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[R-meta] H2 in the context of rma.mv

Thank you Wolfgang!

The reference and rewritten calculation are very helpful.

All the best,
Marlene

-----Oorspronkelijk bericht-----
Van: Viechtbauer, Wolfgang (SP) <wolfgang.viechtbauer at maastrichtuniversity.nl>
Verzonden: vrijdag 18 maart 2022 10:58
Aan: Werner, M.A. (Marlene) <m.a.werner at amsterdamumc.nl>; r-sig-meta-analysis at r-project.org
Onderwerp: RE: H2 in the context of rma.mv

Dear Marlene,

It's not there, but you can calculate it manually. You can find a discussion on computing I^2 for more complex models here:

https://eur04.safelinks.protection.outlook.com/?url=https%3A%2F%2Fwww.metafor-project.org%2Fdoku.php%2Ftips%3Ai2_multilevel_multivariate&amp;data=04%7C01%7Cm.a.werner%40amsterdamumc.nl%7Ce44e912f5cbd4865757008da08c5e292%7C68dfab1a11bb4cc6beb528d756984fb6%7C0%7C0%7C637831943240660937%7CUnknown%7CTWFpbGZsb3d8eyJWIjoiMC4wLjAwMDAiLCJQIjoiV2luMzIiLCJBTiI6Ik1haWwiLCJXVCI6Mn0%3D%7C3000&amp;sdata=eXYGwfpD5lzaehVitYQjgFMIi7nAGEOyvvYUhOPnkMg%3D&amp;reserved=0

Using the same principles discussed there, one can also compute H^2. And yes, your calculation below is doing just that. Using the notation in the page above, H^2 for the Konstantopoulos example would be:

W <- diag(1/res$vi)
X <- model.matrix(res)
P <- W - W %*% X %*% solve(t(X) %*% W %*% X) %*% t(X) %*% W
(sum(res$sigma2) + (res$k-res$p)/sum(diag(P))) / ((res$k-res$p)/sum(diag(P)))

Note that I put sum(res$sigma2) in the numerator instead of just res$sigma2[1]. I think for H^2, that is more sensible, since we are asking how much larger the *total variance* is compared to the amount of variance expected based on sampling variability alone. Just putting part of the variance components in the numerator leads to a rather convoluted interpretation of what the H^2 value represents.

Best,
Wolfgang
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