A review session is given in order to find if the group whoattends it has a better average than who doesn't. The grades areorganized into two independent groups and the data is obtained. Themean and standard deviation of the 29 students who attended thereview session are 70.5 and 11.6 percent respectively. The mean andstandard deviation of the 14 students who did NOT attend the reviewsession are 64.6 and 16.1 percent respectively. In this problem,you will calculate the 90%, 95% and 99% confidence intervals forthe difference in the mean of the two student test groups anddetermine if there is a true difference in the average of the twogroups. We will assume for this part of the problem that weconsider the variances are NOT "equal." Perform the followingsteps:
What is the standard error of the mean for the three confidenceintervals?
What is the margin of error for each of the three confidenceintervals?
Construct the 90%, 95% and 99% confidence intervals for the"true" difference between the test averages of the two groups,showing first the mean +/- the margin of error, and then showingthe range of the interval.
Is there a true "difference" in the means of the review sessionattenders vs. those who did not? On what information provided bythe confidence intervals are you basing your answer?
