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

Answer:

$14.42-1.96\frac{6.95}{\sqrt{36}}=12.150$    

$14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690$    

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".

$\bar X=14.42$ 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:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (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):

$14.42-1.96\frac{6.95}{\sqrt{36}}=12.150$    

$14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690$    

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