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 223 members were looked at and their mean number of visits per week was 2.3 and the standard deviation was 2.8. Round answers to 3 decimal places where possible.a. To compute the confidence interval use a ? distribution.b. With 90% confidence the population mean number of visits per week is between and visits.c. If many groups of 223 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

Question A:

- To compute the confidence interval, use a T-distribution.

Question B:

- The 90% confidence interval is:

`\begin{gathered} CI=\bar{x}\pm Z^*(s)/(√(n)) \\ \\ Z^*=1.645 \\ \bar{x}=2.3 \\ s=2.8 \\ n=223 \\ \\ CI=2.3\pm1.645*(2.8)/(√(223)) \\ \\ CI=2.3\pm0.3084406 \\ \\ CI=[1.992,2.608]\text{ \lparen To 3 decimal places\rparen} \end{gathered}`

Question C:

About 90% of these confidence intervals will contain the true population mean number of visits per week and about 10% will not contain the true population mean number of visits per week