Step 1 :The problem is asking to construct a 99% confidence interval for the population mean. We are given a sample size of 19, a sample mean of 3.41%, and a sample standard deviation of 0.36%. We are also told to assume the interest rates are normally distributed.
Step 2 :Since the sample size is less than 30, we should use the t-distribution to construct the confidence interval. The t-distribution is more appropriate for smaller sample sizes, while the standard normal distribution is more appropriate for larger sample sizes.
Step 3 :The formula for a confidence interval is: \[\bar{x} \pm t \left( \frac{s}{\sqrt{n}} \right)\] where \(\bar{x}\) is the sample mean, \(t\) is the t-score corresponding to the desired level of confidence, \(s\) is the sample standard deviation, and \(n\) is the sample size.
Step 4 :Using the given values, we find that the t-score is approximately 2.878 and the margin of error is approximately 0.238.
Step 5 :Substituting these values into the formula, we find that the lower bound of the confidence interval is approximately 3.17% and the upper bound is approximately 3.65%.
Step 6 :This means that we are 99% confident that the true population mean interest rate lies between 3.17% and 3.65%.
Step 7 :Therefore, the correct interpretation of the results is: With 99% confidence, it can be said that the population mean interest rate is between the bounds of the confidence interval.
Step 8 :Final Answer: \(\boxed{\text{B. With 99% confidence, it can be said that the population mean interest rate is between the bounds of the confidence interval.}}\)