A publisher wants to estimate the mean length of time (in minutes) all adults sp
ID: 3050659 • 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.
12
11
12
12
9
9
6
8
9
8
10
8
10
7
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 ( , ). (Round to one decimal place as needed.)
12
11
12
12
9
9
6
8
9
8
10
8
10
7
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 ( , ). (Round to one decimal place as needed.)
Explanation / Answer
The average = 9.47
For 90%
Standard error of the mean = SEM = S/N = 0.491
t(, N-1) = 1.761
Confidence interval = m +/- (t(, N-1)*SEM)
Mean = 9.47
Lower bound: 8.61
Upper bound: 10.33
For 95%
Standard error of the mean = SEM = S/N = 0.491
t(, N-1) = 2.145
Confidence interval = m +/- (t(, N-1)*SEM)
Mean = 9.47
Lower bound: 8.42
Upper bound: 10.52
95% confidence interval is wider
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