-----Original Message-----
From: Simon Harmel [mailto:sim.harmel at gmail.com]
Sent: Friday, 04 March, 2022 18:13
To: Viechtbauer, Wolfgang (SP)
Cc: R meta
Subject: Re: metafor::vcalc() for multiple sources of dependence
Thanks, Wolfgang. If I only intend to assume that the r_i for the same
repeatedly measured group on the same outcome, as well as on multiple
outcomes is constant (r_i = .6) and at the same time factor in the
shared control group, which arguments should be simultaneously used
for the Brown study below?
I did see the help page but wasn't able to find an example that was
similar to my study below.
Thank you,
Simon
m="
study yi vi group time outcome
Brown -0.10 0.05 1 0 1
Brown 0.24 0.05 1 0 2
Brown -0.11 0.05 1 0 3
Brown -0.01 0.05 1 0 4
Brown -0.12 0.05 2 0 1
Brown 0.38 0.05 2 0 2
Brown -0.19 0.05 2 0 3
Brown 0.30 0.05 2 0 4
Brown 0.45 0.05 1 1 1
Brown 0.66 0.05 1 1 2
Brown 0.27 0.05 1 1 3
Brown 0.13 0.05 1 1 4
Brown 0.08 0.05 1 2 1
Brown 0.25 0.05 1 2 2
Brown -0.34 0.05 1 2 3
Brown -0.06 0.05 1 2 4
Brown 0.48 0.05 2 1 1
Brown 0.28 0.05 2 1 2
Brown 0.10 0.05 2 1 3
Brown 0.25 0.05 2 1 4
Brown 0.11 0.05 2 2 1
Brown 0.65 0.05 2 2 2
Brown 0.39 0.05 2 2 3
Brown 0.13 0.05 2 2 4"
data <- read.table(text=m,h=T)
On Fri, Mar 4, 2022 at 4:45 AM Viechtbauer, Wolfgang (SP)
<wolfgang.viechtbauer at maastrichtuniversity.nl> wrote:
Hi Simon,
Yes, that's what vcalc() is for.
Best,
Wolfgang
-----Original Message-----
From: Simon Harmel [mailto:sim.harmel at gmail.com]
Sent: Thursday, 03 March, 2022 17:35
To: Viechtbauer, Wolfgang (SP); R meta
Subject: metafor::vcalc() for multiple sources of dependence
Hello Wolfgang,
Thank you for adding such an important function! In my studies, I have
quite a few studies that are structured as shown below.
Can I use vcalc() to construct a V matrix for the three sources of
dependence (i.e., multiple groups, times, outcomes) in say the study
below?
Thank you for your guidance,
Simon
m="
study yi vi group time outcome
Brown -0.10 0.05 1 0 1
Brown 0.24 0.05 1 0 2
Brown -0.11 0.05 1 0 3
Brown -0.01 0.05 1 0 4
Brown -0.12 0.05 2 0 1
Brown 0.38 0.05 2 0 2
Brown -0.19 0.05 2 0 3
Brown 0.30 0.05 2 0 4
Brown 0.45 0.05 1 1 1
Brown 0.66 0.05 1 1 2
Brown 0.27 0.05 1 1 3
Brown 0.13 0.05 1 1 4
Brown 0.08 0.05 1 2 1
Brown 0.25 0.05 1 2 2
Brown -0.34 0.05 1 2 3
Brown -0.06 0.05 1 2 4
Brown 0.48 0.05 2 1 1
Brown 0.28 0.05 2 1 2
Brown 0.10 0.05 2 1 3
Brown 0.25 0.05 2 1 4
Brown 0.11 0.05 2 2 1
Brown 0.65 0.05 2 2 2
Brown 0.39 0.05 2 2 3
Brown 0.13 0.05 2 2 4"
read.table(text=m,h=T)