Use the standard normal distribution or the t-distribution to construct a $99 \%$ confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results. In a random sample of 16 mortgage institutions, the mean interest rate was $3.69 \%$ and the standard deviation was $0.33 \%$. Assume the interest rates are normally distributed. Which distribution should be used to construct the confidence interval? A. Use a t-distribution because the interest rates are normally distributed and $\sigma$ is known. B. Use a normal distribution because the interest rates are normally distributed and $\sigma$ is known. C. Use a normal distribution because $n<30$ and the iriterest rates are normally distributed. D. Use a t-distribution because it is a random sample, $\sigma$ is unknown, and the interest rates are normally distributed. E. Cannot use the standard normal distribution or the t-distribution because $\sigma$ is unknown, $n<30$, and the interest rates are not normally distributed. Select the correct choice below and, if necessary, fill in any answer boxes to complete your choice. A. The $99 \%$ confidence interval is ( $\square, \square$ ). (Round to two decimal places as needed.) B. Neither distribution can be used to construct the confidence interval. Interpret the results. Choose the correct answer below.

Step 1 ：The problem provides us with a sample of 16 mortgage institutions with a mean interest rate of 3.69% and a standard deviation of 0.33%. The interest rates are assumed to be normally distributed. We are asked to determine which distribution should be used to construct a 99% confidence interval for the population mean.

Step 2 ：The first step is to decide whether to use the standard normal distribution or the t-distribution. The standard normal distribution is used when the population standard deviation is known, while the t-distribution is used when the population standard deviation is unknown and is estimated from the sample data.

Step 3 ：In this case, we are given the sample standard deviation, not the population standard deviation. Therefore, we should use the t-distribution. This corresponds to option D: 'Use a t-distribution because it is a random sample, \(\sigma\) is unknown, and the interest rates are normally distributed.'

Step 4 ：Next, we calculate the 99% confidence interval using the t-distribution. The formula for a confidence interval is: \[\bar{x} \pm t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}}\] where \(\bar{x}\) is the sample mean, \(t_{\alpha/2, n-1}\) is the t-score for a two-tailed test with \(\alpha\) level of significance and \(n-1\) degrees of freedom, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 5 ：After calculating, we get the lower and upper bounds of the 99% confidence interval.

Step 6 ：The interpretation of the confidence interval is that we are 99% confident that the true population mean interest rate lies within this interval.

Step 7 ：\(\boxed{\text{Final Answer: The correct distribution to use is the t-distribution (Option D). The 99% confidence interval for the population mean is approximately (3.45%, 3.93%). This means we are 99% confident that the true population mean interest rate lies within this interval.}}\)