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I was conducting a meta-analysis of single proportions(i.e. without a control group) and was trying to perform a meta-regression on a moderator in my data set (lable:modb) using two different R packages (meta and metafor) in order to see if they could give me the same results. I used three transformation methods before performing meta-regression:1)no transformation,2)logit transformation, and 3)freeman-tukey double arcsine transformation. Interestingly, when I applied the logit transformation, the two packages gave me exactly the same results (I also tested the data in the Comprehensive Meta-Analysis and I got exactly the same results as I did with the logit transformation). But, when I didn't transform my data or used the double arcsine transformation, the packages gave me slightly different results. I wonder why this happened. Below is my code: The meta package code:

dat=read.table("D:\\...\\Example.csv",header=T,sep=",")
pes=metaprop(cases,total,author,data=dat,sm="PRAW",method.tau="REML",method.ci="CP",incr=0.5,allincr=FALSE,addincr=FALSE,title="") #pes=pooled effect size
mar.modb=metareg(pes,modb,method.tau = pes$method.tau) #mar.modb=meta-regression.moderator b;change "PRAW" to "PFT" if you want to use the double arcsine transformation
mar.modb

The results:

Mixed-Effects Model (k = 10; tau^2 estimator: REML)

tau^2 (estimated amount of residual heterogeneity):     0.0019 (SE = 0.0014)
tau (square root of estimated tau^2 value):             0.0434
I^2 (residual heterogeneity / unaccounted variability): 89.14%
H^2 (unaccounted variability / sampling variability):   9.20
R^2 (amount of heterogeneity accounted for):            0.00%

Test for Residual Heterogeneity: 
QE(df = 8) = 72.2927, p-val < .0001

Test of Moderators (coefficient(s) 2): 
QM(df = 1) = 0.6958, p-val = 0.4042

Model Results:

           estimate      se     zval    pval    ci.lb   ci.ub     
intrcpt    0.9868  0.0535  18.4597  <.0001   0.8820  1.0915  ***
modb      -0.0020  0.0024  -0.8341  0.4042  -0.0067  0.0027     

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

The metafor package code:

dat=read.csv("D:\\...\\Example.csv",header=T,sep=",")
ies=escalc(measure="PR",xi=cases,ni=total,data=dat)#ies=individual effect size;change "PR" to "PFT" if you want to use the double arcsine transformation
mar.modb=rma(yi,vi,data=ies,mods = ~ modb,method="REML",test="z")
mar.modb

The results:

Mixed-Effects Model (k = 10; tau^2 estimator: REML)

tau^2 (estimated amount of residual heterogeneity):     0.0016 (SE = 0.0012)
tau (square root of estimated tau^2 value):             0.0400
I^2 (residual heterogeneity / unaccounted variability): 87.45%
H^2 (unaccounted variability / sampling variability):   7.97
R^2 (amount of heterogeneity accounted for):            0.00%

Test for Residual Heterogeneity: 
QE(df = 8) = 62.1739, p-val < .0001

Test of Moderators (coefficient(s) 2): 
QM(df = 1) = 0.5879, p-val = 0.4432

Model Results:

           estimate      se     zval    pval    ci.lb   ci.ub     
intrcpt    0.9805  0.0502  19.5228  <.0001   0.8820  1.0789  ***
modb      -0.0017  0.0022  -0.7667  0.4432  -0.0061  0.0027     

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

My data is here.

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This small discrepancy is due to different handling of proportions equal to 0 or 1 (there are a couple of 1s in this dataset). The meta package does not adjust the counts for computing the proportions themselves, but it does the usual +1/2 adjustment for computing the sampling variances. The metafor package applies the +1/2 adjustment for computing the proportions and the sampling variances. With the following code, you can get metafor to do the same as the meta package:

ies <- escalc(measure="PR", xi=cases, ni=total, data=dat, add=0)
ies$vi <- escalc(measure="PR", xi=cases, ni=total, data=dat)$vi
mar.modb <- rma(yi, vi, data=ies, mods = ~ modb, method="REML", test="z")
mar.modb
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  • $\begingroup$ Hi sir, the code above works when I don't transform my data. What code should I use to get metafor to do the same as the meta package when I perform the double arcsine transformation? $\endgroup$ – Naike Wang Jun 10 '17 at 1:52
  • $\begingroup$ Just skip the ies$vi <- ... line. $\endgroup$ – Wolfgang Jun 10 '17 at 6:52

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