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 268 members were looked at and their mean number of visits per week was 2.5 and the standard deviation was 2.9. Round answers to 3 decimal places where possible.

a. To compute the confidence interval use a ? t z  distribution.

b. With 98% confidence the population mean number of visits per week is between  and   visits.

c. If many groups of 268 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)

t distribution

b)

sample mean, xbar = 2.5
sample standard deviation, s = 2.9
sample size, n = 268
degrees of freedom, df = n - 1 = 267

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.34

ME = tc * s/sqrt(n)
ME = 2.34 * 2.9/sqrt(268)
ME = 0.415

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (2.5 - 2.34 * 2.9/sqrt(268) , 2.5 + 2.34 * 2.9/sqrt(268))
CI = (2.085 , 2.915)

b. With 98% confidence the population mean number of visits per week is between 2.085 and 2.915 visits.

c. If many groups of 268 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.