[R-meta] Covariance-variance matrix when studies share multiple treatment x control comparison
Ju Lee
juhyung2 @end|ng |rom @t@n|ord@edu
Thu Sep 26 08:13:03 CEST 2019
Dear Wolfgang,
I deeply appreciate your time looking into this issue, and this has been immensely helpful.
I was able to incorporate all possible inter-dependence among effect sizes by adding different layers of non-independence to our dataframe.
I manually calculated hedges'd based on based on Hedges and Olkin (1985), and it generates exactly same value as hedges' g in escalc() "SMD" function. So hopefully I am doing everything right using the equation we've discussed earlier.
I have been also wondering if it is possible to account of this variance-covariance structure that I've constructed when running publication bias analysis, for example, when using fsn() function or modified egger's regression test (looking at intercept term of residual ~ precision meta-regression using rma.mv). I had no luck so far finding information on this, and I would appreciate if you have any suggestions related to this
Thank you for all of your valuable helps!
Best regards,
JU
________________________________
From: Viechtbauer, Wolfgang (SP) <wolfgang.viechtbauer using maastrichtuniversity.nl>
Sent: Tuesday, September 24, 2019 3:59 AM
To: Ju Lee <juhyung2 using stanford.edu>; r-sig-meta-analysis using r-project.org <r-sig-meta-analysis using r-project.org>
Subject: RE: Covariance-variance matrix when studies share multiple treatment x control comparison
Hi Ju,
The equation for the covariances applies to the standardized mean difference for two independent groups (so, what one could call 'classical' Cohen's d or Hedges' g if one uses the bias correction). It does not apply to other types of standardized mean differences, such as those for dependent/paired groups. Also, the equation assumes that the pooled variance 'sdp' (that is used for computing pairwise SMDs, such as (m1 - m3) / sdp and (m2 - m3) / sdp for three groups, with group 3 being the reference group) has been computed by pooling the SDs of all three groups. Doing so isn't standard practice and takes some extra effort to implement.
Just a sidenote: When pasting tables like this, please use a fixed width font (and don't use tabs). Without that, the formating is all off and makes the table hard to read.
Even study 1 is much more complicated. It seems that rows 1 and 2 share a common group (mc=0.68 with nc=6). Then rows 3 and 4 share a common group (mt=0.84 with nt=9). But rows 1 and 4 and 2 and 3 also share a common group each. So, the var-cov matrix for this study would look like this:
[ 0.370 cov12 0 cov14 ]
[ 0.278 cov23 0 ]
[ 0.234 cov34 ]
[ 0.291 ]
In study 2, it would be:
[ 0.056 cov12 0 ]
[ 0.070 cov23 ]
[ 0.052 ]
Leaving aside the issue of whether SDs are getting pooled across all groups within a study or not, the covariance terms should be computable with the equation (19.19) from Gleser and Olkin (2009). I am not 100% sure though if the fact that the reference group is sometimes T and sometimes C matters. In any case, you should treat the group (whether it is actually a 'treatment' or a 'control' group as the 'control' group in equation (19.19)).
As for your ID variable and how to code it -- a single ID variable cannot capture these complexities. You will need 2 ID variables, indicating which group is the T and which is the C group. So, for study 1, it would be:
Study mt mc nt nc TID CID hedges.d var
1 0.92 0.68 6 6 1 3 0.937 0.370
1 0.69 0.68 9 6 2 3 0.038 0.278
1 0.84 0.69 9 9 4 2 0.659 0.234
1 0.84 0.92 9 6 4 1 -0.623 0.291
Best,
Wolfgang
-----Original Message-----
From: Ju Lee [mailto:juhyung2 using stanford.edu]
Sent: Thursday, 19 September, 2019 1:24
To: Viechtbauer, Wolfgang (SP); r-sig-meta-analysis using r-project.org
Subject: Re: Covariance-variance matrix when studies share multiple treatment x control comparison
Dear Wolfgang and all,
Please ignore my previous e-mail, and refer to this one instead:
Thank you very much for your response. I am using Hedges' d and I assumed this code applies to all SMD type of ES including Hedges' d?
I am still little baffled how I can apply this to more complex combinations. Below table looks busy, but I would appreciate if you can answer the following two questions. FYI, C_ID1 is the shared group ID.
1) In study 1, treatment 0.84 is shared in 3th and 4th response. To account for this inter-dependence, should I use "nt" instead of "nc" for calculating cov.d matrix?
2) I am not sure how to proceed if treatments and controls are shared in fashion as study 2. Clearly, 1st/2nd response shares treatment (1.00) and 2nd/3rd shares control (2.00), but 1st and 3rd do not share any experimental unit. I've categorized 2nd and 3rd with same C_ID1 because they share control, but am unclear if I should assign 1st response the same ID as 2nd and 3rd (c3) since it is still technically not independent from 2nd response (shared control) OR do I assign it a complete different control ID?
Study mt mc nt nc C_ID1 hedges.d var
1 0.92 0.68 6 6 c1 0.937 0.370
1 0.69 0.68 9 6 c1 0.038 0.278
1 0.84 0.69 9 9 c2 0.659 0.234
1 0.84 0.92 9 6 c2 -0.623 0.291
2 1.00 0.32 10 20 c? 0.760 0.056
2 1.00 2.00 10 5 c3 0.885 0.070
2 0.61 2.00 15 5 c3 0.209 0.052
I hope I am not asking too much of your time, but if you can answer this specific question, it will clarify all the doubts I have with this stage of analysis.
Thank you very much, and I hope to hear from you.
Best wishes,
JU
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