Question 2 of 14, Step 1 of 1 A professor wants to estimate how many hours per week her students study. A simple random sample of 56 students had a mean of 20 hours of studying per week. Construct a \( 98 \% \) confidence interval for the mean number of hours a student studies per week. Assume that the population standard deviation is known to be 2 hours per week. Round to two decimal places.

To construct a \( 98\% \) confidence interval for the mean number of hours a student studies per week, we can use the formula: \[ \text{Confidence Interval} = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \] Where: - \( \bar{x} = 20 \) (sample mean), - \( z \) is the z-value corresponding to \( 98\% \) confidence, which is approximately \( 2.33 \) (found using z-tables), - \( \sigma = 2 \) (population standard deviation), - \( n = 56 \) (sample size). First, we calculate the standard error (SE): \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{2}{\sqrt{56}} \approx \frac{2}{7.48} \approx 0.2673 \] Now we can calculate the margin of error (ME): \[ ME = z \times SE \approx 2.33 \times 0.2673 \approx 0.6245 \] Finally, the confidence interval is: \[ 20 \pm 0.6245 \implies (19.38, 20.62) \] Thus, the \( 98\% \) confidence interval for the mean number of hours a student studies per week is approximately \( (19.38, 20.62) \).