Can you run another t-test on the differences after your 1st t-test results? Let's say we are measuring the amount of anger (numerically) among males and females before and after watching a short video.




Males
Anger (before video)
Anger (after video)




Aaron
71
75


Bob
68
81


Carl
70
77


...
...
...




Imagine we do the same for females (see the table below).




Females
Anger (before video)
Anger (after video)




Alice
73
75


Bailey
66
78


Carol
72
76


...
...
...




1st t-test (males): I can use a t-test on the 1st table (males) to determine whether the two columns are significantly different, i.e. did the video make them angry.
1st t-test (females): Exactly the same except using the 2nd table (females).
Conclusions: Let's pretend that both males & females had significantly higher levels of anger after watching the short video.
Question: The short video gets males and females angry. More specifically, the males got more angry by an amount (respectively) of 4,13, and 7. The females have similarly calculated differences. My question is whether Males got more angry than females from watching the short video. So, can I perform a t-test on the two columns below (the difference columns)?





Difference (male)
Difference (female)




Subject 1
4
2


Subject 2
13
12


Subject 3
7
4


...
...
...




A significant result would mean that one of the sexes was affected more by the video. A non-significant result would mean that they both were affected about equal. (I know that I am simplifying significance a little bit here)

Speculation with fictitious data: If your real data
are sufficiently similar in key respects to my
fictitious data, then maybe you have your answer.
If not, please use what is below to edit your Question, saying what
possible difficulties you think might arise with your real data.
Suppose you nave male differences w1 and female
differences w2 as summarized below:
summary(w1)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  2.00    5.75    7.00    6.91    8.00   12.00 
length(w1); sd(w1)
[1] 100
[1] 2.165361
summary(w2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  0.000   5.000   6.000   6.007   7.000  10.000 
length(w2); sd(w2)
[1] 150
[1] 2.031618


par(mfrow=c(1,2))
 qqnorm(w1); qqline(w1, col="blue")
 qqnorm(w2); qqline(w2, col="blue")
par(mfrow=c(1,1))


Normal probability plots (quantile-quantile plots)
are roughly linear, so it seems OK to compare
differences for men and women with a two-sample t test.
I will use a Welch version of the test because I
can't be sure the variances of the two populations
are equal. For my fictitious data, the difference
in sample means $6.9$ and $6.0$ is highly significant
with P-value near $0.$
t.test(w1,w2)

        Welch Two Sample t-test

data:  w1 and w2
t = 3.3117, df = 202.87, p-value = 0.001098
alternative hypothesis: 
 true difference in means is not equal to 0
95 percent confidence interval:
 0.3655023 1.4411643
sample estimates:
mean of x mean of y 
 6.910000  6.006667 

The 'flat sequences' in the normal probability plots
indicate numerous ties. But with sample sizes over 100
the implementation of the Wilcoxon rank sum test in R,
handles the ties without giving a warning message.
So it would also be OK to use a Wilcoxon rank sum
test to see if there is a difference in location
of the two populations.
wilcox.test(w1,x2)

        Wilcoxon rank sum test 
        with continuity correction

data:  w1 and x2
W = 0, p-value < 2.2e-16
alternative hypothesis: 
 true location shift is not equal to 0

The shapes of the two samples
is sufficiently similar to say that the medians
of the two populations differ. Also the 'notches'
in the sides of the boxes do not overlap, which
is a further indication of different medians.
boxplot(w1,w2, horizontal=T, 
        col="skyblue2", notch=T)


Note: The fictitious data for this demonstration
were sampled in R as below. The rounding gives rise
to lots ot ties.
set.seed(12)
w1 = round(rnorm(100, 7, 2.5))
w2 = round(rnorm(150, 6, 2))

