6.1-6.2 Homework Question 8, 6.2.18-T HW Score: $75 \%, 6$ of 8 points Part 2 of 3 Points: 0 of 1 Save In a random sample of ten people, the mean driving distance to work was 19.2 miles and the standard deviation was 5.6 miles. Assume the population is normally distributed and use the 1-distribution to find the margin of error and construct a $99 \%$ confidence interval for the population mean $\mu$. Interpret the results. Identify the margin of error. 58 miles (Round to one decimal place as needed) Construct a $99 \%$ confidence interval for the population mean. (Round to one decimal place as needed) Get more help - Clear all Check answer

Step 1 ：Given that the sample size (n) is 10, the sample mean (\(\bar{x}\)) is 19.2 miles, the standard deviation (s) is 5.6 miles, and the Z-score (Z) for a 99% confidence level is approximately 2.576.

Step 2 ：The formula for the margin of error (E) is \(E = Z \times \frac{s}{\sqrt{n}}\). Substituting the given values, we get \(E = 2.576 \times \frac{5.6}{\sqrt{10}}\).

Step 3 ：Calculating the above expression, we find that the margin of error is approximately 4.6 miles.

Step 4 ：The formula for the confidence interval is \(\bar{x} \pm E\). Substituting the given values, we get \((19.2 - 4.6, 19.2 + 4.6)\).

Step 5 ：Calculating the above expression, we find that the 99% confidence interval for the population mean is approximately (14.6, 23.8).

Step 6 ：Interpreting the results, the margin of error of approximately 4.6 miles means that the true population mean could be up to 4.6 miles more or less than our sample mean of 19.2 miles.

Step 7 ：The 99% confidence interval of approximately (14.6, 23.8) means that we can be 99% confident that the true population mean lies somewhere between 14.6 miles and 23.8 miles.

Step 8 ：Final Answer: The margin of error is approximately \(\boxed{4.6}\) miles. The 99% confidence interval for the population mean is approximately \(\boxed{(14.6, 23.8)}\).