A marketing research company desires to know the mean consumption of milk per week among people over age 32. A sample of 1097 people over age 32 was drawn and the mean milk consumption was 3.9 liters. Assume that the population standard deviation is known to be 0.9 liters. Construct the $99 \%$ confidence interval for the mean consumption of milk among people over age 32 . Round your answers to one decimal place. Answer Tables Keypac How to enter your answer (opens in new window) Keyboard Shortcu Lower endpoint: Upper endpoint:

Step 1 ：First, we need to find the z-score that corresponds to a 99% confidence interval. The z-score for a 99% confidence interval is approximately 2.576.

Step 2 ：The formula for a confidence interval is \(CI = \bar{x} ± Z * (\sigma/\sqrt{n})\), where \(\bar{x}\) is the sample mean, Z is the z-score, \(\sigma\) is the population standard deviation, and n is the sample size.

Step 3 ：Substituting the given values into the formula, we get \(CI = 3.9 ± 2.576 * (0.9/\sqrt{1097})\).

Step 4 ：To find the lower endpoint of the confidence interval, we calculate \(3.9 - 2.576 * (0.9/\sqrt{1097}) = 3.9 - 2.576 * 0.0271 = 3.9 - 0.07 = 3.8\) (rounded to one decimal place).

Step 5 ：To find the upper endpoint of the confidence interval, we calculate \(3.9 + 2.576 * (0.9/\sqrt{1097}) = 3.9 + 2.576 * 0.0271 = 3.9 + 0.07 = 4.0\) (rounded to one decimal place).

Step 6 ：So, the 99% confidence interval for the mean consumption of milk among people over age 32 is \(\boxed{(3.8, 4.0)}\) liters.