Question

# A publisher wants to estimate the mean length of time​ (in minutes) all adults spend reading...

A publisher wants to estimate the mean length of time​ (in minutes) all adults spend reading newspapers. To determine this​ estimate, the publisher takes a random sample of 15 people and obtains the results below. From past​ studies, the publisher assumes sigma is 2.1 minutes and that the population of times is normally distributed.

6 10 12 6 6 10 7 7 8 8 11 11 9 10 7

Construct the​ 90% and​ 99% confidence intervals for the population mean. Which interval is​ wider? If​ convenient, use technology to construct the confidence intervals.

The​ 90% confidence interval is ​( ____​, ____ ).​ (Round to one decimal place as​ needed.)

First we need to find the sample mean is: Following is the screen shot of excel for 90% confidence interval: Following is the output generated by excel:

 Confidence interval - mean 90% confidence level 8.53 mean 2.1 std. dev. 15 n 1.645 z 0.892 half-width 9.422 upper confidence limit 7.638 lower confidence limit

The 90% confidence interval is (7.6, 9.4).

---------------------

Following is the output generated by excel for 99% confidence interval:

 Confidence interval - mean 99% confidence level 8.53 mean 2.1 std. dev. 15 n 2.576 z 1.397 half-width 9.927 upper confidence limit 7.133 lower confidence limit

The 99% confidence interval is (7.1, 9.9).