A publisher wants to estimate the mean length of time (in minutes) all adults sp

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 is 1.6 minutes and that the population of times is normally distributed. 12 9 9 8 10 9 11 7 8 6 10 7 11 11 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 99% confidence interval

The 90% confidence interval

Explanation / Answer

AT 90% CI
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=9
Standard deviation( sd )=1.8127
Sample Size(n)=15
Confidence Interval = [ 9 ± t a/2 ( 1.8127/ Sqrt ( 15) ) ]
= [ 9 - 1.761 * (0.468) , 9 + 1.761 * (0.468) ]
= [ 8.176,9.824 ]

AT 99% CI
Confidence Interval = [ 9 ± t a/2 ( 1.8127/ Sqrt ( 15) ) ]
= [ 9 - 2.977 * (0.468) , 9 + 2.977 * (0.468) ]
= [ 7.607,10.393 ]

AT 99% CI is wider

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