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 decision should she make? A) Reject H₀. B) Accept H₀. C) Fail to reject H₀. D) We cannot tell what her decision should be from the information given.

Question

Quizplus · Accepted Answer

We need to conduct a one-sample t-test to test the null hypothesis H₀: μ = 30 against the alternative hypothesis H1: μ > 30 (one-tailed test, since we want to know if the mean age is over 30). The test statistic is given by: t = (x̄ - μ) / (s / sqrt(n)) = (30.45 - 30) / (5 / sqrt(250)) ≈ 1.91 (using a calculator or software). We have n - 1 = 249 degrees of freedom (df). The p-value for t = 1.91 with df = 249 is about 0.028 (using a t-distribution table or software). Since this is a right-tailed test, we reject the null hypothesis if t > tα,df, where α = 0.01 (from the 99% confidence level) and tα,df is the critical t-value with df degrees of freedom and area α to the right of it (which can be found using a t-distribution table or software). For df = 249 and α = 0.01, we get tα,df ≈ 2.33. Since t = 1.91 < tα,df = 2.33, we fail to reject the null hypothesis. Therefore, we do not have enough evidence to conclude that the mean age of the nightclub customers is over 30 at the 99% confidence level. Hence, the owner should not alter the entertainment to appeal to an older crowd based on this sample.

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