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

Answer:

$150-2.797\frac{35}{\sqrt{25}}=130.421$    

$150+2.797\frac{35}{\sqrt{25}}=169.579$    

Step-by-step 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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 0.99 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):

$150-2.797\frac{35}{\sqrt{25}}=130.421$    

$150+2.797\frac{35}{\sqrt{25}}=169.579$