Question

A fitness center is interested in finding a 99% 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 253 members were looked at and their mean number of visits per week was 2.4 and the standard deviation was 2.8. Round answers to 3 decimal places where possible.

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

b. With 99% confidence the population mean number of visits per week is between  and   visits.

c. If many groups of 253 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

n = 253

sample mean = 2.4

sample sd = 2.8

(a) As the population sd is not given, we will use t distribution.

(b) CI = mean +/- E

t value = TINV(0.01, 252) = 2.595

CI =(1.94 , 2.86)

c)

If many groups of 253 randomly selected members are studied, then a different confidence interval would be produced from each group. About 99% percent of these confidence intervals will contain the true population mean number of visits per week and about 1% percent will not contain the true population mean number of visits per week.