[R] Metafor -can't calculate heterogeneity with non-positive sampling variances

[Owen, Branwen]

Thank you very much for your quick reply Wolfgang. The 0 does make sense - I'm working on some behavioural data and in one study none reported that particular behaviour. Up til now I have been calculating I2 without that study, as it's on the small side I don't think it makes much difference. I'm just beginning to wonder if there is another way. 

I'll think about it some more
best wishes
Branwen
________________________________________
From: Viechtbauer Wolfgang (STAT) [wolfgang.viechtbauer at maastrichtuniversity.nl]
Sent: 27 August 2014 17:30
To: Owen, Branwen; r-help at r-project.org
Subject: RE: [R] Metafor -can't calculate heterogeneity with non-positive sampling variances

The warning message pretty much says it: When one of the variances is zero, then the I^2 statistic (and various other things) cannot be computed, at least if one sticks to the usual equations/methods. So, if you think the 0 sampling variances really make sense and you really want to get something like I^2, you will have to come up with a creative solution.

On the metafor package website, I explain how I^2 is computed (for the random-effects model):

http://www.metafor-project.org/doku.php/faq#how_are_i_2_and_h_2_computed_i

The crux of the problem is how to compute the 'typical' within-study variance (s^2). With any vi=0, you get division by zero in the equation given. So, you will have to compute s^2 in a different way. You could leave out the studies where vi=0, but this doesn't seem quite right, because this will inflate s^2. You could just take the simple average of the vi values and use that for s^2, but then it's not really I^2 anymore (it's I^2-like).

My question would be: How come you have studies where the sampling variance is estimated to be zero and does that really make sense? Maybe the solution is not to fix the computation of I^2, but to consider if vi=0 is really sensible.

Best,
Wolfgang

> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
> On Behalf Of Owen, Branwen
> Sent: Wednesday, August 27, 2014 13:48
> To: r-help at r-project.org
> Subject: [R] Metafor -can't calculate heterogeneity with non-positive
> sampling variances
>
> Hi, I'm doing a meta-analysis in metafor. All is fine except when there
> are 0s in the values that i'm pooling, then i get a pooled estimate but
> not the I2 that i am also interested in.
> for example:
>
> summary(rma.1<-
> rma(yi,vi,data=mix,method="ML",knha=F,weighted=F,intercept=T))
> (where yi are the study outcomes, one of which is 0, and vi is the
> variance of the study outcomes)
>
> Random-Effects Model (k = 17; tau^2 estimator: ML)
>
>   logLik  deviance       AIC       BIC      AICc
>  13.0539       Inf  -22.1077  -20.4413  -21.2506
>
> tau^2 (estimated amount of total heterogeneity): 0.0119 (SE = 0.0043)
> tau (square root of estimated tau^2 value):      0.1089
>
> Model Results:
>
> estimate       se     zval     pval    ci.lb    ci.ub
>   0.1837   0.0274   6.7154   <.0001   0.1301   0.2374      ***
>
> ---
> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
>
> Warning messages:
> 1: In rma(yi, vi, data = mix, method = "ML", knha = F, weighted = F,  :
>   There are outcomes with non-positive sampling variances.
> 2: In rma(yi, vi, data = mix, method = "ML", knha = F, weighted = F,  :
>   Cannot compute Q-test, I^2, or H^2 with non-positive sampling
> variances.
>
> Is there any way around this?
> thanks
> Branwen
> ________________________________________
> From: r-help-bounces at r-project.org [r-help-bounces at r-project.org] on
> behalf of r-help-owner at r-project.org [r-help-owner at r-project.org]
> Sent: 27 August 2014 13:36
> To: Owen, Branwen
> Subject: Metafor -can't calculate heterogeneity with non-positive
> sampling variances
>
> Message rejected by filter rule match
>
>
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