Questions: Question 14 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 231 members were looked at and their mean number of visits per week was 2.3 and the standard deviation was 1.9. Round answers to 3 decimal places where possible. a. To compute the confidence interval use a ? distribution. b. With 98% confidence the population mean number of visits per week is between ? and ? visits. c. If many groups of 231 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.

Solution Steps

To compute the confidence interval for the mean number of visits per week, we use a normal distribution.

The margin of error is calculated using the formula:

$$\text{Margin of Error} = \frac{Z \times \sigma}{\sqrt{n}}$$

where:

• $Z = 2.326$ (Z-Score for 98% confidence level)
• $\sigma = 1.9$ (standard deviation)
• $n = 231$ (sample size)

Substituting the values, we find:

$$\text{Margin of Error} = \frac{2.326 \times 1.9}{\sqrt{231}} \approx 0.291$$

With the margin of error calculated, we can determine the confidence interval for the population mean number of visits per week:

$$\text{Lower Bound} = \bar{x} - \text{Margin of Error} = 2.3 - 0.291 \approx 2.009$$

$$\text{Upper Bound} = \bar{x} + \text{Margin of Error} = 2.3 + 0.291 \approx 2.591$$

Thus, with 98% confidence, the population mean number of visits per week is between $2.009$ and $2.591$.

If many groups of 231 randomly selected members are studied, about $98.0\%$ of these confidence intervals will contain the true population mean number of visits per week, while about $2.0\%$ will not.

Final Answer