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The researchers recorded expenses per full-time equivalent employee for each in a sample of 1,751 army hospitals. The sample yielded the following summary statistics: $\stackrel{¯}{x}=\$6,563$and$s=\$2,484$.

Assume that the distribution of expenses per full-time equivalent employee is normally distributed.

Here, the confidence coefficient is 0.90.

Therefore,

$\begin{array}{rcl}\left(1-\mathrm{Î\pm }\right)& =& 0.90\\ \mathrm{Î\pm }& =& 0.10\\ \frac{\mathrm{Î\pm }}{2}& =& 0.05\\ & & \end{array}$

From table, the required value for 90% coefficient level is 1.645.

The 90% coefficient interval is obtained below:

$\begin{array}{rcl}\stackrel{¯}{x}+z_{\frac{\mathrm{Î\pm }}{2}}\left(\frac{\mathrm{σ}}{\sqrt{n}}\right)& =& 6563Â\pm 1.645×\left(\frac{2484}{\sqrt{1751}}\right)\\ & =& 6563Â\pm 97.65\\ & =& \left(6563-97.65,6563+97.65\right)\\ & =& \left(6465.35,6660.65\right)\\ & & \end{array}$

Therefore, the 90% confidence interval is ($6565.35, $6660.65).

There is 90% confident that the true mean expense per full-time equivalent employee of all U.S. Army hospitals$\left(\mathrm{μ}\right)$ lies between $6565.35 and $6660.65.