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The following is a random sample of the annual salaries of high school counselors in the United States. Assuming that the distribution of salaries is approximately normal, construct a $99 \%$ confidence interval for the mean salary of high school counselors across the United States. Round to the nearest dollar.
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline$\$ 63,670$ & $\$ 35,250$ & $\$ 61,780$ & $\$ 32,370$ & $\$ 32,290$ & $\$ 60,250$ & $\$ 60,590$ \\
\hline
\end{tabular}
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So, the 99% confidence interval for the mean salary of high school counselors across the United States is \(\boxed{[\$35774, \$63141]}\).

Steps

Step 1 :Given the salaries of high school counselors in the United States, we are asked to construct a 99% confidence interval for the mean salary.

Step 2 :The given salaries are: \$63670, \$35250, \$61780, \$32370, \$32290, \$60250, \$60590.

Step 3 :First, we calculate the mean and standard deviation of the given salaries. The mean is \$49457.14 and the standard deviation is \$14055.04.

Step 4 :Next, we use the formula for the confidence interval which is mean ± (z-score * (standard deviation / sqrt(n))), where n is the number of samples, and the z-score for a 99% confidence interval is approximately 2.576.

Step 5 :Substituting the values into the formula, we get the margin of error as \$13683.59.

Step 6 :Finally, we subtract and add the margin of error from the mean to get the lower and upper endpoints of the confidence interval respectively.

Step 7 :The lower endpoint is \$35774 and the upper endpoint is \$63141.

Step 8 :So, the 99% confidence interval for the mean salary of high school counselors across the United States is \(\boxed{[\$35774, \$63141]}\).

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