Question

A random sample of the costs of repair jobs at a large muffler repair shop produces a mean of $127.95. and a standard deviation of $24.03. If the size of this sample is 40, which of the following is an approximate 90 percent confidence interval for the average cost of a repair at this repair shop

Answer 1

Answer:

$127.95-1.685\frac{24.03}{\sqrt{40}}=121.55$    

$127.95 +1.685\frac{24.03}{\sqrt{40}}=134.35$    

So on this case the 90% confidence interval would be given by (121.55;134.35)    

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".

$\bar X=127.95$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s=24.03 represent the sample standard deviation

n 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:

$df=n-1=40-1=39$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,39)".And we see that $t_{\alpha/2}=1.685$

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

$127.95-1.685\frac{24.03}{\sqrt{40}}=121.55$    

$127.95 +1.685\frac{24.03}{\sqrt{40}}=134.35$    

So on this case the 90% confidence interval would be given by (121.55;134.35)