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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.

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nobillionaireNobley
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]
JeanaShupp

Answer: (69.44, 69.96).

Step-by-step explanation:

Given : Sample size : n= 772

Significance level : [tex]\alpha: 1-0.99=0.01[/tex]

Critical value : [tex]z_{\alpha/2}=\pm2.576[/tex]

Sample mean : [tex]\overline{x}=69.7[/tex]

Standard deviation: [tex]\sigma=2.8[/tex]

The confidence interval for population mean is given by :-

[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]

i.e [tex]69.7\pm(2.576)\dfrac{2.8}{\sqrt{772}}[/tex]

i.e [tex]\approx69.7\pm0.26=(69.7-0.26,\ 69.7+0.26)=(69.44,\ 69.96)[/tex]

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).

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