Ages of students: A simple random sample of 110 U.S. college students had a mean age of 23.02 years. Assume the population standard deviation is o years. Construct a $98 \%$ confidence interval for the mean age of U.S. college students, Round the answers to two decimal places. A $98 \%$ confidence interval for the mean age of U.S. college students is Save For Later Check

Step 1 ：We are given a simple random sample of 110 U.S. college students with a mean age of 23.02 years. We are asked to construct a 98% confidence interval for the mean age of U.S. college students. However, the population standard deviation is not provided.

Step 2 ：The formula for a confidence interval is \(\bar{x} \pm Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(Z_{\alpha/2}\) is the Z-score corresponding to the desired confidence level, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

Step 3 ：Without the population standard deviation, we cannot proceed with the calculation. If we had this information, we would substitute the given values into the formula and calculate the confidence interval.

Step 4 ：The Z-score for a 98% confidence interval can be found from a standard normal distribution table. The Z-score for a 98% confidence interval is approximately 2.33.

Step 5 ：Assuming the standard deviation to be 0, the margin of error is also 0. Therefore, the lower and upper bounds of the confidence interval are both equal to the sample mean, which is 23.02.

Step 6 ：Final Answer: Since the standard deviation is assumed to be 0, the confidence interval for the mean age of U.S. college students is \(\boxed{(23.02, 23.02)}\) at a 98% confidence level. However, this result is based on the assumption that the standard deviation is 0, which is not realistic in a real-world context. In reality, the standard deviation would not be 0, and we would need this information to calculate a more accurate confidence interval.