Step 1 :The question is asking for the 99% confidence interval for the population mean, given a sample mean and standard deviation. The formula for a confidence interval is: \[\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\] where \(\bar{x}\) is the sample mean, \(Z\) is the Z-score corresponding to the desired level of confidence, \(\sigma\) is the standard deviation, and \(n\) is the sample size.
Step 2 :In this case, the sample mean is 35.2 hours, the standard deviation is 7.4 hours, and the sample size is 464. The Z-score for a 99% confidence interval is approximately 2.576.
Step 3 :We can plug these values into the formula to find the lower and upper bounds of the confidence interval.
Step 4 :Calculate the margin of error: \[Z \frac{\sigma}{\sqrt{n}} = 2.576 \frac{7.4}{\sqrt{464}} \approx 0.885\]
Step 5 :Calculate the lower bound of the confidence interval: \[\bar{x} - \text{margin of error} = 35.2 - 0.885 = 34.315\]
Step 6 :Calculate the upper bound of the confidence interval: \[\bar{x} + \text{margin of error} = 35.2 + 0.885 = 36.085\]
Step 7 :Final Answer: The $99 \%$ confidence interval for the population mean is approximately \(\boxed{34.315}\) to \(\boxed{36.085}\) hours.