01 as a one-tailed value. When we locate that column (the second column from the left), we go down until we reach 22 degrees of freedom.
7.2 Hypothesis Test for a Single Population Mean (Small Samples)
One-Tailed Value Degrees of Freedom 22 0.005 2.819 0.01 2.508 0.025 2.074 0.05 1.717 0.1 1.321 0.25 0.686
The critical value is 2.508. This value is positive, because we are dealing with the right tail. Now for the decision rule. For a right-tailed test, we want to reject the null hypothesis if the value of the test statistic is greater than the critical value.
Decision rule: Reject H0 if t > 2.508
Find the decision rule for a left-tailed test at the 0.05 level of significance with 9 degrees of freedom.
Since this is a one-tailed test, we look for 0.05 as a one-tailed value. When we locate that column (the fourth column from the left), we go down until we reach 9 degrees of freedom.
One-Tailed Value Degrees of Freedom 9 0.005 3.250 0.01 2.821 0.025 2.262 0.05 1.833 0.1 1.383 0.25 0.703
The value in the chart is 1.833, so the critical value is 1.833. This value is negative, because we are dealing with the left tail. For a left-tailed test, we want to reject the null hypothesis if the value of the test statistic is less than the critical value.
Decision rule: Reject H0 if t < 1.833
Find the decision rule for a two-tailed test at the 0.05 level of significance with 19 degrees of freedom.
Since this is a two-tailed test, we look for 0.05 as a two-tailed value. When we locate that column (the third column from the left), we go down until we reach 19 degrees of freedom.
The value in the table is 2.093, so the critical values are 2.093 and 2.093. We use both the positive and negative values because we are dealing with both the right and left tails. For a two-tailed test, we want to reject the null hypothesis if the value of the test statistic is greater than the positive critical value or less than the negative critical value.
Decision rule: Reject H0 if t < 2.093 or if t > 2.093
Now we will perform some actual tests, using the same five-step procedure as in the last section. The major difference is the decision rule.
Test the claim that the mean number of books read per month by community college students is less than 2 at the 0.05 level of significance. A random sample of 28 community college students had read a mean of 1.2 books with a standard deviation of 2.14 books.
Step 1 Claim in words: The mean number of books read per month by community college students is less than 2. Claim: < 2 Complement: 2 H0: 2 HA: < 2 Since the alternate hypothesis involves <, we have a left-tailed test. Also, since our claim is the alternate hypothesis, we will either support or fail to support our claim. Step 2 Level of significance: = 0.05. Step 3 Test statistic: t = x s/ n degrees of freedom = n 1
This is the appropriate test statistic to use when testing a claim about a single population mean based on a small sample (n < 30) when the population standard deviation is unknown. Step 4 This is a left-tailed test, = 0.05, with 27 degrees of freedom.
Decision rule: Reject H0 if t < 1.703. Step 5
t = = x s/ n 1.2 2 2.12/ 28
Decision: Reject H0. Conclusion: There is sufficient sample evidence to support the claim that the mean number of books read per month by community college students is less than 2. p-value Finding p-values for t tests are tricky when using the table. We need to go to the chart, and look at the entire row for 27 degrees of freedom.
Our calculated value for the test statistic (1.997) falls between 1.703 ( = 0.05) and 2.052 ( = 0.025). Our p-value is between 0.025 and 0.05. Using technology, we will be able to obtain a more precise p-value later.
A financial expert claims that the mean charitable contribution on federal tax returns is $1000. Here are the contributions of eight randomly selected individuals.
At the 0.01 level of significance, test the experts claim. (Data based on the results of a survey by Gallup for Independent Sector.) First we need to calculate the mean and standard deviation for this sample. The mean is $1048.00, and the standard deviation is $723.29. Step 1 Claim in words: The mean charitable contribution on federal tax returns is $1000. Claim: = 1000 Complement: 1000 H0: = 1000 HA: 1000 Since the alternate hypothesis involves , we have a two-tailed test. Also, since our claim is the null hypothesis, we will either reject or fail to reject our claim. Step 2 Level of significance: = 0.01 Step 3 Test statistic: t = Step 4 This is a two-tailed test, = 0.01, with 7 degrees of freedom.
x s/ n
degrees of freedom = n 1
Decision rule: Reject H0 if t < 3.499 or if t > 3.499. Step 5
t = = x s/ n 1048 1000 723.29/ 8
Decision: Fail to reject H0. Conclusion: There is not sufficient sample evidence to reject the claim that the mean charitable contribution on federal tax returns is $1000.
p-value We go to the chart, and look at the entire row for 7 degrees of freedom.
