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Chapter 8: Problem 57

Jack'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 \(99 \%\) confidence interval for the population mean of such waiting times. Assume that such waiting times for the population follow a normal distribution.

Short Answer

Expert verified
The 99% confidence interval for the population mean of waiting times is approximately from 18.91 minutes to 25.09 minutes.

Step by step solution

01

Find the z-score

First, you need to find the z-score for the 99% confidence level. Since the population distribution is normal, you can use a z-table or a calculator for finding the z-score associated with a 99% confidence level. The z-score for a 99% confidence level is approximately \(2.576\).
02

Calculate the standard error

The standard error is the standard deviation divided by the square root of the sample size. In this instance, the standard deviation is 6 minutes and the sample size is 25. So, the standard error would be \( \frac{6}{\sqrt{25}} \) which equals to \(1.2\) minutes.
03

Construct the confidence interval

Now, multiply the z-score by the standard error. You obtain \(2.576 \times 1.2 = 3.0912\) minutes. This is the margin of error. To find the confidence interval, add and subtract this margin of error from the sample mean. So, the confidence interval is \( 22 - 3.0912 \) to \( 22 + 3.0912 \), or approximately from \( 18.91 \) minutes to \( 25.09 \) minutes.

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Most popular questions from this chapter

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