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

127.95 ± 6.25

Explanation:

Answer 2

`127.95-1.685(24.03)/(√(40))=121.55`

`127.95 +1.685(24.03)/(√(40))=134.35`

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

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)(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=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(24.03)/(√(40))=121.55`

`127.95 +1.685(24.03)/(√(40))=134.35`

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