Final answer:
The hypothesis test involves assessing if the exercise program helps improve running times. The test statistic is approximately -3.24 with 29 degrees of freedom. If the p-value is below 0.05, we reject the null hypothesis, thus concluding that the program is effective.
Step-by-step explanation:
Hypothesis Testing in Statistics:
The question involves conducting a hypothesis test to assess the effectiveness of a particular exercise program in helping people run a mile faster. The null hypothesis (H0) for this study would be that the mean time to run a mile before and after the exercise program is the same, i.e., μ1 = μ2. The alternative hypothesis (Ha) would be that the mean time to run a mile after the program is less than before, i.e., μ1 > μ2. A matched-pairs t-test is suitable here since the same subjects are being compared before and after the intervention.
Given the sample mean difference (x = -5), sample standard deviation of the differences (sa = 6), and the number of pairs (n = 30), we can calculate the test statistic t using the formula:
t = (x - 0) / (sa / √n)
For this case:
t = (-5 - 0) / (6 / √30) ≈ -3.24
The degrees of freedom for this test are n - 1, which is 29. When the significance level (alpha) is 0.05 and we observe a p-value less than 0.05, we reject the null hypothesis. Therefore, if the calculated p-value is less than 0.05, we can conclude that there is sufficient evidence to suggest the exercise program has helped to improve running speed.