[R-meta] Multilevel / Specification of levels

[Dudenhöffer, Tina]

Dear all,

I am running a multilevel model using the rma.mv function.

res4 <- rma.mv(Hedges_g, Hedges_v, method=“REML", random = list( ~ 1 | Study), data=Data)
res4

The question I have is regarding the random argument. If I understand correctly, this is where the levels are specified. My effect sizes (Hedges_g) are nested in studies (Study). The variable Study refers to the unique study ID in my data set. When reading the documentation, it says that the ~ 1 | id is a grouping id and outcomes with the same id receive the same random effect. Hence, in my example, all outcomes from the same study would receive the same random effect.

However, I have also seen other models that seem to include the unique effect size ID and if I translate this to my example it would look like this:

res4 <- rma.mv(Hedges_g, Hedges_v,  method=“REML", random = list(~ 1 | EScumulative, ~ 1 | Study), data=Data)
res4

As you can see the difference is the ~ 1 | EScumulative. EScumulative in my data set refers to each unique effect size. Based on the documentation and the notion that the id in the random argument refers to grouping variables, I don’t believe that I should include the EScumulative in the model, because none of them share the same value (unlike the id for Study, where some effect sizes share the same study ID. However, there are some examples (and opinions) that suggest otherwise.

That said: My question is, which of the two models is correct?

I am now working using the second model (including the EScumulative):

The variance components  look like this: so to me this almost seems like there is no variation on the study level and I wonder if I should even run multilevel?

                   estim    sqrt      nlvls  fixed        factor
sigma^2.1     0.2283  0.4778     62     no  EffectSizeID
sigma^2.2     0.0000  0.0000     28     no         Study


I am new to Meta Analysis, new to Multilevel and new to R - excuses in case some of this is not clear.

Thanks for your help and kind regards,

Tina




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