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I have a two-way factorial design, with factors treatment (A vs. B) and time (T1 vs. T2). The same subject was measured twice, at T1 and T2, so I have repeated measurements. I am interested to see whether there is interaction between treatment and time on the response variable $x$. I am applying a t-test. $H_0$ is: $(\mu_{A,T2}-\mu_{A,T1})-(\mu_{B,T2}-\mu_{B,T1})=0$, that is, there is no differences in the average change (from T1 to T2) between the treatment A and B groups.

To conduct the t-test, I computed the differences of $x$ between T2 and T1, this gives me $\Delta x_{A}$ and $\Delta x_{B}$, and then I can apply a t-test to see whether the average of $\Delta x$ are significantly different between the A and B groups.

My question is:

  1. Is this a statistically sound way to test the hypothesis?
  2. What's the difference between this method and the classical ANOVA method, which computes the mean square and uses the F-test? Which one is more powerful?
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The two approaches (t-test on “gain scores” and F-test for the interaction term in the ANOVA) give exactly the same result, as you can easily verify by running both on the same data set.

Whether this is sound is subject to some debate, other approaches (e.g. ANCOVA on T2 measurements, with group as independent variable and T1 measurements as covariate) might be more powerful in some cases. This will in particular depend on the magnitude of the correlation between T1 and T2 measures from the same participants.

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  • $\begingroup$ Thanks for the reply. I am going to verify it. So I should verify the t-test result with a repeated ANOVA result? Now I am wondering is this t-test on "gain scores" also applicable on non-repeated case (in case T1 and T2 are conducted on independent samples)? Thanks $\endgroup$ Commented May 3, 2013 at 8:36
  • $\begingroup$ You don't need to verify anything, you can just check it for yourself if you want. You should also really follow the link provided by @Jeromy Anglim to find more information and references regarding the debate I alluded to $\endgroup$
    – Gala
    Commented May 3, 2013 at 8:44
  • $\begingroup$ Regarding the non-repeated measures case, I don't really understand how you would compute a “gain score”; scores need to be paired to be able to subtract T2 scores from T1 scores. If you measure a completely new sample of participants at T2, you only have two bunch of scores with no obvious way to match them to the T1 measures. $\endgroup$
    – Gala
    Commented May 3, 2013 at 8:46

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