A publisher wants to estimate the mean length of time (in minutes) all adults sp
ID: 3050256 • Letter: A
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 is 1.9
minutes and that the population of times is normally distributed.
11
6
8
12
8
12
9
7
9
6
7
10
10
9
11
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 ____, ____
11
6
8
12
8
12
9
7
9
6
7
10
10
9
11
Explanation / Answer
Ans:
From above data:
sample mean=135/15=9
population standard deviation=1.9
z value for 90% confidence level is 1.645.
90% confidence interval for population mean
=9+/-1.645*1.9/sqrt(15)
=9+/-0.807
=(8.193,9.807)
z value for 99% confidence interval is 2.58.
99% confidence interval for population mean
=9+/-2.58*1.9/sqrt(15)
=9+/-1.266
=(7.734,10.266)
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