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. If she wants to be 99% confident in her decision, what conclusion can she make?
A) There is not sufficient evidence that the mean age of her customers is over 30.
B) There is sufficient evidence that the mean age of her customers is over 30.
C) There is not sufficient evidence that the mean age of her customers is not over 30.
D) There is sufficient evidence that the mean age of her customers is not over 30.

Question

Quizplus · Accepted Answer

We can use a one-sample t-test with the null hypothesis being that the mean age is 30 and the alternative hypothesis being that the mean age is greater than 30. Since we want to be 99% confident in our decision, we use a significance level of 0.01 (or alpha = 0.01).

Using a t-test calculator or a t-table, we find the t-statistic to be:

t = (30.45 - 30) / (5 / sqrt(250)) = 2.756

The degrees of freedom for this test is 249 (n-1).

The critical t-value for a one-tailed test with alpha = 0.01 and 249 degrees of freedom is 2.653.

Since our calculated t-value (2.756) is greater than the critical t-value (2.653), we would reject the null hypothesis and conclude that there is sufficient evidence that the mean age of her customers is over 30. However, we should note that we cannot say with 100% certainty that the mean age is over 30, only that we are 99% confident in our conclusion.

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