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

sigmaσ

is

2.22.2

minutes and that the population of times is normally distributed.

 99 88 99 1111 66 99 99 77 1212 77 1212 66 99 1212 99

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

​(8.18.1​,9.99.9​).

​ (Round to one decimal place as​ needed.)The​ 99% confidence interval is

​(7.57.5​,10.510.5​).

​ (Round to one decimal place as​ needed.)

Which interval is​ wider?

The​ 90% confidence interval

The​ 99% confidence interval

Solution:- The​ 99% confidence interval is wider than 90% confidence interval.

90% confidence intervals for the population mean is C.I = (8.1, 9.9)

C.I = 9 + 1.645 × 0.56804

C.I = 9 + 0.934

C.I = (8.1, 9.9)

99% confidence intervals for the population mean is C.I = (7.5, 10.5).

C.I = 9 + 2.576 × 0.56804

C.I = 9 + 1.46427

C.I = (7.5, 10.5)

The​ 99% confidence interval is wider than 90% confidence interval.

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