You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures:
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline 43.1 & 44.9 & 52.9 & 47.1 & 78.5 & 55.2 & 56.3 & 53.5 & 67.9 & 60.9 & 69.6 & 79.3 \\
\hline
\end{tabular}
Find the 99\% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. Assume the data is from a normally distributed population.
\[
99 \% \text { C.I. }=
\]

Step 1 ：Given the sample temperatures in degrees Fahrenheit: 43.1, 44.9, 52.9, 47.1, 78.5, 55.2, 56.3, 53.5, 67.9, 60.9, 69.6, 79.3.

Step 2 ：We need to find the 99% confidence interval for the mean temperature. The formula for the confidence interval is \(\bar{x} \pm z \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(z\) is the z-score (for a 99% confidence interval, \(z = 2.576\)), \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 3 ：First, calculate the sample mean (\(\bar{x}\)) and the sample standard deviation (\(s\)). The sample mean is approximately 59.1 and the sample standard deviation is approximately 12.32.

Step 4 ：The sample size (\(n\)) is 12, as there are 12 temperatures in the sample.

Step 5 ：Substitute these values into the formula for the confidence interval: \(59.1 \pm 2.576 \frac{12.32}{\sqrt{12}}\).

Step 6 ：Solving this gives the 99% confidence interval for the mean temperature as approximately \((49.938, 68.262)\).

Step 7 ：\(\boxed{\text{Final Answer: The 99% confidence interval for the mean temperature is } (49.938, 68.262)}\)