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--- tips:comp_two_independent_estimates [2020/07/03 10:23] – Wolfgang Viechtbauer
+++ tips:comp_two_independent_estimates [2020/07/03 10:28] – Wolfgang Viechtbauer
@@ Line 73: / Line 73: @@
 While we find that studies using random assignment obtain on average larger (i.e., more negative) effects than studies not using random assignment ($b_1 = -0.490$, $SE = 0.351$), the difference between the two estimates is not significant ($z = -1.395$, $p = .163$).
-The test of the difference between the two estimates is really just a Wald-type test, given by the equation $$z = \frac{\hat{\mu}_1 - \hat{\mu}_2}{\sqrt{SE[\hat{\mu}_1]^2 + SE[\hat{\mu}_2]^2}},$$ where $\hat{\mu}_1$ and $\hat{\mu}_2$ are the two estimates and $SE[\hat{\mu}_1]$ and $SE[\hat{\mu}_2]$ the corresponding standard errors. The test statistics can therefore also be computed with:
+The test of the difference between the two estimates is really just a [[https://en.wikipedia.org/wiki/Wald_test|Wald-type test]], given by the equation $$z = \frac{\hat{\mu}_1 - \hat{\mu}_2}{\sqrt{SE[\hat{\mu}_1]^2 + SE[\hat{\mu}_2]^2}},$$ where $\hat{\mu}_1$ and $\hat{\mu}_2$ are the two estimates and $SE[\hat{\mu}_1]$ and $SE[\hat{\mu}_2]$ the corresponding standard errors. The test statistics can therefore also be computed with:
 <code rsplus>
 with(dat.comp, round(c(zval = (estimate[1] - estimate[2])/sqrt(stderror[1]^2 + stderror[2]^2)), 3))