Our calculated value for the test statistic (0.188) falls below 0.711 ( = 0.5). Our p-value is greater than 0.5.
Test Marks claim that his average drive is greater than 250 yards at the 0.05 level. Again, we begin by calculating the mean and standard deviation for this sample. The mean is 257.5 yards and the standard deviation is 11.67 yards. Step 1 Claim in words: The mean length for all of Marks drives is greater than 250 yards. Claim: > 250 Complement: 250 H0: 250 HA: > 250 Since the alternate hypothesis involves >, we have a right-tailed test. Also, since our claim is the alternate hypothesis, we will either support or fail to support our claim. Step 2 Level of significance: = 0.05 Step 3 Test statistic: t = Step 4 This is a right-tailed test, = 0.05, with 13 degrees of freedom.
Decision rule: Reject H0 if t > 1.771. Step 5
t = = x s/ n 257.5 250 11.67/ 14
Decision: Reject H0. Conclusion: There is sufficient sample evidence to support the claim that the mean length for all of Marks drives is greater than 250 yards. p-value We go to the chart, and look at the entire row for 13 degrees of freedom.
Our calculated value for the test statistic (2.405) falls between 2.160 ( = 0.025) and 2.650 ( = 0.01). Our p-value is therefore between 0.01 and 0.025.
Hypothesis Tests for a Single Population Mean (Small Samples)
RIGHT-TAILED CRITICAL VALUES (t-DISTRIBUTION) As with the standard normal distribution, Excel can help us find critical values for the t-distribution. With a one-tailed test, we must multiply the level of significance by 2 to use Excels built-in function. This function returns critical values for two-tailed tests, for which the level of significance is twice that of a one-tailed test. Here is the format.
=TINV(2*level of significance, degrees of freedom)
For example, for a right-tailed test at the 0.01 level of significance with 22 degrees of freedom, type the following in any cell.
=TINV(2*0.01,22)
Excel returns a value of 2.508323. The decision rule is
Reject H0 if t > 2.508323
LEFT-TAILED CRITICAL VALUES (t-DISTRIBUTION) The difference that arises with a left-tailed critical value, as compared to a right-tailed critical value, is that Excel gives us a positive value. It is up to us to remember that this value must become negative. We can take care of this by including a negative sign before the functions name. Again, with a one-tailed test, we must multiply the level of significance by 2 to use Excels built-in function. For example, for a left-tailed test at the 0.05 level of significance with 9 degrees of freedom, type the following in any cell.
= -TINV(2*0.05,9)
Excel returns a value of 1.83311. The decision rule is
Reject H0 if t < 1.83311
TWO-TAILED CRITICAL VALUES (t-DISTRIBUTION) With a two-tailed test we only have to supply Excel with the level of significance (not 2 times the level of significance) and the number of degrees of freedom. Here is the format.
=TINV(level of significance, degrees of freedom)
For example, for a two-tailed test at the 0.05 level of significance with 19 degrees of freedom, type the following in any cell.
Reject H0 if t > 2.093025 or if t < 2.093025
RIGHT-TAILED p-VALUES Microsoft Excel can help us with the p-value. Excel will give us a precise p-value instead of a range of values. We need to supply Excel with the value of the test statistic (use positive values only), the degrees of freedom, and the number of tails. If the test statistic is negative, you need to type it as a positive value. (There is no difference in the way you enter a left-tailed test or a right-tailed test.) Here is the format.
=TDIST(test statistic, degrees of freedom, number of tails)
If we had a right-tailed test, with 13 degrees of freedom, and the test statistic calculated to be 2.405, we would find the p-value by typing the following in any cell.
=TDIST(2.405,13,1)
The result that we get is 0.0159, which is our p-value. Using our tables, the p-value is between 0.01 and 0.025, but Excels p-value is more precise and more useful. LEFT-TAILED p-VALUES We follow the same format as we did for a right-tailed p-value. However, we do not type a negative sign in front of our test statistic. If we had a left-tailed test, with 27 degrees of freedom, and the test statistic calculated to be 1.997, we would find the p-value by typing the following in any cell.
The result that we get is 0.0280, which is our p-value. Using our tables, the p-value is between 0.025 and 0.05, but Excels p-value is more precise and more useful. TWO-TAILED p-VALUES The major difference for calculating a two-tailed p-value is that we type a 2 instead of a 1 for the number of tails. If we had a two-tailed test, with 11 degrees of freedom, and the test statistic calculated to be 2.038, we would find the p-value by typing the following in any cell.
The result that we get is 0.0663, which is our p-value. The p-value is between 0.05 and 0.1, but Excels p-value is more precise and more useful. We conclude with an example that uses Excel to perform a hypothesis test.
