The u.s. department of health and human services collected sample data for 772 males between the ages of 18 and 24. that sample group has a mean height of 69.7 inches with a standard deviation of 2.8 inches. find the 99% confidence interval for the mean height of all males between the ages of 18 and 24.

The 99% confidence interval for the true mean is given by:

[tex]\bar{x}\pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}} [/tex]

where: 

[tex]\bar{x}[/tex] is the sample mean = 69.7
[tex]\sigma[/tex] is the standard deviation = 2.8
[tex]z_{\alpha/2}[/tex] is the test statistics = 2.58 for for 99% confidentce interval.
n is the sample size = 772.

Therefore, the 99% confidence interval is:

[tex]69.7\pm2.58\left( \frac{2.8}{\sqrt{772}} \right)=69.7\pm2.58(0.1008) \\ \\ =69.7-0.26\leq\mu\leq69.7+0.26 \\ \\ =69.44\leq\mu\leq69.96[/tex]