[R-meta] Weighted r values in a meta-analysis

Oliver Clark oliver.clark3 at stu.mmu.ac.uk
Mon Aug 14 17:48:45 CEST 2017

Dear Michael,

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=.16,rw =.21

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)

Best wishes,

On 14 Aug 2017, at 16:28, Michael Dewey <lists at dewey.myzen.co.uk<mailto:lists at dewey.myzen.co.uk>> wrote:

Dear Oliver

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:
Hi all,

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.

Many thanks!

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