Question 1120471: 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.
7
12
7
12
11
9
12
10
8
10
6
8
7
9
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.)
The 99% confidence interval is?
(Round to one decimal place as needed.)
Which interval is wider?
The 90% confidence interval?
The 99% confidence interval?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The mean is 9
Assuming the sd is known is consistent with using a z-test
the 90% interval will be mean+/-1.645, the z value for 90%*2.1/sqrt(15). The 99% interval will be using 2.576 in place of 1.645
The intervals are (8.1, 9.9) for a 90% CI. To be more confident, one needs a wider interval and the 99%CI is (7.6, 10.4). Units are minutes.
|
|
|
