text: # Question #7f13f
  
  Apr 23, 2017
  
  The population mean may be 66.
  
  #### Explanation:
  
  Our initial hypothesis is
  ${H}_{0} : \mu = 66$
  Since we do not claim to have knowledge either way, our alternative hypothesis is
  ${H}_{1} : \mu \ne 66$
  $n = 10$ is the sample size.
  
  Since we do not know the population standard deviation, although the population is normal, we must use a t-test with 9 = 10-1 degrees of freedom.
  The sample mean is $\overline{x} = 67.8$ .
  The sample standard deviation is $s \approx 3.011$ .
  Our test statistic is $t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$.
  In this case t is approximately $1.8904$ .
  
  If we consult a t-table, we see that the value of 67.8 deviates enough from 66 for us to reject the null hypothesis if our tolerance is $\alpha = 0.1$, but that is a relatively weak significance level. For the more commonly-used tolerance specification of $\alpha = 0.05$, we would be unable to reject the null hypothesis and would have to conclude that $\mu$ could be equal to 66.
  
  If you are doing this work on a TI-83/84 calculator, the T-Test will reveal that the area of the rejection region is approximately $P \approx 0.0913$.
explanation: The document contains a detailed explanation of a hypothesis test to determine if the population mean is 66. It includes the null and alternative hypotheses, sample statistics, the test statistic, and a conclusion based on different significance levels. The document also mentions how to use a t-table and a TI-83/84 calculator to find the p-value.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Question #7f13f
  
  Apr 23, 2017
  
  The population mean may be 66.
  
  #### Explanation:
  
  Our initial hypothesis is
  ${H}_{0} : \mu = 66$
  Since we do not claim to have knowledge either way, our alternative hypothesis is
  ${H}_{1} : \mu \ne 66$
  $n = 10$ is the sample size.
  
  Since we do not know the population standard deviation, although the population is normal, we must use a t-test with 9 = 10-1 degrees of freedom.
  The sample mean is $\overline{x} = 67.8$ .
  The sample standard deviation is $s \approx 3.011$ .
  Our test statistic is $t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$.
  In this case t is approximately $1.8904$ .
  
  If we consult a t-table, we see that the value of 67.8 deviates enough from 66 for us to reject the null hypothesis if our tolerance is $\alpha = 0.1$, but that is a relatively weak significance level. For the more commonly-used tolerance specification of $\alpha = 0.05$, we would be unable to reject the null hypothesis and would have to conclude that $\mu$ could be equal to 66.
  
  If you are doing this work on a TI-83/84 calculator, the T-Test will reveal that the area of the rejection region is approximately $P \approx 0.0913$.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a detailed explanation of a hypothesis test to determine if the population mean is 66. It includes the null and alternative hypotheses, sample statistics, the test statistic, and a conclusion based on different significance levels. The document also mentions how to use a t-table and a TI-83/84 calculator to find the p-value.
  
  Label: question-and-answer