Question

A fitness center is interested in finding a 95% confidence interval for the mean number of days per week that Americans who are members of a fitness club go to their fitness center. Records of 240 members were looked at and their mean number of visits per week was 2.3 and the standard deviation was 1.7. Round answers to 3 decimal places where possible. a. To compute the confidence interval use a ? v distribution. b. With 95% confidence the population mean number of visits per week is between and visits. c. If many groups of 240 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of visits per week and about | percent will not contain the true population mean number of visits per week.

Answer 1

ofvisits per week is between 12.084 9x=5%, it means of about 5% population mean will not contain in the confidence inteova). n=240 Slution > Given the info omation, Ce confidence level = 95% = 0.95 x = sample mean number of visits per week = 2:3 5- sample standaod deviation = 1.7 @hede population standaod deviation is unknown, but sample standaod deviation is known sowe use t-disto ibution To compute the confidence inteoval use a (t-distribution The formula for 95% confidence interval for population mean iu is, (-E) LULE) E = tikidof.) b *5 To dof. = degree of foredom = (1-1) dof. = 240-1 duf 239 x = 1-C = level of significanie X=1-0.95 K = 0.05 trid.f.) - toos, 239) t60.05,239 ) = 1.9699 E - 1.9699*107 1240 E-0.216 X-E = 2.3-0.216 = 2:084 TE = 2:3+ 0.216 = 2:516 Hith 95% confidence the population mean number of and[ 2.516 © Here confidence level is as iets means we are 95% confident that papulation mean is lies in the confidence interva)

then If many gooups 240 Tondomly selected numbers age studied a diffegent confidence interval woulb would be Pooduced from each group. About 195] percent of these confidence integral will contain the roue population mean number of visits per week and about 15 peocent will not contain the toue population mean number of visit per week.