Question

Joe's Auto Insurance Company customers sometimes have to wait a long time to speak to a

customer service representative when they call regarding disputed claims. A random sample
of 25 such calls yielded a mean waiting time of 22 minutes with a standard deviation of 6
minutes. Construct a 95% and 99% confidence interval for the population mean of such
waiting times. Explain what these interval means.

Answer 1

The 95% confidence interval for the population mean of such waiting times is between 19.5 and 24.5 minutes. This means that we are 95% sure that the true mean waiting time of all calls for this company is between 19.5 and 24.5 minutes.

The 99% confidence interval for the population mean of such waiting times is between 18.6 and 25.4 minutes. This means that we are 99% sure that the true mean waiting time of all calls for this company is between 18.6 and 25.4 minutes.

Explanation:

We have the standard deviation for the sample, which means that the t-distribution is used 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 = 25 - 1 = 24

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.95)/(2) = 0.975`. So we have T = 2.0639

The margin of error is:

`M = T(s)/(√(n)) = 2.0639(6)/(√(25)) = 2.5`

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 22 - 2.5 = 19.5 minutes

The upper end of the interval is the sample mean added to M. So it is 22 + 2.5 = 24.5 minutes

The 95% confidence interval for the population mean of such waiting times is between 19.5 and 24.5 minutes. This means that we are 95% sure that the true mean waiting time of all calls for this company is between 19.5 and 24.5 minutes.

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.99)/(2) = 0.995`. So we have T = 2.797

The margin of error is:

`M = T(s)/(√(n)) = 2.797(6)/(√(25)) = 3.4`

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 22 - 3.4 = 18.6 minutes

The upper end of the interval is the sample mean added to M. So it is 22 + 3.4 = 25.4 minutes

The 99% confidence interval for the population mean of such waiting times is between 18.6 and 25.4 minutes. This means that we are 99% sure that the true mean waiting time of all calls for this company is between 18.6 and 25.4 minutes.