Question

10

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 o is 23 minutes and that the population of times is normally distributed 12 9 10 7 9 8 7 12 12 6 11 9 9 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 (D). (Round to one decimal place as needed.) The 99% confidence intervalis 10 (Round to one decimal place as needed) Which interval is wider? O The 90% confidence interval The 99% confidence interval

Answer 1

Here we have n = 15 Sample meanin) = Exi n 135 15 = 9 standard deviation (3) - Van Elxi- 132 1112-914.... 1150 +19-93 n-1=15-12 14 = 20702 Ał 90% confidence interval de 0.90- 0.10 ..012=0.05 degree of frudom Id}) Fcrdicas = 10.0514=1.761 90% confidence interval for population mean Ž + tarihicas * coun Vn 2.6702 = 9 1.761 * 15. = 9+0.9413 80587 9.9413 =10.10, 9.9) Round to one decimal placa Ał 99.4. confidence interval, d=1-0.99 = 0.01 ..012 0.005 d = 14 Puncal = 40.005, 14 40.005,14 = 2.977

* s 99% confidence interval for population mean =ł z bcritical 9 + 2.977* 2.0702 vñ 715 foto = 9 1,5913 = 7.4087 10.5913 -17.410.67 Round to me decimal a Places

99% confidence interval is wider.

Thanks.