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.
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Answer

The $95 %$ confidence interval of the mean length is $(17.1, 20.5)$

Steps

Step 1 ：Calculate the mean length of the sample data: $\overset{―}{x} = \frac{\sum x_{i}}{n}$

Step 2 ：Calculate the standard deviation of the sample data: $s = \sqrt{\frac{\sum (x_{i} - \overset{―}{x})^{2}}{n - 1}}$

Step 3 ：Find the critical t-value for a 95% confidence level with $n - 1$ degrees of freedom: $t_{α / 2, n - 1}$

Step 4 ：Calculate the margin of error: $E = t_{α / 2, n - 1} \cdot \frac{s}{\sqrt{n}}$

Step 5 ：Calculate the 95% confidence interval of the mean length: $\overset{―}{x} \pm E$

Step 6 ：The $95 %$ confidence interval of the mean length is $(17.1, 20.5)$