# Why is Stouffer's method often used with one-tailed $p$-values rather than two-tailed $p$-values?

Why is Stouffer's method so often performed on z's that correspond to one-tailed $p$-values when the mathematics allows for z's that correspond to two-tailed $p$-values?

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Can you clarify your question a little bit, maybe even providing an example (or a reference to one) that you're thinking of? In particular, it's not clear what you mean by "one-tailed $p$-values" and "two-tailed $p$-values" here since Stouffer's method may be combining $p$-values for tests that have nothing to do with $t$-tests. –  cardinal Jan 17 '12 at 15:09
@Joel: I was wondering if you might be interested in an alternate answer to your question. My (perhaps overly earnest) hope was that the current answer would be updated, but that doesn't appear to have happened yet. –  cardinal Feb 5 '12 at 19:42

Suppose the null hypothesis $\mu = 0$ is considered in two $2$-tailed studies. Suppose that one study rejects the null hypothesis because all data are strongly positive (supporting the alternative hypothesis $\mu > 0$ as well as the alternative hypothesis $\mu \neq 0$), while the other study rejects the null because all the data are strongly negative (supporting the alternative hypothesis $\mu < 0$ as well as the alternative hypothesis $\mu \neq 0$). Clearly, if the data from the two studies were combined, the null hypothesis would be rejected because all the data differ significantly from $0$ and thus support the alternative hypothesis $\mu \neq 0$ corresponding to a $2$-tailed study. However, the $z$'s from the studies will be positive and negative respectively, and Stouffer's method will add the two $z$ scores to get an answer close to $0$ and thus say that the null hypothesis should not be rejected at all.

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I think there may be either (a) a misunderstanding of the OP's question and/or (b) a misunderstanding of what Stouffer's method is doing here. In particular, if you are combining two $p$-values from two-tailed tests using Stouffer's method, you would get a highly positive value in the example you provide since $\Phi^{-1}(1-p_i)$ would be large in both cases. –  cardinal Jan 17 '12 at 15:14
On the other hand, it is not clear that "if the data from the two studies were combined, the null hypothesis would be rejected". As a slight application of reductio ad absurdum, suppose $X_i = -c \quad\forall i$ in study 1 and $Y_i = c \quad\forall i$ in study 2 where the sample sizes were the same. –  cardinal Jan 17 '12 at 15:14
Maybe I am misunderstanding your answer then which discusses two-tailed studies which I interpreted as using a $p$-value from a two-tailed test which would involve the absolute value of the respective means. –  cardinal Jan 17 '12 at 16:16
I think @cardinal is correct Dilip. Most of your reply looks good, but this part seems mistaken: "However, the z's from the studies will be positive and negative respectively." This presumes the z's would be calculated as if the studies performed one -tailed tests. With a correct two-tailed test, the z's will both be positive, not "positive and negative respectively" as you claim. There's still some thinking to do here: although Stouffer's method will conclude there is a significant difference, it cannot tell us in what direction! –  whuber Jan 17 '12 at 19:46
Would you care to discuss this in chat sometime, perhaps this weekend? I would be both interested and willing. Cheers. :) –  cardinal Jan 21 '12 at 3:52