Question

1. The General Social Survey polled a sample of 209 people aged 18-30 in the year 2000, asking them how many hours per week they spend on the internet. The sample mean was 6.75 with a standard deviation of 7.71. A second sample of 541 people aged 18-30 was taken in 2006. For this sample, the mean was 7.34 and standard deviation of 10.93. Assume these are simple random samples from populations of people aged 18-30. Can you conclude that the mean number of hours per week spent on the internet increased between 2000 and 2006?

a. Do the hypothesis test with α=0.05 significance level.

b. Construct a 95% confidence interval for the difference in mean number of hours spend per week on the Internet between 2000 and 2006.

c. Calculate the Margin of Error for 95% confidence level.

Answer 1

Part a)

To Test :-

H0 :- µ1 = µ2
H1 :- µ1 < µ2

Test Statistic :-


t = -0.83

Test Criteria :-
Reject null hypothesis if t < -t(α, DF)


DF = 532
Critica value t(α, DF) = t( 0.05 , 532 ) = 1.648
t < -t(α, DF) = -0.83 > -1.648
Result :- Fail to Reject Null Hypothesis

part b)

Confidence interval :-

t(α/2, DF) = t(0.05 /2, 532 ) = 1.964

Lower Limit =
Lower Limit = -1.986
Upper Limit =
Upper Limit = 0.806
95% Confidence interval is ( -1.986 , 0.806 )

Part c)

Margin of Error = = 1.396