To test the belief that sons are taller than their fathers, a student randomly selects 13 fathers who have adult male children. She records the height of both the father and son in inches and obtains the following data. Are sons taller than their fathers? Use the $\alpha=0.10$ level of significance. Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outtiers. Click the icon to view the table of data. Which conditions must be met by the sample for this test? Select all that apply. A. The sampling method results in a dependent sample. B. The sample size is no more than $5 \%$ of the population size. C. The differences are normally distributed or the sample size is large. D. The sampling method results in an independent sample. E. The sample size must be large. Let $d_{i}=X_{1}-Y_{i}$. Write the hypotheses for the test. \[ \begin{array}{l} H_{0}: \mu_{d}=0 \\ H_{1}: \mu_{d}<0 \end{array} \] Calculate the test statistic. $t_{0}=[I]$ (Round to two decimal places as needed.)
Solution
Step 1 ：The conditions that must be met by the sample for this test are that the sampling method results in a dependent sample (A) and the differences are normally distributed or the sample size is large (C).
Step 2 ：The hypotheses for the test are \( H_{0}: \mu_{d}=0 \) and \( H_{1}: \mu_{d}<0 \).
Step 3 ：The test statistic is calculated as \( t_{0} = \frac{\text{mean of differences}}{\text{standard deviation of differences} / \sqrt{\text{sample size}}} \).
Step 4 ：Using the given data, the mean of the differences is 1.0 and the standard deviation of the differences is approximately 2.415.
Step 5 ：The test statistic is \( t_{0} = \frac{1.0}{2.415 / \sqrt{13}} \).
Step 6 ：The calculated test statistic is approximately 1.49.
Step 7 ：The final answer is \(\boxed{1.49}\).
