# What's the statistical method of comparing the amount of pre-post difference scores of between groups?

I am currentnly doing a meta-analysis of the effects of social cognitive training on social cognitive function of patients of schizophrenia.

There are wo conditions: treatment group receving social cognitive traning and control group receving TAU, and the participants were assessed at baseline and post-treatment using a number of different measures.

Most of the previous research studies found post treatment effect size between groups.

However, I wonder if there is a way to find effect size of between groups in the amount of change from pre-test to post-test.

• You tagged this meta-analysis. Does that mean you want some equivalent of standardised mean difference? – mdewey Jul 15 at 12:35
• It is a case of second order statistics. The net effect . No need for meta analysis. – Subhash C. Davar Jul 20 at 14:50
• great go for difference in difference analysis ? – Subhash C. Davar Jul 21 at 12:20
• Let me explain my study agian. I am currently doing a meta-analysis study to assess the efficacy of cognitive programs designed to improve cognitive function in schizophrenia. As most of the meta-analysis treatment studies find d by comparing between groups, I will mainly focus on calculating the effect size(cohen's d) of the between groups, comparing the scores of cognitive function between treatment and control group after treatment. However, my professor also wants to see if we can compare the amount of pre-post difference scores between groups and find an effect size. – Joohee Lee Jul 22 at 9:52
• Assess pre -post difference for treatment group as well as for control group. Then, compute the difference between treatment group and control group. How many experiments you have for meta analysis – Subhash C. Davar Jul 25 at 17:42

First, consider just the control group. Suppose you want to see if there is a difference between baseline and post-treatment. If data are nearly normal and you have more than just a few subjects, then you can do a 'paired' t test. (This amounts to doing a one-sample t test on the 'post minus pre; differences.)

For example, if you have $$n = 30$$ subjects, then 'pre' and 'post' scores might be as follows:

      [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]  [,8]  [,9] [,10]
pre  50.71 52.64 53.12 56.73 53.56 46.60 52.03 55.22 52.44 50.34
post 54.07 58.28 58.33 62.01 58.18 53.11 55.89 61.29 57.32 55.76
[,11] [,12] [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20]
pre  55.09 43.46 49.27 48.88 47.27 47.57 49.59 47.78 50.42 54.89
post 59.41 47.69 53.09 53.26 50.92 52.31 54.66 52.92 57.05 60.20
[,21] [,22] [,23] [,24] [,25] [,26] [,27] [,28] [,29] [,30]
pre  54.00 48.46 48.40 50.63 50.12 54.25 52.85 46.45 54.66 49.67
post 59.78 53.40 54.89 54.18 55.81 61.07 58.16 51.72 59.09 54.50


Then summaries and improvement scores imp from R would be as follows:

summary(pre); sd(pre)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
43.46   48.56   50.52   50.90   53.45   56.73
[1] 3.18757
summary(post); sd(post)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
47.69   53.30   55.78   55.95   58.32   62.01
[1] 3.451662
imp = post - pre
summary(imp); sd(imp)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
3.360   4.393   5.105   5.042   5.585   6.820
[1] 0.9280126


We wish to test $$H_0: \mu_2 = \mu_1$$ against $$H_a: \mu_2 > \mu_1.$$ In R, a one-sided, paired t test can be done in either of two ways. Full results are shown for the first. The second method uses t.test(imp, alt="gr"), giving essentially the same result. Because the P-value is essentially equal to $$0$$ the difference is significant at the 5% level, the 1% level, and any other reasonable significance level.

t.test(post, pre, pair=T, alt="gr")

Paired t-test

data:  post and pre
t = 29.756, df = 29, p-value < 2.2e-16
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
4.753782      Inf
sample estimates:
mean of the differences
5.041667


Second, if you want to see whether improvement scores are different between treatment and control groups, you would do a two-sample t test. I have no idea what TAU is, but I assume there is a similar improvement scor, which we might call 'imp.c'.

imp.c
[1] 4.57 2.76 3.74 5.41 6.71 4.85 4.85 4.87 5.17 5.77
[11] 3.13 3.11 3.96 3.36 4.35 3.74 4.07 5.86 5.81 5.16
[21] 4.90 5.33 4.13 5.02 2.33 3.81 3.01 3.17 3.94 4.65
summary(imp.c); sd(imp.c)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
2.330   3.740   4.460   4.385   5.125   6.710
[1] 1.055687


Because it is reasonable to suppose that imp and imp.c may come from populations with different variances, it is best to use a Welch two-sample test. Also, it seems best to do a two-sided test because the purpose of the study is to see whether the treatment method differs from the control. There is a significant difference at the 5% level, but not (quite) at the 1% level.

t.test(imp, imp.c)

Welch Two Sample t-test

data:  imp and imp.c
t = 2.5602, df = 57.062, p-value = 0.01313
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.1431308 1.1708692
sample estimates:
mean of x mean of y
5.041667  4.384667


Here are boxplots of improvement scores:

boxplot(imp, imp.c, col="skyblue2", names=c("Treatment","Control"))


Note: If data seem far from normal, you might want to use a nonparametric one-sample Wilcoxon Signed-Rank test to judge whether improvement scores are significantly above $$0,$$ and a nonparametric two-sample Wilcoxon Rank-Sum test to judge whether Treatment and Control scores differ.

• TAU = treatment as usual. It is now slightly dated and replaced by the more general phrase standard of care. – mdewey Jul 15 at 12:33