[R-meta] Weighted r values in a meta-analysis
Viechtbauer Wolfgang (SP)
wolfgang.viechtbauer at maastrichtuniversity.nl
Tue Aug 15 14:07:06 CEST 2017
It seems you are asking about this paper:
I can't figure out what exactly they did. The authors do write: "If there was more than one condition representing a level of agency (e.g., if an avatar was controlled by a friend in one condition but a stranger in another), we combined them using averages weighted by the respective sample size." That would usually imply:
sum(ni*ri) / sum(ni)
However, as far as I can tell, that's not what underlies the distinction between 'r' and 'r_w' in Table/Figure A1. Also, sometimes r_w is larger than r, sometimes smaller, sometimes the same. Quite mysterious. Maybe contact the authors directly?
From: R-sig-meta-analysis [mailto:r-sig-meta-analysis-bounces at r-project.org] On Behalf Of Oliver Clark
Sent: Monday, August 14, 2017 17:49
To: Michael Dewey
Cc: r-sig-meta-analysis at r-project.org; Oliver Clark
Subject: Re: [R-meta] Weighted r values in a meta-analysis
Thank you very much for your reply.
They did not specify this. From their procedure it seems they weighted before Z-transforming:
'All of the calculated r values were weighted by sample size to ensure that studies with more participants were given stronger weight (Hunter et al., 1982). Subsequently, weighted r values were transformed to Fisher’s Z values. '
A couple of examples from their paper are:
N = 90
r =.15,rw =.20
They also mention the use of Zero (citing Rosenthal (1991) and Max coded (citing Shery 2001) r values but give no details of the equations used.
They provide r and rw values for each DV in each study (total n of 170). When I try to aggregate their rw values using the MAc agg function I get a few cases of r>1 which breaks escalc, which is also strange.
I’ve perhaps given more information than required, but sticking to the original question, does this weighting of r values seem like standard practice? Multiplying by n-3 for each r seems sensible but I don’t know what the denominator would be - using the mean of all ns doesn’t reproduce their values)
On 14 Aug 2017, at 16:28, Michael Dewey <lists at dewey.myzen.co.uk<mailto:lists at dewey.myzen.co.uk>> wrote:
Not sure if I understand exactly what happened here but the variance of a z-transformed r value is (n-3)^-1 so the weight is (n-3). Is that what they did?
On 14/08/2017 16:14, Oliver Clark wrote:
I was wondering if anyone could think of a situation in which individual Pearson’s Rho values would be weighted by sample size in a meta-analysis?
If so, what the equation for doing so would be?
I am trying to reproduce a published meta-analysis and rw[i] values are given, but the equation and justification are not.
My guess was (r[i] * n[i]) / (mean(n[k]) but the numbers were different, and in fact gave r > 1.
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