Exhibit 11-2.The manager of a video library would like the variance of the waiting times of the customers not to exceed 2.30 minutes-squared.He would like to add an additional billing counter if the variance exceeds the cut-off.He checks the recent sample data.For a random sample of 24 customer waiting times,he arrives at a sample variance of 3.8 minutes-squared.The manager assumes the waiting times to be normally distributed. Refer to Exhibit 11-2.Approximate the p-value.
A)p-value lies between 0.005 and 0.010
B)p-value lies between 0.010 and 0.025
C)p-value lies between 0.050 and 0.10
D)p-value lies between 0.025 and 0.05

Question

Quizplus · Accepted Answer

To approximate the p-value, we need to calculate the test statistic, which is given by:

t = (n-1) * s^2 / σ^2

where n is the sample size, s^2 is the sample variance, and σ^2 is the hypothesized population variance.

In this case, n = 24, s^2 = 3.8, and σ^2 = 2.30.

So, t = (24 - 1) * 3.8 / 2.30 = 9.04

Next, we need to find the p-value associated with this test statistic. Since the manager is testing for a variance that exceeds the cut-off, this is a one-tailed test with the alternative hypothesis being σ^2 > 2.30.

Using a t-distribution table with 23 degrees of freedom (df = n-1), we find that the p-value associated with a test statistic of 9.04 is between 0.025 and 0.05.

Therefore, the best choice is D.