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.

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

Step-by-step 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".

represent the sample mean

population mean (variable of interest)

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:

  (1)

Since the Confidence is 0.95 or 95%, the value of and , 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

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

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