Final answer:
The correct rejection region for this hypothesis test is t > 1.761 for a t-distribution with 14 degrees of freedom at the 0.05 significance level, which is an option (a). This is because the sample size is less than 30, making the t-distribution the appropriate choice.
Step-by-step explanation:
To determine the rejection region for testing the hypothesis that the mean fee for car washers (including tips) is greater than $12, we must consider whether the sample size is large enough to justify the use of the z-distribution. Since the sample size in this case is 15, which is smaller than 30, the appropriate distribution to use is the t-distribution. This is because the central limit theorem typically requires a larger sample size to justify the normal approximation, particularly when working with means.
Given that the significance level (α) is 0.05 and the sample size is less than 30, we should use the t-distribution. The degrees of freedom (df) for this test is n - 1, where n is the sample size. Thus, df = 15 - 1 = 14. We are conducting a one-tailed test because we want to test if the mean is greater than $12, not equal to or not equal to a specific value.
Using the t-distribution table or calculator, we find the critical t-value for df = 14 at a significance level of 0.05 for a one-tailed test is 1.761 (t0.05,14). Thus, our rejection region is t > 1.761. Therefore, the correct rejection region is:
Rejection region: t > t0.05,14 = 1.761
This corresponds to option (a). Any t-statistic greater than 1.761 would lead us to reject the null hypothesis that the mean fee is $12 or less, and conclude that it is significantly greater than $12 at the 5% significance level.