Question

A fitness center is interested in finding a 90% 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 247 members were looked at and their mean number of visits per week was 2.1 and the standard deviation was 2.9. Round answers to 3 decimal places where possible.

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

b. With 90% confidence the population mean number of visits per week is between  and   visits.

c. If many groups of 247 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 because sigma is unknown

b)

sample mean, xbar = 2.1
sample standard deviation, s = 2.9
sample size, n = 247
degrees of freedom, df = n - 1 = 246

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.651

ME = tc * s/sqrt(n)
ME = 1.651 * 2.9/sqrt(247)
ME = 0.305

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (2.1 - 1.651 * 2.9/sqrt(247) , 2.1 + 1.651 * 2.9/sqrt(247))
CI = (1.795 , 2.405)

With 90% confidence the population mean number of visits per week is between 1.795 and 2.405 visits.

c. If many groups of 247 randomly selected members are studied, then a different confidence interval would be produced from each group. About 90 percent of these confidence intervals will contain the true population mean number of visits per week and about 10 percent will not contain the true population mean number of visits per week.