Question

The Nielsen Company reported that U.S. residents aged 18 to 24 years spend an average of 32.5 hours per month using the Internet on a computer.13 You wonder if this it true for students at your large university because so many students use their smartphones to access the Internet. You collect an SRS of n=75 students and obtain ¯x=28.5 hours with s=23.1 hours.
Required:
a. Report the 95% confidence interval for μ, the average number of hours per month that students at your university use the Internet on a computer.
b. Use this interval to test whether the average time for students at your university is different from the average reported by Nielsen. Use the 5% significance level. Summarize your results.

Answer 1

Answer:

a) $28.5-1.993\frac{23.1}{\sqrt{75}}=23.18$    

$28.5+1.993\frac{23.1}{\sqrt{75}}=33.82$    

b) For this case since the value 32.5 is in the confidence interval obtained then we can't conclude that the statement by Nielsen is wrong

Step-by-step explanation:

Information given

$\bar X=28.5$ represent the sample mean

$\mu$ population mean (variable of interest)

s=23.1 represent the sample standard deviation

n=75 represent the sample size  

Part a

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:

$df=n-1=75-1=74$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ and the critical value would be $t_{\alpha/2}=1.993$

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

$28.5-1.993\frac{23.1}{\sqrt{75}}=23.18$    

$28.5+1.993\frac{23.1}{\sqrt{75}}=33.82$    

Part b

For this case since the value 32.5 is in the confidence interval obtained then we can't conclude that the statement by Nielsen is wrong