Answer Happy

A survey of students’ use of hours per day using the internet
for their coursework found
that 149 students in the U.S. spent an average of 1.76 hours with a
standard deviation
of 1.52, and 306 students in India spent an average of .73 hours
with a standard
deviation of .79.
a. Construct and interpret a 99% confidence interval for the
difference
between students in the U.S. and India.
b. Conduct a 5-step test of significance to test if there is a
difference
between students in the U.S. and India, using a = .01 and df =
453.
c. Compare the results of the significance test and the
confidence interval.
Table 1 ttest chldidel==3 if race==1 One-sample t test Variable Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] chldidel 1,082 3.051756 .0566308 1.862801 2.940637 3.162875 mean = mean(chldidel) Ho: mean = 3 t = degrees of freedom = 0.9139 1081 Ha: mean < 3 Pr(t < t) = 0.8195 Ha: mean != 3 Pr(|T| > |t|) = 0.3610 Ha: mean > 3 Pr(t > t) = 0.1805 =