The Nielsen Company reported that U.S. residents aged 18 to 24 years spend an av
ID: 3386720 • Letter: T
Question
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.
Explanation / Answer
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.