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[R-meta] Metareg: Significant reduction in tau-squared but no change in i-squared

Dear Richard,

See comments below.

Best,
Wolfgang
Not sure where you get the 23% from. The R^2 of 12% means that (an estimated) 12% of the heterogeneity was accounted for (by the moderator(s) included in the model).
First of all, all these things (I^2 in a model without moderators, I^2 in a meta-regression model, and R^2) are estimates, so they can simply be 'off'. For example, it can happen that tau^2 or I^2 actually increase when including a moderator in a model, although in theory that doesn't make sense.

Also, I^2 in a meta-regression model has a different interpretation than in a model without moderators. In a model without moderators, I^2 is an estimate how much of the total variability (which is composed of heterogeneity and sampling variability) is due to heterogeneity. However, in a meta-regression model, I^2 is an estimate how much of the *unaccounted for variability* (which is composed of residual / unaccounted for heterogeneity and sampling variability) is due to the residual / unaccounted for heterogeneity. That's different than asking how much of the *total variability* is due to residual / unaccounted for heterogeneity. So I^2 in a meta-regression model is a bit of a strange statistic anyway and I am not sure how many people actually grasp its correct interpretation.

As a more technical point: The way I^2 is computed in meta-regression models is also a bit strange. The way the 'typical' sampling variance is computed under such models actually depends on the moderators included in the model, which one could argue is a bit odd. But given that people typically compute I^2 with (Q-df)/df (where Q could be the test for heterogeneity or the test for residual heterogeneity) this is what happens.
In meta-regression models, I would report R^2 and not report I^2. You can also report the test for residual heterogeneity, so if somebody feels the need to compute I^2 based on that, they always can.