# Meta-analysis: is there a way of including dependent effect sizes without averaging across them

I'm running a meta-analysis with a binary outcome (succeed/fail). I am primarily interested on the effect of predictors related to the study (e.g. age of participants, type of task) on the binary outcome, rather than the actual outcome of the study (that is i'm doing a meta-regression).

In some cases studies report multiple outcomes from the same population. The intro meta-analysis textbooks (e.g. Bornstein, 2009) only give information about how to find the weighted average of multiple effect sizes and to use these.

However, I am also interested in the different outcomes because they provide interesting (different) information. I can find lots of papers on the best way to find the average of multiple dependent effect sizes, however the only guide to this (in Cooper's 2009 handbook) is pretty undecipherable for me. Can anyone explain how to do this? If i use the same population twice, can i just divide the weight given to the study by the number of times it is included in the meta-analysis?

I've read about incorporating a third level into the meta-analysis, however in most cases i've only got one observation from a single population so i'm not sure if that would work in that case.

Thanks

• I highly recommend reading both papers, referenced in my answer to a related question. – Aleksandr Blekh May 7 '15 at 1:46
• Just to clarify: You are analyzing data from individual groups, correct? So, the observed outcomes are proportions (or some transformation thereof), not some contrast between two groups, such as (log) odds ratios. – Wolfgang May 7 '15 at 5:36
• @AleksandrBlekh I skimmed through both articles you reference in that answer, but didn't find anything that addresses the problem the OP is asking about -- or did I miss the relevant part(s)? – Wolfgang May 7 '15 at 7:04
• @Wolfgang: I recommended those papers to the OP mostly as nice general tutorials on the topic, namely determining and reporting effect sizes (ES) in context of meta-analysis (hence the comment). They also discuss alternative to Cohen's $d$ ES measures as well as aspects of determining ES for some research designs. Finally, the paper by Durlak discusses specifics of determining and reporting ES for cases of dichotomous outcomes, which corresponds to the OP's mention of "binary outcome". – Aleksandr Blekh May 7 '15 at 9:52
• @Wolfgang: I was curious enough to visit your website. Needless to say that I am very impressed. If I knew that I'm replying to the author of the metafor package, perhaps, I'd make my comment shorter to not look that stupid :-). – Aleksandr Blekh May 7 '15 at 10:14

If I understand your description correctly, you have a data structure that looks like this:

   study group obs ni xi mean.age
1      1     1   1 32  6     24.6
2      1     2   1 28  6     37.7
3      1     3   1 38 20     23.5
4      2     1   1 22  4     21.8
5      2     1   2 22  9     21.8
6      2     2   1 46 21     26.2
7      2     2   2 46 21     26.2
8      3     1   1 25  8     22.1
9      4     1   1 28 15     25.3
10     5     1   1 34 14     24.2
11     5     2   1 21  3     33.8
12     6     1   1 37  2     27.9
13     6     1   2 37  5     27.9
14     6     1   3 37 13     27.9
15     7     1   1 33  7     22.9
16     8     1   1 51  9     24.4
17     9     1   1 43 11     24.1
18     9     1   2 43 10     24.1
19    10     1   1 43 11     26.3
20    10     2   1 30 14     22.6


So, you have multiple studies, within studies there may be one or more groups, and for some groups you have multiple observations. For example, study 1 included 3 different groups, each observed once. Study 2 included 2 different groups, each observed twice. And so on. Each observation consists of the number of subjects that pass a task (xi) and the corresponding group size (ni).

And you have some predictors/covariates, such as the mean age of the group (so a variable at the group level) and possibly also variables at the study and/or observation level. And the goal is to examine how these predictors/covariates are related to the chances of passing.

The first major issue are groups that are observed more than once. What ultimately gave rise to the data are subject-level observations -- in other words, the same subjects were assessed more than once. And the chances of a particular subject passing under multiple assessments are likely to be correlated. So, to properly account for such dependence, you really would need the subject-level data. A standard approach to account for the dependent observations would then be to add random effects at the subject level to the model.

