Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. To accomplish this, the records of 276 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 18.2 seats and the standard deviation is 4 seats.

Construct a $99 \%$ confidence interval for the population mean number of unoccupied seats per flight during the past year.

Round your answers to one decimal.

Step 1 ：Given that the sample mean (\(\bar{x}\)) is 18.2, the sample standard deviation (s) is 4, and the sample size (n) is 276.

Step 2 ：We are asked to construct a 99% confidence interval for the population mean. The z-score corresponding to a 99% confidence level is approximately 2.576.

Step 3 ：We can use the formula for a confidence interval: \(\bar{x} \pm z \frac{s}{\sqrt{n}}\)

Step 4 ：Substitute the given values into the formula: \(18.2 \pm 2.576 \frac{4}{\sqrt{276}}\)

Step 5 ：Calculate the margin of error: \(2.576 \frac{4}{\sqrt{276}} = 0.6201868162221142\)

Step 6 ：Subtract and add the margin of error from the sample mean to get the confidence interval: \((18.2 - 0.6201868162221142, 18.2 + 0.6201868162221142) = (17.6, 18.8)\)

Step 7 ：Final Answer: The 99% confidence interval for the population mean number of unoccupied seats per flight during the past year is \(\boxed{(17.6, 18.8)}\)