[R-meta] Question regarding Generalized Linear Mixed-effects Model for Meta-analysis

Akifumi Yanagisawa ayanagis at uwo.ca
Tue Jan 2 21:48:41 CET 2018

Thank you for your reply, Wolfgang.

Your guess is right. I do not have a single count out of a total number of trials in each study. What I am using is the mean proportion and SD among the proportions.

I am sad to hear that I cannot use the binomial distribution in glmer() in this case, and weights argument cannot be used as usual weights.

Do you have any ideas on how to deal with this type of data?

Thank you very much.

Best regards,


On Jan 2, 2018, at 4:03 AM, Viechtbauer Wolfgang (SP) <wolfgang.viechtbauer at maastrichtuniversity.nl<mailto:wolfgang.viechtbauer at maastrichtuniversity.nl>> wrote:

Dear Aki,

Before I could even suggest a modeling approach, I would need to better understand your dependent variable. You say that you have 'proportional data', but then also mention 'means' and 'SDs'. So, it seems to me that you do not have proportions per se (that is, you do not have a single count out of a total number of trials in each study -- which we could indeed model using a binomial GLMM with a logit link).

Maybe you have studies where each participant conducted a number of trials, so that there is a proportion per participant and what is reported is the mean proportion and the SD among the proportions. But now I am just guessing.

In either case, your glmer() syntax doesn't make sense. For a binomial GLMM, the 'weights' argument is used to give the number of trials when the response is the proportion of successes, but you are using 1/vi as weights.


-----Original Message-----
From: R-sig-meta-analysis [mailto:r-sig-meta-analysis-bounces at r-project.org] On Behalf Of Akifumi Yanagisawa
Sent: Wednesday, 20 December, 2017 19:57
To: r-sig-meta-analysis at r-project.org<mailto:r-sig-meta-analysis at r-project.org>
Subject: [R-meta] Question regarding Generalized Linear Mixed-effects Model for Meta-analysis

Dear all,

I am having some difficulty dealing with proportional data; the dependent variable is learning gain from an activity, in which means and SDs are converted into proportion. The learning gains are nested in each article; each article examined the learning gains from different types of activities and measured the learning gain at different timing (i.e., immediate post and delayed post). The main thing I would like to do is to get the estimated learning gain percentage and its confidence interval for each activity.

Using the rma.mv() function, I noticed that estimation values go over 100% sometimes; then I thought I should use generalized linear mixed effects model. On the metafor’s webpage (http://www.metafor-project.org/doku.php/todo), I found that the rma.glmm() command does not support Multilevel Models so far and suggested using the LME4 package. I have been trying to figure out how to do this by myself, but I am not sure if I am doing this right. I would appreciate it if you could see if my approach is appropriate and answer to some of my questions.

(1) The approach I tried was, (1) calculated variance from means, SDs, and the numbers of participants by using the escalc function, and (2) then I tried ‘results <- glmer (learning_gain ~ ACTIVITY * TEST_TYPE * TEST_TIMING + (1|article_number/participant_group) + (1|TEST_TIMING:participant_group), weights = 1/vi, family = binomial (link = logit))’. I use the sjPlot package for plotting and the emmeans package to get estiamted learning gain percentages.  Does this sound like the proper approach? Are there other options should I add?

(2) Is it possible for me to get I^2 and H^2 values? I would like to know the proportion of variance explained by each the moderator.

(3) Is there anyway I can conduct (a) Test for Residual Heterogeneity and (b) Test of Moderators? If so, which R package would you recommend? I noticed that the anova function does not provide p-values for the test, and the LmerTest package does not work with the glmer function, either.

Any suggestions and comments will be greatly appreciated. Thank you for your help.


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