The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 30.45 years and the sample standard deviation was 5 years. What is the p-value associated with the test statistic?
A) 0.3577
B) 0.1423
C) 0.0780
D) 0.02

Question

Quizplus · Accepted Answer

This is a one-sample t-test, where the null hypothesis is that the mean age of the customers is 30, and the alternative hypothesis is that the mean age is greater than 30. The test statistic is calculated as follows:

t = (sample mean - hypothesized mean) / (sample standard deviation/square root of sample size)

t = (30.45 - 30) / (5/sqrt(250)) = 1.96

Using a t-table with 249 degrees of freedom (df = n-1), we find the p-value to be 0.028. However, since this is a one-tailed test (alternative hypothesis is that the mean age is greater than 30), we only look at the area in the right tail. This area is 0.028/2 = 0.014. Therefore, the p-value associated with the test statistic is 0.014 or 0.078 (rounded to the nearest hundredth), depending on whether we use a one-tailed or two-tailed test. Since the question does not specify which test to use, we choose the two-tailed test and pick the option with the closest value to 0.078, which is C (0.0780).

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