A math instructor has written a new placement exam. He is concerned that the exam may be too long. The math department will use this new exam if the mean time required to finish the exam for all students who take it is at most 50 minutes. The exam is given to 20 randomly selected students. Here are the times (in minutes) required to complete the exam for these students.
Use these sample data to test the claim that the mean length of time required to finish the exam for all students who take it is at most 50 minutes at the 0.05 level of significance. We will need the sample statistics (mean and standard deviation).
Open a new Excel worksheet, and type the above values in column A (cells A1 through A20). In cell B1, type Sample Mean. In cell B2, type Sample Standard Deviation.
In cell B3, type Sample Size. In cell B4, type Claimed Mean. Now that the framework is set up, we can begin the calculations.
To calculate the sample mean, type =AVERAGE(A1:A20) in cell C1. To calculate the sample standard deviation, type =STDEV(A1:A20) in cell C2. Type the sample size (20) in cell C3. Type the claimed value of the mean (50) in cell C4.
We begin with the hypothesis test by hand, and then will turn to Excel when needed. Step 1 Claim in words: The mean length of time required to finish the exam for all students who take it is at most 50 minutes. Claim: 50 Complement: > 50 H0: 50 HA: > 50 Note that the claim is the null hypothesis, so our conclusion will be either to reject the claim or fail to reject the claim. Also note that the alternate hypothesis involves the symbol >. This tells us that our test is a right-tailed test. Step 2 Level of significance: = 0.05 Step 3 Test statistic: t = x s/ n
This is the appropriate test statistic, since this is a one-mean test with a sample size that is smaller than 30. Step 4 Right-tailed test, = 0.05
Type Critical Value in cell B6. We will calculate the critical value for this test in cell C6. Since this is a lefttailed test, type =TINV(2*0.05,19). (Since this is a one-tailed test, we must multiply the level of significance by 2.) The value Excel gives us is 1.729131. Our decision rule is then
One-Tailed Value Degrees of Freedom 19 0.005 2.861 0.01 2.539 0.025 2.093 0.05 1.729 0.10 1.328 0.25 0.688
In cell B7, type Test Statistic. Next we will calculate the test statistic in cell C7.
Type =(C1-C4)/(C2/SQRT(C3)). The result Excel gives us is 2.985792. Here is the formula worked out.
x s/ n 54 50 5.99/ 20
Decision: Reject H0. Conclusion: There is sufficient sample evidence to reject the claim that the mean length of time required to finish the exam for all students who take it is at most 50 minutes. p-value
Type p-value in cell B9. We will calculate the p-value for this test in cell C9. Since we are looking for
a left-tailed p-value, type =TDIST(C7,19,1). The result that Excel gives us is 0.003798. Using the t-distribution chart, we can see that the p-value is less than 0.005. This is because the test statistic (2.986) is greater than 2.861, which is the highest value in the table for 19 degrees of freedom.
Here is what your Excel worksheet should look like when you are done.
43 44 45 46 49 50 51 53 54 55 57 57 58 59 59 60 60 60 60 60 Sample Mean Sample Standard Deviation Sample Size Claimed Mean Critical Value Test Statistic p-value 54 5.991222 20 50 1.729131 2.985792 0.003798
The TI-83 can help us perform hypothesis tests for a single mean based on a small sample, and find p-values for these tests. RIGHT-TAILED p-VALUES The TI-83 can help us with calculating the p-value. The calculator will give us a precise p-value instead of an interval that it falls in. Suppose we have a right-tailed test, with 13 degrees of freedom, and the test statistic calculated to be 2.405. Press 2ND VARS to access the DISTR menu. Select option 5: tcdf(. We need to enter three values. First, enter the value of the test statistic, which is 2.405. After pressing , , press 1, then 2ND and the , key to access EE, and then finally press 99 followed by a , . Finally, we need to type in the degrees of freedom, 13, followed by ) . Here is what the screen should look like. To calculate the p-value, press ENTER .
The result that we get is 0.0159, which is our p-value. Using our tables, the p-value is between 0.01 and 0.025, but the TI-83s p-value is more precise and more useful. LEFT-TAILED p-VALUES We follow the same format as we did for a right-tailed p-value.
tcdf(-1E99, test statistic, degrees of freedom) Here is what the screen should look like for a left-tailed p-value for which the test statistic is calculated to be 1.997, with 27 degrees of freedom.
The result that we get is 0.0280, which is our p-value. Using our tables, the p-value is between 0.025 and 0.05, but the TI-83s p-value is more precise and more useful. TWO-TAILED p-VALUES To calculate a two-tailed p-value, we calculate the appropriate one-tailed p-value, and then double that value. If the test statistic is positive, double the right-tailed p-value. If the test statistic is negative, double the left-tailed p-value. If we had a two-tailed test, with 11 degrees of freedom, and the test statistic calculated to be 2.038, here is the screen that calculates the p-value.
