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 237 members were looked at and their mean number of visits per week was 2.5 and the standard deviation was 1.6.

a. To compute the confidence interval use a

distribution.

b. With 95% confidence the population mean number of visits per week is between answer ______ and answer _________ visits.

c. If many groups of 237 randomly selected members are studied, then a different confidence interval would be produced from each group. About answer _________ percent of these confidence intervals will contain the true population mean number of visits per week and about answer ________ percent will not contain the true population mean number of visits per week.

Answer 1

Answer :

Given data is :

sample size n = 237

Mean = = 2.5

Standard deviation = = 1.6

a)Here we are using normal distribution to calculate confidence interval .

b)GIven confidence interval = 95%

Z value at 95% is 1.96

therefore,

CI =

Substitute the value sin above formula,

CI =

= 2.5 +/- 1.96(1.6 / 15.395)

= 2.5 +/- 1.96(0.104)

= 2.5 +/- 1.140

= (2.5 - 1.140 , 2.5 + 1.140)

= (1.396, 3.604)

CI = (1.396 , 3.964)

With 95% confidence the population mean number of visits per week is between 1.396 and 3.964

c) About 95% percent of these confidence intervals will contain the true population mean number of visits per week and about 5% percent will not contain the true population mean number of visits per week.