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.

a)    

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

represent the sample mean

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:

  (1)

The degrees of freedom are given by:

Since the Confidence is 0.95 or 95%, the value of and and the critical value would be

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

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

207/8 is the right