Question 3 Assume that a sample is used to estimate a population mean $\mu$. Find the $99 \%$ confidence interval for a sample of size 72 with a mean of 68.4 and a standard deviation of 13.3. Enter your answer as an openinterval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place). \[ 99 \% \text { C.I. }= \]

Step 1 ：We are given a sample size of 72, a sample mean of 68.4, and a standard deviation of 13.3. We are asked to find the 99% confidence interval for the population mean.

Step 2 ：The formula for a confidence interval is given by \(\bar{x} \pm z \frac{\sigma}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(z\) is the z-score corresponding to the desired confidence level, \(\sigma\) is the standard deviation of the sample, and \(n\) is the sample size.

Step 3 ：In this case, we have \(\bar{x} = 68.4\), \(\sigma = 13.3\), and \(n = 72\). The z-score corresponding to a 99% confidence level is approximately 2.576.

Step 4 ：Substituting these values into the formula, we get the confidence interval as \((64.36259355155744, 72.43740644844257)\).

Step 5 ：However, the question asks for the answer accurate to one decimal place. Therefore, we need to round these values to one decimal place, giving us a confidence interval of \((64.4, 72.4)\).

Step 6 ：Final Answer: The 99% confidence interval for a sample of size 72 with a mean of 68.4 and a standard deviation of 13.3 is \(\boxed{(64.4, 72.4)}\).