Question

The proprietor of a boutique in New York wanted to determine the average age of his customers. A random sample of 25 customers revealed an average age of 28 years with a standard deviation of 10 years. Determine a 95% confidence interval for the average all of all his customers. Specifically provide the lower limit and upper limit of the confidence interval to one decimal.

Answer 1

`28-2.064(10)/(√(25))=23.872`

`28+2.064(10)/(√(25))=32.128`

So on this case the 95% confidence interval would be given by (23.9;32.1)

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=28` represent the sample mean

`\mu` population mean (variable of interest)

s=10 represent the sample standard deviation

n=25 represent the sample size

Solution to the problem

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

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=25-1=24`

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: "=-T.INV(0.025,24)".And we see that
`t_(\alpha/2)=2.064`

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

`28-2.064(10)/(√(25))=23.872`

`28+2.064(10)/(√(25))=32.128`

So on this case the 95% confidence interval would be given by (23.9;32.1)