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. Which of the following is the correct approximation of the p-value used to conduct this test?
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 conduct a hypothesis test, the manager can set up the following hypotheses:

H0: σ2 ≤ 2.30 
Ha: σ2 > 2.30

where σ2 is the population variance of the waiting times.

Since the sample size n = 24 is large (n > 20), the manager can use the chi-square distribution to test the hypothesis. The test statistic is given by:

χ2 = (n - 1) * S^2 / σ^2

where S^2 is the sample variance and σ^2 is the hypothesized population variance under the null hypothesis.

Under the null hypothesis, the test statistic follows a chi-square distribution with degrees of freedom df = n - 1 = 23. The p-value is the probability of observing a test statistic as extreme or more extreme than the observed one, assuming the null hypothesis is true.

The observed test statistic is:

χ2obs = (24 - 1) * 3.8 / 2.3 = 39.65

Using a chi-square distribution table with df = 23 and a significance level of α = 0.05 (one-tailed test), the critical value is χ2crit = 35.172.

Since χ2obs > χ2crit, we reject the null hypothesis and conclude that the variance of the waiting times exceeds 2.30 at a significance level of 0.05.

To find the p-value, we can use a chi-square distribution table or a calculator. The p-value is the area to the right of the observed test statistic under the chi-square distribution curve with df = 23.

Using a chi-square distribution table or a calculator, we find that the p-value is between 0.025 and 0.05. Therefore, the correct answer is D.

The Manager of a Video Library Would Like the Variance