But it is probably safe to assume that you do not have the subject-level data, that is, you have data of the form shown above. So, for example, you do not know for the first group of study 2 whether the 4 subjects that passed under the first assessment are part of the 9 subjects that passed under the second assessment and so on.

So, the next best thing you can do would be to add random effects at the group level to the model, as a very rough way of accounting for dependence in multiple observations at the group level. In addition, you would probably want to add random effects at the study level and also at the observation level -- the latter is the standard way of accounting for heterogeneity in meta-analytic data.

You did not specify what outcome measure you really want to use for the meta-analysis. It is probably not a good idea to analyze the proportions directly. The more typical approach is to use logit-transformed proportions (log odds) for the meta-analysis.

You also did not mention anything about software, but I'll go with R from now on. So, to follow along, you can recreate the dataset above in R with:

dat <- structure(list(study = c(1, 1, 1, 2, 2, 2, 2, 3, 4, 5, 5, 6,
6, 6, 7, 8, 9, 9, 10, 10), group = c(1, 2, 3, 1, 1, 2, 2, 1,
1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2), obs = c(1L, 1L, 1L, 1L,
2L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 3L, 1L, 1L, 1L, 2L, 1L, 1L
), ni = c(32, 28, 38, 22, 22, 46, 46, 25, 28, 34, 21, 37, 37,
37, 33, 51, 43, 43, 43, 30), xi = c(6L, 6L, 20L, 4L, 9L, 21L,
21L, 8L, 15L, 14L, 3L, 2L, 5L, 13L, 7L, 9L, 11L, 10L, 11L, 14L
), mean.age = c(24.6, 37.7, 23.5, 21.8, 21.8, 26.2, 26.2, 22.1,
25.3, 24.2, 33.8, 27.9, 27.9, 27.9, 22.9, 24.4, 24.1, 24.1, 26.3,
22.6)), .Names = c("study", "group", "obs", "ni", "xi", "mean.age"
), row.names = c(NA, -20L), class = "data.frame")


(just copy-paste to R). Next, install and load the metafor package:

install.packages("metafor")
library(metafor)


Next, we use the escalc() command to compute the logit-transformed proportions and corresponding sampling variances:

dat <- escalc(measure="PLO", xi=xi, ni=ni, data=dat)


The dataset now looks like this:

   study group obs ni xi mean.age      yi     vi
1      1     1   1 32  6     24.6 -1.4663 0.2051
2      1     2   1 28  6     37.7 -1.2993 0.2121
3      1     3   1 38 20     23.5  0.1054 0.1056
.      .     .   .  .  .        .       .      .
20    10     2   1 30 14     22.6 -0.1335 0.1339


Variables yi and vi are the logit-transformed proportions and corresponding sampling variances. You can then fit a model with random effects at the study, group, and observations level to these data with:

res <- rma.mv(yi, vi, random = ~ 1 | study/group/obs, data=dat)
res


The output:

Multivariate Meta-Analysis Model (k = 20; method: REML)

Variance Components:

estim    sqrt  nlvls  fixed           factor
sigma^2.1  0.0000  0.0005     10     no            study
sigma^2.2  0.1395  0.3735     15     no      study/group
sigma^2.3  0.1643  0.4053     20     no  study/group/obs

Test for Heterogeneity:
Q(df = 19) = 56.9399, p-val < .0001

Model Results:

estimate       se     zval     pval    ci.lb    ci.ub
-0.8402   0.1638  -5.1302   <.0001  -1.1612  -0.5192      ***

---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


So, you get estimates of the variances of the random effects at each level, the usual test for heterogeneity, and the estimated average log odds (the estimate) with corresponding SE, z-value, p-value, and CI bounds. Most of the heterogeneity is found at the observation level, followed by group level, and next to none at the study level. For easier interpretation, you can back-transform the estimated average log odds by applying the inverse logit transformation:

predict(res, transf=transf.ilogit)


This yields:

   pred  ci.lb  ci.ub  cr.lb  cr.ub
0.3015 0.2384 0.3730 0.1227 0.5712


So, the estimated chances of passing are on average around 30% (with 95% CI: 24% to 37%). The next two values are the bounds of a 95% credibility/prediction interval for the true passing chance within an individual group.

