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)`

Explanation:

The confidence interval for population mean is given by :-

`\overline{x}\pm z_(\alpha/2)(\sigma)/(√(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)(6.95)/(√(36))\\\\\approx14.42\pm2.98\\\\=(14.42-2.98,\ 14.42+2.98)=(11.44,\ 17.4)`