Question

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.

Answer 1

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

$\bar{x}\pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

where: 

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

Therefore, the 99% confidence interval is:

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

Answer 2

Answer: (69.44, 69.96).

Step-by-step explanation:

Given : Sample size : n= 772

Significance level : $\alpha: 1-0.99=0.01$

Critical value : $z_{\alpha/2}=\pm2.576$

Sample mean : $\overline{x}=69.7$

Standard deviation: $\sigma=2.8$

The confidence interval for population mean is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

i.e $69.7\pm(2.576)\dfrac{2.8}{\sqrt{772}}$

i.e $\approx69.7\pm0.26=(69.7-0.26,\ 69.7+0.26)=(69.44,\ 69.96)$

Hence, the confidence interval for the mean height ( in inches) of all males between the ages of 18 and 24 is (69.44 , 69.96).