Question

The average score of 100 teenage boys playing a computer game was 80,000 with a population standard deviation of 20,000. What is the 95% confidence interval for the true mean score of all teenage boys?

Answer 1

The 95% confidence interval would be given by (76080;83920)

Explanation:

1) 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 =80000` represent the sample mean for the sample

`\mu` population mean (variable of interest)

`\sigma=20000` represent the population standard deviation

n=100 represent the sample size

2) 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):

`80000-1.96(20000)/(√(100))=76080`

`80000+1.96(20000)/(√(100))=83920`

So on this case the 95% confidence interval would be given by (76080;83920)