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 251 members were looked at and their mean number of visits per week was 3.1 and the standard deviation was 2.8. Round answers to 3 decimal places where possible.

a. To compute the confidence interval use a Z or T distribution?

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

c. If many groups of 251 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

Solution :

Given that,

Point estimate = sample mean = = 3.1

sample standard deviation = s = 2.8

sample size = n = 251

Degrees of freedom = df = n - 1 = 251 - 1 = 249

a) T distribution

b) At 98% confidence level

= 1 - 98%

=1 - 0.98 =0.02

/2 = 0.01

t/2,df = t0.01,249 = 2.341

Margin of error = E = t/2,df * (s /n)

= 2.341 * (2.8 / 251)

Margin of error = E = 0.41

The 98% confidence interval estimate of the population mean is,

  ± E  

= 3.1  ± 0.41

= ( 2.69, 3.51 )

c) If many groups of 251 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.