Question

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 σ is 2.1 minutes and that the population of times is normally distributed. 6 9 6 11 6 7 10 12 8 6 8 9 10 10 9

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

Answer 1

Solution:- Given that 6 9 6 11 6 7 10 12 8 6 8 9 10 10 9

X = 8.4667 , σ = 2.1 , n = 15

90% confidence interval for Z = 1.645

99% confidence interval for Z = 2.576

90% confidence interval for the population mean = X +/- Z*s/sqrt(n)
= 8.4667 +/- 1.645*2.1/sqrt(15)
= 7.5748 , 9.3586
= 7.6 , 9.4 (rounded)

99% confidence interval for the population mean = X +/- Z*s/sqrt(n)
= 8.4667 +/- 2.576*2.1/sqrt(15)
= 7.0699 , 9.8634
= 7.1 , 9.9

=> 99% confidence interval is the wider