Question

A random sample of 130 students is chosen from a population of 4,500 students. The mean IQ in the sample is 120, with a standard deviation of 5. Using a margin of error of 1.13%, what is the 99% confidence interval for the students' mean IQ score?

Answer 1

Answer:

$120-2.614\frac{5}{\sqrt{130}}=118.85$    

$120+2.614\frac{5}{\sqrt{130}}=121.15$    

Step-by-step explanation:

Information given

$\bar X= 120$ represent the sample mean

$\mu$ population mean (variable of interest)

s=5 represent the sample standard deviation

n=130 represent the sample size  

For this case we can't set a margin of error just with a % since they not specify 1.13% respect to something for this case we can omit this value

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)

The degrees of freedom are given by:

$df=n-1=130-1=129$

The Confidence level is 0.99 or 99%, the significance would be $\alpha=0.01$ and $\alpha/2 =0.005$, and the critical value would be $t_{\alpha/2}=2.614$

And replacing we got:

$120-2.614\frac{5}{\sqrt{130}}=118.85$    

$120+2.614\frac{5}{\sqrt{130}}=121.15$