[R-meta] Meta-analysis with proportion and mean data

Viechtbauer Wolfgang (SP) wolfgang.viechtbauer at maastrichtuniversity.nl
Tue Aug 29 15:14:58 CEST 2017


Dear M,

I am not sure I fully understand. My interpretation of what you describe is this. The first group reports:

100 * (mean_post_treatment - mean_pre_treatment) / mean_pre_treatment
100 * (mean_post_control - mean_pre_control) / mean_pre_control

or 

100 * (mean_post_treatment - mean_pre_treatment) / mean_pre_treatment - 100 * (mean_post_control - mean_pre_control) / mean_pre_control

while the second group reports:

mean_post_treatment
mean_post_control

or 

mean_post_treatment - mean_post_control

Is that correct? If not, please write our in some form what exactly is reported.

Best,
Wolfgang

-- 
Wolfgang Viechtbauer, Ph.D., Statistician | Department of Psychiatry and    
Neuropsychology | Maastricht University | P.O. Box 616 (VIJV1) | 6200 MD    
Maastricht, The Netherlands | +31 (43) 388-4170 | http://www.wvbauer.com    

-----Original Message-----
From: R-sig-meta-analysis [mailto:r-sig-meta-analysis-bounces at r-project.org] On Behalf Of M West
Sent: Monday, August 28, 2017 23:09
To: r-sig-meta-analysis at r-project.org
Subject: [R-meta] Meta-analysis with proportion and mean data

Hello,

I am gathering data for a meta-analysis. So far, most of the studies report
the results as the percentage reduction in the focal trait (i.e., percent
weight loss) in the control group vs. the percentage reduction in the focal
trait (percent weight loss) in the treatment group (and the SE). However, a
small subset of studies report the actual means of the control (i.e., mean
weight +/- SE) and the treatment (mean weight +/- SE).

My question is, should I convert the smaller subset into proportions so
that all of the data are the same (or v.v.,)? i.e., is that ok to do (i.e.,
since I am using other people's data)? Or would it be better to analyze the
data separately? (my apologies, I know this seems like a rather silly
question, but I feel a bit stumped).

Many thanks in advance for your helpful (and patient) answers!

Best,
~M



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