Statistical Inference – Quiz 7

Quiz Question Explanations

Question 1

Compare the average Year 1 test scores for EG and non-EG students using the ‘summarize’ command with conditional statements:

sum total_ely1 if treatment == 0
sum total_ely1 if treatment == 1

The Y1 average for EG students is 9.606 while the Y1 average for non-EG students is 9.557.

Question 2

Compare the median Year 1 test scores for EG and non-EG students using the ‘summarize’ command with conditional statements, this time including the ‘detail’ option:

sum total_ely1 if treatment == 0, det
sum total_ely1 if treatment == 1, det

The Y1 median for both EG and non-EG students is 10.

Question 3

Perform a t-test comparing Year 1 endline scores for EG and non-EG students:

ttest total_ely1, by(treatment)

The p-value for a two-sided test is 0.64 > 0.05. Thus we cannot reject the null hypothesis that the difference in means between EG and non-EG students is 0.

Question 4

First create the pass variable, assigning a 1 to students who scored 11 points and above on the Year 1 test and a 0 otherwise, and omitting students who did not take the Y1 test:

gen pass = (total_ely1 >= 11) if total_ely1 != .

Now perform a proportions test on this variable for EG vs non-EG students:

prtest pass, by(treatment)

The p-value from a two-sided test is 0.09 > 0.05, so we cannot reject the null hypothesis of no difference in EG and non-EG pass rates at the 5% level. However, note that we could reject this difference at the 10% significance level, whereas we would not be able to reject the difference in raw scores at the 10% significance level. This highlights the need to pre-specify your main outcome of interest, since with enough fishing, you can prove almost anything.