Answer to Question #117016 in Statistics and Probability for Terri Gans

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 278 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 95% confidence the population mean number of visits per week is between ? and ? visits?

c. If many groups of 278 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

$\bar x =2.1$

s=2.9

n=278

a) To compute the confidence interval, we use t distribution because the population standard deviation is not provided.

b) 95% confidence interval

$CI=\bar x ±t_{(n-1, \frac {\alpha}{2})}*\frac{s}{\sqrt n}$

=$2.1±1.96×\frac {2.9}{\sqrt{278}}$

=(1. 759, 2.441)

With 95% confidence the population mean number of visits per week is between 1.759 and 2.441 visits

c)If many groups of 278 randomly selected members are studied, then a different confidence interval would be produced from each group. 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