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

Answer:

$28-2.064\frac{10}{\sqrt{25}}=23.872$    

$28+2.064\frac{10}{\sqrt{25}}=32.128$    

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

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".

$\bar X=28$ 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:

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

$28-2.064\frac{10}{\sqrt{25}}=23.872$    

$28+2.064\frac{10}{\sqrt{25}}=32.128$    

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