[R-meta] Multi-level meta-analysis: large sigma^2

James Pustejovsky jepu@to @end|ng |rom gm@||@com
Wed Nov 24 18:08:47 CET 2021

Hi Juan,

I would guess that you probably have one or more outliers in your data
that need to be checked. Plot your data using a forest plot or even
just a histogram of the logRR values to find the outliers. Then double
check that the values actually make sense (or whether there might be
typos or something).


On Wed, Nov 24, 2021 at 7:33 AM Juan Gallego Zamorano
<j.gallego.zamorano using gmail.com> wrote:
> Dear meta-analysis list subscribers,
> I am conducting a multi-level meta-analysis for which I am using the log
> response ratio (logRR or ratio of means) as an outcome effect. I have
> several comparisons within a common control for some studies so I
> calculated a variance-covariance matrix and included it in the V argument.
> Because my dataset is hierarchical, I included the individual IDs for the
> logRR nested within the ID of the Source (papers) in the random effect
> structure to account for between observation variability. The model runs
> well and the profile plots look good, however, I'm a bit surprised by the
> value of the sigma^2 for the individual logRR which is extremely high
> compared to the Source one (see below the summary of the model) and the I^2
> are also very high (~99% for the Source/RowID and ~1% for the Source only).
> Therefore, I would like to ask:
> 1) Is this normal? I feel that this is way too high and maybe there is
> something that I am missing and needs correction but I do not know what.
> 2) If it is not normal, how can I check what is going on? As I said I run
> the profile plots and they are fine but I am not sure what else I can check.
> Multivariate Meta-Analysis Model (k = 1839; method: REML)
> Variance Components:
>             estim    sqrt  nlvls  fixed        factor
> sigma^2.1  0.0515  0.2269     62     no        Source
> sigma^2.2  5.0483  2.2468   1839     no  Source/RowID
> Test for Heterogeneity:
> Q(df = 1838) = 100501.9124, p-val < .0001
> Model Results:
> estimate      se    zval    pval   ci.lb   ci.ub
>   0.2772  0.0720  3.8518  0.0001  0.1361  0.4182  ***
> Thanks a lot in advance!
> Best,
> Juan
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