Question

g Brian Vanecek, VP of Operations at Portland Trust Bank, is evaluating the service level provided to walk-in customers. Accordingly, his staff recorded the waiting times for 64 randomly selected walk-in customers, and determined that their mean waiting time was 15 minutes and that the standard deviation was 4 minutes. The 88% confidence interval for the population mean of waiting times is __________.

Answer 1

The 88% confidence interval for the population mean of waiting times is between 7.34 minutes and 22.66 minutes.

Explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 64 - 1 = 63

88% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 63 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.88)/(2) = 0.94`. So we have T = 1.9153

The margin of error is:

M = T*s = 1.9153*4 = 7.66.

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 15 - 7.66 = 7.34 minutes

The upper end of the interval is the sample mean added to M. So it is 15 + 7.66 = 22.66 minutes.

The 88% confidence interval for the population mean of waiting times is between 7.34 minutes and 22.66 minutes.