A researcher must estimate the avereage height (in inches) with the following sample heights:
\begin{tabular}{|r|}
\hline 65.8 \\
\hline 66.9 \\
\hline 73 \\
\hline 69.7 \\
\hline 65.3 \\
\hline 62.7 \\
\hline 71.5 \\
\hline 73.3 \\
\hline
\end{tabular}

Assuming the population of heights are normally distributed, find the $90 \%$ confidence interval. Round the answers to two decimal places. Enter your answer as an interval of the form (LB,UP).

The researcher is $90 \%$ confident that the population average height is within the interval

Step 1 ：Given the sample heights of [65.8, 66.9, 73, 69.7, 65.3, 62.7, 71.5, 73.3], we first calculate the mean and standard deviation of these heights.

Step 2 ：The mean of the sample heights is calculated as \( \frac{65.8 + 66.9 + 73 + 69.7 + 65.3 + 62.7 + 71.5 + 73.3}{8} = 68.525 \)

Step 3 ：The standard deviation of the sample heights is calculated as \( \sqrt{\frac{(65.8-68.525)^2 + (66.9-68.525)^2 + (73-68.525)^2 + (69.7-68.525)^2 + (65.3-68.525)^2 + (62.7-68.525)^2 + (71.5-68.525)^2 + (73.3-68.525)^2}{8-1}} = 3.918 \)

Step 4 ：Next, we calculate the margin of error for a 90% confidence interval. The Z-score for a 90% confidence interval is approximately 1.645.

Step 5 ：The margin of error is calculated as \( 1.645 \times \frac{3.918}{\sqrt{8}} = 2.278 \)

Step 6 ：Finally, we calculate the 90% confidence interval for the average height. The lower bound is calculated as \( 68.525 - 2.278 = 66.247 \) and the upper bound is calculated as \( 68.525 + 2.278 = 70.803 \)

Step 7 ：So, the 90% confidence interval for the average height is approximately \(\boxed{(66.25, 70.80)}\)