A professor samples 9 random TCC students and record their commute times. She finds that the average commute time of the 9 students is 14.08 minutes with a standard deviation of 4.08 minutes. She also finds that the commute times look like they come from a normal distribution.

Find a $99 \%$ confidence interval for the mean commute time of all TCC students.
Round to three decimal places.

Step 1 ：We are given that the sample mean commute time, denoted as \(\bar{x}\), is 14.08 minutes, the sample standard deviation, denoted as \(s\), is 4.08 minutes, and the sample size, denoted as \(n\), is 9 students.

Step 2 ：We are asked to find a 99% confidence interval for the mean commute time of all TCC students. The z-score corresponding to a 99% confidence level is approximately 2.576.

Step 3 ：The formula for the confidence interval for a mean is \(\bar{x} \pm z \frac{s}{\sqrt{n}}\).

Step 4 ：Substituting the given values into the formula, we get \(14.08 \pm 2.576 \frac{4.08}{\sqrt{9}}\).

Step 5 ：Solving the above expression, we get the margin of error to be approximately 3.50336 minutes.

Step 6 ：Subtracting this margin of error from the sample mean, we get the lower bound of the confidence interval to be approximately 10.57664 minutes.

Step 7 ：Adding this margin of error to the sample mean, we get the upper bound of the confidence interval to be approximately 17.58336 minutes.

Step 8 ：Rounding to three decimal places, we get the 99% confidence interval for the mean commute time of all TCC students to be approximately \(\boxed{(10.577, 17.583)}\) minutes.