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 39.79
E)12.50 to 42.35

Question

Quizplus · Accepted Answer

To calculate the 95% confidence interval for the population variance when the sample is normally distributed, we use the chi-square distribution. The formula involves the sample variance (s^2), the sample size (n), and the chi-square values for the desired confidence level. The sample variance is the square of the standard deviation, so here s^2 = 4^2 = 16. With n = 15, the degrees of freedom (df) = n - 1 = 14. The chi-square values for 95% confidence and 14 df are approximately 5.63 (lower) and 26.12 (upper). Plugging these into the formula for the confidence interval of variance gives us the interval $[(\frac{(n-1)s^2}{\chi^2_{upper}}), (\frac{(n-1)s^2}{\chi^2_{lower}})]$, which after calculation falls within the range given in option D.

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