Question

A fitness center is interested in finding a 98% 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 213 members were looked at and their mean number of visits per week was 3.5 and the standard deviation was 1.8. Round answers to 3 decimal places where possible. a. To compute the confidence interval use a distribution. b. With 98% confidence the population mean number of visits per week is between and visits. c. If many groups of 213 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

a)

use t distribution

b)

sample mean, xbar = 3.5
sample standard deviation, s = 1.8
sample size, n = 213
degrees of freedom, df = n - 1 = 212

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.344

ME = tc * s/sqrt(n)
ME = 2.344 * 1.8/sqrt(213)
ME = 0.2891

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (3.5 - 2.344 * 1.8/sqrt(213) , 3.5 + 2.344 * 1.8/sqrt(213))
CI = (3.211 , 3.789)

With 98% confidence the population mean number of visits per week is between 3.211 and 3.789visits

. c. If many groups of 213 randomly selected members are studied, then a different confidence interval would be produced from each group. About 98 percent of these confidence intervals will contain the true population mean number of visits per week and about 2 percent will not contain the true population mean number of visits per week.