Question

The percent body fat in a random sample of 36 men aged 20 to 29 has a sample mean of 14.42. Find a 99% confidence interval for the mean percent body fat in all men aged 20 to 29. Assume that percent body fat follows a Normal distribution, with a standard deviation of 6.95.
A) (0.8, 28.04)B) (12.15, 16.69)C) (12.51, 16.33)D) (12.07, 16.77)

Answer 1

Answer: $(11.44,\ 17.4)$

Step-by-step explanation:

The confidence interval for population mean is given by :-

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

Given : Sample size : $n=36$

Sample mean : $\ovreline{x}=14.42$

Standard deviation : $\sigma=6.95$

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

Critical value : $z_{\alpha/2}=z_{0.005}=2.576$

Now, the 99% confidence interval for the mean percent body fat in all men aged 20 to 29 will be :-

$14.42\pm (2.576)\dfrac{6.95}{\sqrt{36}}\\\\\approx14.42\pm2.98\\\\=(14.42-2.98,\ 14.42+2.98)=(11.44,\ 17.4)$