Use the following sample to estimate a population mean $\mu$.
\begin{tabular}{|c|}
\hline 29.5 \\
\hline 12.1 \\
\hline 59.9 \\
\hline 23.8 \\
\hline 43.7 \\
\hline
\end{tabular}
Assuming the population is normally distributed, find the $99.5 \%$ confidence interval about the population mean. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places.
\[
99.5 \% \text { C.1. }=
\]
Question Help: D post to forum

Step 1 ：First, we calculate the sample mean and the sample standard deviation from the given data. The data is [29.5, 12.1, 59.9, 23.8, 43.7].

Step 2 ：The sample mean \(\bar{x}\) is calculated as the sum of all the data points divided by the number of data points. In this case, \(\bar{x} = 33.8\).

Step 3 ：The sample standard deviation \(s\) is calculated using the formula \(s = \sqrt{\frac{1}{N-1} \sum_{i=1}^{N} (x_i - \bar{x})^2}\), where \(x_i\) are the data points, \(\bar{x}\) is the sample mean, and \(N\) is the number of data points. In this case, \(s = 18.497297099846776\).

Step 4 ：The sample size \(n\) is the number of data points. In this case, \(n = 5\).

Step 5 ：The z-score \(z\) corresponding to the desired confidence level of 99.5% is approximately 2.807.

Step 6 ：We can now calculate the confidence interval for the population mean using the formula \(\bar{x} \pm z \frac{s}{\sqrt{n}}\).

Step 7 ：The lower bound of the confidence interval is calculated as \(\bar{x} - z \frac{s}{\sqrt{n}} = 12.49211473637326\).

Step 8 ：The upper bound of the confidence interval is calculated as \(\bar{x} + z \frac{s}{\sqrt{n}} = 55.107885263626734\).

Step 9 ：Thus, the 99.5% confidence interval for the population mean is approximately \((12.49, 55.11)\).

Step 10 ：\(\boxed{\text{Final Answer: } (12.49, 55.11)}\)