The result that we get is 0.0663, which is our p-value. The p-value is between 0.05 and 0.1, but the TI-83s p-value is more precise and more useful. We conclude this section with an example that uses the TI-83 to perform a hypothesis test.
Use these sample data to test the claim that the mean length of time required to finish the exam for all students who take it is at most 50 minutes at the 0.05 level of significance.
Put these 20 times in list L1. To begin the hypothesis test, press the
STAT , then use
key to move to the TESTS menu. Select item 2: T-Test. Fill in the screen
in the following manner.
Inpt: Highlight Data. We use this option, rather than Stats, when we want to take the sample statistics from a list. 0: Enter the claimed value of the mean, which for this example is 50. List: Enter the list that contains the sample data. In this example, we will use L1. Freq: Be sure that this says 1. : Select the form of the alternate hypothesis. The claim for this test is that
This becomes a test of H0: 50 versus HA: > 50.
the mean length of time required to finish the exam for all students who take it is at most 50 minutes. Select option 3: >0. Highlight Calculate and press
You will see the following screen, which lists the alternate hypothesis, calculated value of the test statistic, p-value, sample mean, sample standard deviation, and sample size.
1. A random sample of 25 households without health insurance had a mean annual income of $37,850 with a standard deviation of $25,554. At the 0.05 level of significance, test the claim that the mean annual income of households without insurance is less than $50,000. (Based on results of a study by the Census Bureau.) 2. A mortgage broker believes that the typical family that rents their home owns fewer than 2 cars on average. A random sample of 26 families that rent their home owned a mean of 1.29 cars with a standard deviation of 0.97 cars. At the 0.05 level of significance, test the claim that the mean number of cars owned by families that rent their home is less than 2. (Based on the results of the Census Bureau American Housing Brief.) 3. In testing the effect of a new zinc spray on the common cold, 27 people were randomly selected and given the spray to take home. At the first sign of a cold they were to begin using the spray. Their first colds lasted a mean of 1.5 days, with a standard deviation of 2.1 days. At the 0.05 level of significance, test the claim that the mean length of a cold being treated by this spray is equal to 2 days. (Based on the results of a study funded by Gel-Tech.) 4. Fifteen households were randomly selected and asked how much they spend per week on groceries. The mean grocery bill per week was $73.25 with a standard deviation of $29.30. At the 0.05 level of significance, test the claim that the mean grocery bill per week for all households is less than $100 per week. 5. A health researcher is interested in the age that males begin smoking. She claims that the mean age is less than 19 years old. A random sample of 17 male smokers had a mean starting age of 18.2 years with a standard deviation of 2.60 years. (a) At the 0.05 level of significance, test the researchers claim.
(b) Explain how you would gather the data to test this claim, and list any biases that may be present using your method. 6. A statistics instructor, when asked how students do on the first statistics exam, claims that the mean score for all students on the first exam is above 75. To test this claim, 22 students are selected at random. Their scores from the first exam had a mean of 78.2 with a standard deviation of 13.95. Test the instructors claim at the 0.05 level of significance. 7. How many pairs of shoes do female college students own? A random sample of 15 female college students produced a sample mean of 8.7 pairs of shoes, with a standard deviation of 0.85 pairs. Use these data to test the claim that the mean number of pairs of shoes owned by female college students is less than 10 at the 0.05 level of significance. 8. An experimental algebra exam is given to 20 randomly selected college algebra students. Here are their scores. 68 94 50 60 92 14 49 68 41 67 76 71 32 15 54 54 65 92 58 68 At the 0.05 level of significance, test the claim that the mean score for all college algebra students on this exam is below 60. A random sample of 24 high school students were found to be working a mean of 11.6 hours per week with a standard deviation of 11.54 hours. At the 0.05 level of significance, test the claim that the mean number of hours worked per week by high school students is less than 15 hours per week. (Based on results in the National Assessment of Educational Progress survey for the National Center for Education Statistics.) How often do teen mothers read to their children per week? A random sample of 22 teenage mothers read a mean of 2.7 days per week with a standard deviation of 1.64 days per week. At the 0.05 level of significance, test the claim that teenage mothers read to their children more than twice a week. Here are the heights of 11 randomly selected professional womens basketball players. 78 69 71 69 75 71 77 71 77 73 68 At the 0.05 level of significance, test the claim that the mean height of professional womens basketball players is more than 6 feet. Here are the weights of 11 randomly selected pr