The Nielsen Company reported that US residents aged 18 to 24

The Nielsen Company reported that U.S. residents aged 18 to 24 years spend an average of 46.2 hours per month using the Internet on a computer.
You think this is quite low compared with the amount of time that students at your university spend using the Internet on a computer, and you decide to do a survey to verify this. You collect an SRS of n=40 students and obtain x¯¯=40.1 hours with s=28.6 hours

a) Report the 95% confidence interval for , the average number of hours per month that students at your university use the Internet on a computer.

Use software to calculation.

95% confidence interval (±0.001) is from  hours to  hours

(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.

Solution

a)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    40.1          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    28.6          
n = sample size =    40          
              
Thus,              
Margin of Error E =    8.863068962          
Lower bound =    31.23693104          
Upper bound =    48.96306896          
              
Thus, the confidence interval is              
              
(   31.23693104   ,   48.96306896   ) [ANSWER]

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b)

As 46.2 is inside the confidence interval, then there is no significant evidence at 0.05 level that the true mean is different from the reported average by Nielsen.