You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures: \begin{tabular}{|l|} \hline 67.3 \\ \hline 71.9 \\ \hline 63.6 \\ \hline 58.3 \\ \hline 60.2 \\ \hline 68.8 \\ \hline 48.2 \\ \hline 83.5 \\ \hline 83.8 \\ \hline \end{tabular} Find the $99 \%$ confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places. \[ 99 \% \text { C.I. }= \]

Step 1 ：Calculate the mean (average) of the sample data. Mean = \(\frac{67.3 + 71.9 + 63.6 + 58.3 + 60.2 + 68.8 + 48.2 + 83.5 + 83.8}{9} = 65.51\)

Step 2 ：Calculate the standard deviation of the sample data. First, calculate the variance. Variance = \(\frac{(67.3-65.51)^2 + (71.9-65.51)^2 + (63.6-65.51)^2 + (58.3-65.51)^2 + (60.2-65.51)^2 + (68.8-65.51)^2 + (48.2-65.51)^2 + (83.5-65.51)^2 + (83.8-65.51)^2}{9-1} = 119.04\). Then, take the square root of the variance to get the standard deviation. Standard deviation = \(\sqrt{119.04} = 10.91\)

Step 3 ：Calculate the 99% confidence interval. Confidence interval = \(65.51 \pm (2.576 \times \frac{10.91}{\sqrt{9}}) = 65.51 \pm 9.35\)

Step 4 ：\(\boxed{\text{So, the 99% confidence interval is (56.16, 74.86).}}\)