This is a pretty complex model and I hope you have more data to work with than my little made-up dataset above. At any rate, you will want to check that all variance components of the model are actually identifiable. You can do this by profiling the restricted log-likelihood for each component. This can be done with:

par(mfrow=c(3,1))
profile(res, sigma2=1, xlim=c(0,.6))
profile(res, sigma2=2, xlim=c(0,.6))
profile(res, sigma2=3, xlim=c(0,.6))


For these data, this will look as follows (you may have to adjust the x-axis limits depending on your results): Most importantly, all profiles are peaked at the respective parameter estimates, which is what you hope to see.

You can add covariate/predictors to the model via the mods argument. For example, to add the mean age of the subjects to the model, you would use:

res <- rma.mv(yi, vi, mods = ~ mean.age, random = ~ 1 | study/group/obs, data=dat)
res


(output not shown).

You can also take another approach with analyzing these data. In particular, you can use a mixed-effects logistic regression model, using the same random effects structure. The lme4 package will allow you to fit such a model:

install.packages("lme4")
library(lme4)
res <- glmer(cbind(xi,ni-xi) ~ 1 + (1 | study/group/obs), family='binomial', data=dat)
summary(res)


The results are:

Generalized linear mixed model fit by maximum likelihood (Laplace
Approximation) [glmerMod]
Family: binomial  ( logit )
Formula: cbind(xi, ni - xi) ~ 1 + (1 | study/group/obs)
Data: dat

AIC      BIC   logLik deviance df.resid
125.3    129.3    -58.6    117.3       16

Scaled residuals:
Min      1Q  Median      3Q     Max
-1.4216 -0.4647 -0.1065  0.4460  0.8126

Random effects:
Groups            Name        Variance  Std.Dev.
obs:(group:study) (Intercept) 2.256e-01 4.750e-01
group:study       (Intercept) 1.145e-01 3.384e-01
study             (Intercept) 2.981e-09 5.459e-05
Number of obs: 20, groups:  obs:(group:study), 20; group:study, 15; study, 10

Fixed effects:
Estimate Std. Error z value Pr(>|z|)
(Intercept)  -0.8924     0.1693   -5.27 1.36e-07 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


(if you first set options(scipen=10), you won't get the scientific notation, which you may find easier to read). At any rate, the results are quite similar and can be interpreted analogously.

Final note: Again, this approach is only an approximation as the ideal analysis would make use of the subject-level data. However, averaging multiple observations (proportions, log odds, or whatever your outcome measure) for the same group is certainly even less appropriate and wastes a lot of information. So, my suggestion would be to go with the type of analysis described above and just address this issue as a limitation in your discussion.

• I should add: If you want to replicate those analyses, you will have to download the development version of metafor (see here: metafor-project.org/doku.php/installation#development_version) or wait a day or two before the next version becomes available via CRAN (I just submitted a new version). – Wolfgang May 7 '15 at 21:18
• Excellent, comprehensive answer, as expected (+1). – Aleksandr Blekh May 8 '15 at 16:43
• Thanks alot. R is perfect too (i'm not great at it, but i know the basics), and i've playing around with metafor before, I wasn't really hugely aware of this kind of analysis before though. Is this a three-level meta-analysis (as described by this paper? docdro.id/zrnq). I'll do some more homework on it. Your spot on with how my data looks like btw! – user3084100 May 8 '15 at 21:51
• Does the structure of the data matter? I think i've read that multi-level analysis works best when clusters have more than 5 observations. When I've finished i should have around 40 papers, with around 150 unique groups, with around 300 observations from those groups. Most groups will only have one observation, but some will have 1-5 observations (typically 2/3). – user3084100 May 8 '15 at 21:56
• The more observations you have for a level, themore precisely the corresponding variance component can be estimated. But with 40 papers, 150 groups, and 300 observations, you should be okay. And yes, the idea is similar to the three-level model described by Mike Cheung in that paper, except that this is now a four-level model. Another paper that describes the three-level model is Konstantopoulos (2011), which may be another useful reference. – Wolfgang May 9 '15 at 8:11