You must estimate the mean temperature (in degrees Fahrenhelt) with the following sample temperatures: \begin{tabular}{|l|} \hline 73.8 \\ \hline 75.7 \\ \hline 81.3 \\ \hline 75.1 \\ \hline 75.2 \\ \hline 52.8 \\ \hline 67.8 \\ \hline 57.7 \\ \hline 64.1 \\ \hline 87.7 \\ \hline 84.3 \\ \hline 82.5 \\ \hline \end{tabular} Find the $98 \%$ confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place). \[ 98 \% \text { C. } .= \]

Step 1 ：Given the sample temperatures are 73.8, 75.7, 81.3, 75.1, 75.2, 52.8, 67.8, 57.7, 64.1, 87.7, 84.3, 82.5.

Step 2 ：Calculate the sample mean, which is the sum of all sample temperatures divided by the number of samples. The sample mean is approximately 73.17.

Step 3 ：Calculate the sample standard deviation, which is a measure of the amount of variation or dispersion of the set of values. The sample standard deviation is approximately 10.74.

Step 4 ：The sample size, denoted as \(n\), is the number of observations in a sample. In this case, \(n = 12\).

Step 5 ：Find the Z-score for a 98% confidence level. The Z-score is a measure of how many standard deviations an element is from the mean. The Z-score for a 98% confidence level is approximately 2.33.

Step 6 ：Calculate the confidence interval using the formula: \((\text{mean} - \text{Z-score} \times \frac{\text{standard deviation}}{\sqrt{n}}, \text{mean} + \text{Z-score} \times \frac{\text{standard deviation}}{\sqrt{n}})\). The confidence interval is approximately \((65.96, 80.38)\).

Step 7 ：\(\boxed{\text{The 98% confidence interval for the mean temperature is approximately (65.96, 80.38).}}\)