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kirill115 [55]
3 years ago
7

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 interv

al 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.
Mathematics
1 answer:
Rina8888 [55]3 years ago
5 0

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)    

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