Question

The percentage of body fat of a random sample of 36 men aged 20 to 29 found a sample mean of 14.42. Find a 95% confidence interval for the mean percentage body fat of all men aged 20 to 29. Assume that percentages of body fat follow a normal distribution with a standard deviation of 6.95.

Answer 1

`14.42-1.96(6.95)/(√(36))=12.150`

`14.42+ 1.96(6.95)/(√(36))=16.690`

So on this case the 95% confidence interval would be given by (12.150;16.690)

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=14.42` represent the sample mean

`\mu` population mean (variable of interest)

`\sigma=6.95` represent the population standard deviation

n=36 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that
`z_(\alpha/2)=1.96`

Now we have everything in order to replace into formula (1):

`14.42-1.96(6.95)/(√(36))=12.150`

`14.42+ 1.96(6.95)/(√(36))=16.690`

So on this case the 95% confidence interval would be given by (12.150;16.690)