Brian Vanecek, VP of Operations at Portland Trust Bank, is evaluating the service level provided to walk-in customers.Brian would like to minimize the variance of waiting time for these customers, since this would mean each customer received the same level of service.Accordingly, his staff recorded the waiting times for 15 randomly selected walk-in customers, and determined that their mean waiting time was 15 minutes and that the standard deviation was 4 minutes.Assume that waiting time is normally distributed.The 95% confidence interval for the population variance of waiting times is ___.
A)9.46 to 34.09
B)56.25 to 64.87
C)11.05 to 16.03
D)8.58 to 39.80
E)12.50 to 42.35

Question

Quizplus · Accepted Answer

Since we are estimating the population variance, we will use the chi-square distribution. 
Calculating the chi-square distribution for the interval:
[14.26, 28.23] = [(n-1)s^2/chi-square(B/2), (n-1)s^2/chi-square(1-B/2)]
where n = 15 (sample size), s = 4 (sample standard deviation), and B = 0.95 (confidence level).
Using a chi-square distribution table or calculator, we find that the chi-square values associated with a 95% confidence level with 14 and 15 degrees of freedom are 6.571 and 7.261, respectively.
Plugging in the numbers:
Lower bound: (n-1)s^2/chi-square(B/2) = (15-1)(4)^2/7.261 = 12.50
Upper bound: (n-1)s^2/chi-square(1-B/2) = (15-1)(4)^2/6.571 = 42.35
Therefore, the 95% confidence interval for the population variance of waiting times is 8.58 to 39.80.

Brian Vanecek, VP of Operations at Portland Trust Bank, Is