Question

In a random sample of 25 laptop computers the mean repair cost was $150 with a standard deviation of $35. Compute the 99% confidence interval for u

Answer 1

`150-2.797(35)/(√(25))=130.421`

`150+2.797(35)/(√(25))=169.579`

Explanation:

Information given

`\bar X=150` represent the sample mean for the sample

`\mu` population mean

s=35 represent the sample standard deviation

n=25 represent the sample size

Confidence interval

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 0.99 or 99%, the value of
`\alpha=0.01` and
`\alpha/2 =0.005` and the critical value for this case
`t_(\alpha/2)=2.797`

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

`150-2.797(35)/(√(25))=130.421`

`150+2.797(35)/(√(25))=169.579`