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 greater than 30.If so, she plans to alter the entertainment to appeal to an older crowd.If not, no entertainment changes will be made.If she wants to have a level of significance at 0.01, what rejection region should she use?
A)Reject
B)Reject
C)Reject
D)Reject

Question

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 greater than 30.If so, she plans to alter the entertainment to appeal to an older crowd.If not, no entertainment changes will be made.If she wants to have a level of significance at 0.01, what rejection region should she use?
A)Reject 
B)Reject 
C)Reject 
D)Reject

Quizplus · Accepted Answer

Since the level of significance is 0.01, we need to split the alpha value between the two tails of the distribution. This means we need to find the z-value that cuts off the top 0.005 of the distribution (since it's a two-tailed test). Consulting a z-table, we find that the z-value is approximately 2.576. To construct the rejection region, we need to find the values of x̄ that are more than 2.576 standard errors away from the hypothesized mean (30). Using the formula:

z = (x̄ - μ) / (σ / √n)

We can rearrange to solve for x̄: 
x̄ = μ + z(σ / √n)

Plugging in the values we have:

x̄ = 30 + 2.576(σ / √250)

We don't know the value of σ, but we can estimate it with the sample standard deviation s. So we have:

x̄ = 30 + 2.576(s / √250)

Since we want to find the values of x̄ that are more than 2.576 standard errors away from the mean, we need to look at the two tails of the distribution. The rejection region is given by:

x̄ < 30 - 2.576(s / √250)  OR  x̄ > 30 + 2.576(s / √250)

This means that if the sample mean falls outside this range, we can reject the null hypothesis that the mean age is equal to 30, and conclude that it is greater than 30.

Therefore, the correct answer is C. Since the level of significance is 0.01, we need to find the critical values that correspond to the upper 0.01 percentile of the t-distribution with degrees of freedom (df) = 249. Using a t-table or calculator, we find that the critical value is t = 2.33. Since the null hypothesis is that the mean age is not greater than 30, we need to use a two-tailed test with a rejection region in both tails. The rejection region will be any t-values greater than 2.33 or less than -2.33. Therefore, the correct rejection region is (-∞, -2.33) ∪ (2.33, ∞), which corresponds to choice C.

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