A sample of the length in inches for newborns is given below. Assume that lengths are normally distributed. Find the $95 \%$ confidence interval of the mean length. Round your answers to at least one decimal, and do not round between steps.
\begin{tabular}{|c|}
\hline Length \\
\hline 22.5 \\
\hline 20.3 \\
\hline 15.8 \\
\hline 19.4 \\
\hline 18.3 \\
\hline 20.8 \\
\hline 17.5 \\
\hline 15.6 \\
\hline 16.9 \\
\hline 21.3 \\
\hline
\end{tabular}
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Answer
The $95\%$ confidence interval of the mean length is \(\boxed{(17.1, 20.5)}\)
Step 1 ：Calculate the mean length of the sample data: \( \overline{x} = \frac{\sum x_i}{n} \)
Step 2 ：Calculate the standard deviation of the sample data: \( s = \sqrt{\frac{\sum (x_i - \overline{x})^2}{n-1}} \)
Step 3 ：Find the critical t-value for a 95% confidence level with \( n-1 \) degrees of freedom: \( t_{\alpha/2, n-1} \)
Step 4 ：Calculate the margin of error: \( E = t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} \)
Step 5 ：Calculate the 95% confidence interval of the mean length: \( \overline{x} \pm E \)
Step 6 ：The $95\%$ confidence interval of the mean length is \(\boxed{(17.1, 20.5)}\)
