Step 1 :The problem provides us with a sample size of 19, a sample mean of 3.41%, and a sample standard deviation of 0.36%. We are also informed that the interest rates are normally distributed.
Step 2 :To construct a confidence interval, we need to decide which distribution to use. 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 but the sample size is large enough (usually n > 30).
Step 3 :In this case, we do not know the population standard deviation, but we do know the sample standard deviation and the sample size is less than 30. Therefore, we should use the t-distribution.
Step 4 :Given that n = 19, the mean = 3.41, and the standard deviation = 0.36, we can calculate the t-score as 2.878440472713585 and the standard error as 0.08258966419340222.
Step 5 :Using these values, we can calculate the lower bound of the confidence interval as 3.172270567957887 and the upper bound as 3.647729432042113.
Step 6 :Thus, the $99 \%$ confidence interval is \(\boxed{(3.17\%, 3.65\%)}\). This means that we are $99\%$ confident that the true population mean interest rate lies between $3.17\%$ and $3.65\%$.