Question 3, 6.2.18-T Part 1 of 3 HW Score: $40 \%, 40$ of 100 points Points: 0 of 20 Save In a random sample of ten people, the mean driving distance to work was 20.9 miles and the standard deviation was 4.9 miles. Assume the population is normally distributed and use the t-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. $\square$ miles (Round to one decimal place as needed.)

Step 1 ：We are given a random sample of ten people, with a mean driving distance to work of 20.9 miles and a standard deviation of 4.9 miles. We are asked to assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 99% confidence interval for the population mean.

Step 2 ：The margin of error for a confidence interval can be calculated using the formula: Margin of Error = t * (s/√n), where t is the t-score, s is the sample standard deviation, and n is the sample size.

Step 3 ：For a 99% confidence interval and 9 degrees of freedom (n-1), we can look up the t-score in a t-distribution table or use a calculator or statistical software to find it. In this case, the t-score is approximately 3.25.

Step 4 ：Substituting the given values into the formula, we get: Margin of Error = 3.25 * (4.9/√10) = 5.0 miles.

Step 5 ：\(\boxed{5}\) is the margin of error for the population mean driving distance to work, with a 99% confidence level. This means we are 99% confident that the true population mean falls within 5 miles of our sample mean of 20.9 miles.