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 \(\boxed{(17.6, 18.8)}\)