text: Vous êtes sur la page 1sur 5
  
  # Hypothesis Test for a Mean
  
  This lesson explains how to conduct a hypothesis test of a mean, when the following conditions
  are met:
  
  ## The sampling method is simple random sampling.
  
  The sample is drawn from a normal or near-normal population.
  
  Generally, the sampling distribution will be approximately normally distributed if any of the
  following conditions apply.
  
  ## The population distribution is normal.
  
  The sampling distribution is symmetric, unimodal, without outliers, and the sample size
  is 15 or less.
  The sampling distribution is moderately skewed, unimodal, without outliers, and the
  sample size is between 16 and 40.
  The sample size is greater than 40, without outliers.
  
  This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3)
  analyze sample data, and (4) interpret results.
  
  ## (1) State the Hypotheses
  
  Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis.
  The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true,
  the other must be false; and vice versa.
  
  The table below shows three sets of hypotheses. Each makes a statement about how the
  population mean is related to a specified value M. (In the table, the symbol means " not equal
  to ".)
  
  ## Set Null hypothesis Alternative hypothesis Number of tails
  
  1 =M M 2
  2 >M <M 1
  3 <M >M 1
  
  The first set of hypotheses (Set 1) is an example of a two-tailed test, since an extreme value on
  either side of the sampling distribution would cause a researcher to reject the null hypothesis.
  The other two sets of hypotheses (Sets 2 and 3) are one-tailed tests, since an extreme value on
  only one side of the sampling distribution would cause a researcher to reject the null hypothesis.
  
  ## (2) Formulate an Analysis Plan
  
  The analysis plan describes how to use sample data to accept or reject the null hypothesis. It
  should specify the following elements.
  Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or
  0.10; but any value between 0 and 1 can be used.
  
  Test method. Use the one-sample t-test to determine whether the hypothesized mean
  differs significantly from the observed sample mean.
  
  ## (3) Analyze Sample Data
  
  Using sample data, conduct a one-sample t-test. This involves finding the standard error, degrees
  of freedom, test statistic, and the P-value associated with the test statistic.
  
  Standard error. Compute the standard error (SE) of the sampling distribution.
  
  ## SE = s * sqrt{ ( 1/n ) * ( 1 - n/N ) * [ N / ( N - 1 ) ] }
  
  Where s is the standard deviation of the sample, N is the population size, and n is the
  sample size. When the population size is much larger (at least 10 times larger) than the
  sample size, the standard error can be approximated by:
  
  SE = s / sqrt( n )
  
  Degrees of freedom. The degrees of freedom (DF) is equal to the sample size (n) minus
  one. Thus, DF = n - 1.
  
  Test statistic. The test statistic is a t-score (t) defined by the following equation.
  
  t = (x - ) / SE
  
  Where x is the sample mean, is the hypothesized population mean in the null
  hypothesis, and SE is the standard error.
  
  P-value. The P-value is the probability of observing a sample statistic as extreme as the
  test statistic. Since the test statistic is a t-score, use the t Distribution Calculator to assess
  the probability associated with the t-score, given the degrees of freedom computed above.
  (See sample problems at the end of this lesson for examples of how this is done.)
  
  ## (4) Interpret Results
  
  If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null
  hypothesis. Typically, this involves comparing the P-value to the significance level, and rejecting
  the null hypothesis, when the P-value is less than the significance level.
  
  ## Problem A: Two-Tailed Test
  
  An inventor has developed a new, energy-efficient lawn mower engine. [Null Hypothesis]: He
  claims that the engine will run continuously for 5 hours (300 minutes) on a single gallon of
  regular gasoline. Suppose a simple random sample of 50 engines is tested. The engines run for an
  average of 295 minutes, with a standard deviation of 20 minutes. Test the null hypothesis that the
  mean run time is 300 minutes against the alternative hypothesis that the mean run time is not 300
  minutes. Use a 0.05 level of significance. (Assume that run times for the population of engines
  are normally distributed.)
  
  Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an
  analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps
  below:
  
  1) State the hypotheses. The first step is to state the null hypothesis and an alternative
  hypothesis.
  
  ## Null hypothesis: = 300
  
  Alternative hypothesis: 300
  
  Note that these hypotheses constitute a two-tailed test. The null hypothesis will be
  rejected if the sample mean is too big or if it is too small.
  
  2) Formulate an analysis plan. For this analysis, the significance level is 0.05. The test
  method is a one-sample t-test.
  3) Analyze sample data. Using sample data, we compute the standard error (SE),
  degrees of freedom (DF), and the t-score test statistic (t).
  
  ## SE = s / sqrt(n) = 20 / sqrt(50) = 20/7.07 = 2.83
  
  DF = n - 1 = 50 - 1 = 49
  t = (x - ) / SE = (295 - 300)/2.83 = 1.77
  
  Where s is the standard deviation of the sample, x is the sample mean, is the
  hypothesized population mean, and n is the sample size.
  
  Since we have a two-tailed test, the P-value is the probability that the t-score having 49
  degrees of freedom is less than -1.77 or greater than 1.77.
  
  We use the t Distribution Calculator to find P(t < -1.77) = 0.04, and P(t > 1.75) = 0.04.
  Thus, the P-value = 0.04 + 0.04 = 0.08.
  
  Interpret results. Since the P-value (0.08) is greater than the significance level (0.05),
  we cannot reject the null hypothesis.
  
  Note: Specifically, the approach is appropriate because the sampling method was simple random
  sampling, and the population was normally distributed.
  
  ## Problem B: One-Tailed Test
  
  Bon Air Elementary School has 300 students. [Null Hypothesis]: The principal of the school
  thinks that the average IQ of students at Bon Air is at least 110. To prove her point, she
  administers an IQ test to 20 randomly selected students. Among the sampled students, the
  average IQ is 108 with a standard deviation of 10. Based on these results, should the principal
  accept or reject her original hypothesis? Assume a significance level of 0.01.
  
  Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an
  analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps
  below:
  
  1) State the hypotheses. The first step is to state the null hypothesis and an alternative
  hypothesis.
  
  ## Null hypothesis: >= 110
  
  Alternative hypothesis: < 110
  
  Note that these hypotheses constitute a one-tailed test. The null hypothesis will be
  rejected if the sample mean is too small.
  
  2) Formulate an analysis plan. For this analysis, the significance level is 0.01. The test
  method is a one-sample t-test.
  3) Analyze sample data. Using sample data, we compute the standard error (SE),
  degrees of freedom (DF), and the t-score test statistic (t).
  
  ## SE = s / sqrt(n) = 10 / sqrt(20) = 10/4.472 = 2.236
  
  DF = n - 1 = 20 - 1 = 19
  t = (x - ) / SE = (108 - 110)/2.236 = -0.894
  
  Where s is the standard deviation of the sample, x is the sample mean, is the
  hypothesized population mean, and n is the sample size.
  
  Since we have a one-tailed test, the P-value is the probability that the t-score having 19
  degrees of freedom is less than -0.894.
  
  We use the t Distribution Calculator to find P(t < -0.894) = 0.19. Thus, the P-value is
  0.19.
  
  4) Interpret results. Since the P-value (0.19) is greater than the significance level (0.01),
  we cannot reject the null hypothesis.
  
  Note: Specifically, the approach is appropriate because the sampling method was simple random
  sampling, and the population was normally distributed.
  
  References
  Website:
  
  http://www.stattrek.com/hypothesis-test/mean.aspx
metadata: {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}