omework: Chapter 6 core: 0 of 1 pt 6.1.45 12 of 25 (3 complete HW Score: 12%, 3
ID: 3318993 • Letter: O
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omework: Chapter 6 core: 0 of 1 pt 6.1.45 12 of 25 (3 complete HW Score: 12%, 3 of 25 pts Question Help * 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 22 minutes and that the population of times is normally distributed 0 10 10 12 12 6 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) Enter your answer in the edit tields and then click Check Answer 2Explanation / Answer
The sample average is 8.733
The sample deviation is 2.2
n = 15
t value for df = n-1 = 14 and 90% CI is 1.761
t value for df = n-1 = 14 and 95% CI is 2.145
So, The Ci is given by Xbar +/- t*s/sqrt(n).
For a 90% CI
= 8.733 +/- 1.761*2.2/sqrt(15) = 7.7 to 9.7
For a 95% CI
= 8.733 +/- 2.145*2.2/sqrt(15) = 7.5 to 10.0
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