[R-meta] 3 candidate random structures

[Timothy MacKenzie]

Dear Colleagues,

I just wanted to clarify that I fully understand that the 3 models that I
suggested need to be empirically tested with my data. However, my question
is that:

Given that I'm new to the `rma.mv()` function in metafor, I was wondering
if the candidate syntax that I have used (below) matches the general
definition of my data structure (i.e., studies>groups>outcomes>time)?

Thank you for your assistance, Tim

list(~ outcome | study, ~ time | interaction(study,outcome)), struct
=c("HCS","HAR")

list(~ outcome | study, ~ time | interaction(study,gr,outcome)), struct
=c("HCS","HAR")

list(~ outcome | interaction(study,gr), ~ time |
interaction(study,gr,outcome)), struct =c("HCS","HAR")

# Data structure:
study gr time outcome
1       1   0     1
1       1   0     2
1       1   1     1
1       1   1     2
1       1   2     1
1       1   2     2
1       2   0     1
1       2   0     2
1       2   1     1
1       2   1     2
1       2   2     1
1       2   2     2
2       1   0     1
2       1   0     2
2       2   0     1
2       2   0     2

On Mon, Jul 26, 2021 at 1:06 PM Timothy MacKenzie <fswfswt using gmail.com> wrote:

> Dear Meta-Analysis Experts,
>
> Several of my studies (15 out of 50) have a data structure similar to what
> I'm showing below. That is, each such study has multiple experimental
> groups ("gr"), tested several times, on 2 or more outcomes.
>
> I'm considering 3 different random structures (below). But I was wondering
> which one better aligns to such studies' data structure?
>
> Thank you for your expertise,
> Tim
>
> list(~ outcome | study, ~ time | interaction(study,outcome)), struct
> =c("HCS","HAR")
>
> list(~ outcome | study, ~ time | interaction(study,gr,outcome)), struct
> =c("HCS","HAR")
>
> list(~ outcome | interaction(study,gr), ~ time |
> interaction(study,gr,outcome)), struct =c("HCS","HAR")
>
> # Data structure:
> study gr time outcome
> 1       1   0     1
> 1       1   0     2
> 1       1   1     1
> 1       1   1     2
> 1       1   2     1
> 1       1   2     2
> 1       2   0     1
> 1       2   0     2
> 1       2   1     1
> 1       2   1     2
> 1       2   2     1
> 1       2   2     2
> 2       1   0     1
> 2       1   0     2
> 2       2   0     1
> 2       2   0     2
>

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