Dear Patricia, I was recently asked the exact same question, so the answer is yes, but it takes a little bit of extra work and the newest version of the metafor package (version 1.5-0, which was submitted to CRAN yesterday and which should hopefully be ready for download in a few days).
I'll briefly describe the additional difficulty that comes with this transformation and the solution. The inverse of the Freeman Tukey double arcine transformation is described in: Miller, J. J. (1978). The inverse of the Freeman-Tukey double arcsine transformation. American Statistician, 32(4), 138. It's a bit messy, but it works just fine: > xi <- c( 0, 1, 12, 38, 40) > ni <- c(10, 10, 20, 40, 40) > > dat <- cbind(xi, ni, pi = xi/ni) > > dat <- escalc(measure="PFT", xi=xi, ni=ni, data=dat, append=TRUE, add=0) > dat xi ni pi yi vi 1 0 10 0.00 0.1531387 0.02380952 2 1 10 0.10 0.3733940 0.02380952 3 12 20 0.60 0.8813361 0.01219512 4 38 40 0.95 1.3224900 0.00617284 5 40 40 1.00 1.4923885 0.00617284 > > ### check back-transformation for individual outcomes > transf.ipft(dat$yi, dat$ni) [1] 0.00 0.10 0.60 0.95 1.00 Okay, so far so good. Note that in order to back-transform a Freeman Tukey transformed value, we not only need the transformed value itself (dat$yi), but also the corresponding n (dat$ni). This is why the transf.ipft() function takes two arguments. Now we fit a fixed- (or random-) effects model to these data and then want to back-transform the estimated average, so we get an estimate in the original units (i.e., as a proportion). But what n should we use? After all, the n that applies to the estimated average is not the same as the n for any individual transformed value. Miller (1978) makes a suggestion here and proposes to use the harmonic mean of the n's. So, let's use that: > res <- rma(yi, vi, method="FE", data=dat) > res Fixed-Effects Model (k = 5) Test for Heterogeneity: Q(df = 4) = 96.4471, p-val < .0001 Model Results: estimate se zval pval ci.lb ci.ub 1.1233 0.0452 24.8643 <.0001 1.0347 1.2118 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 > ### back-transformation of the estimated average using the harmonic mean of > the n's > pred <- predict(res, transf=transf.ipft.hm, targs=list(ni=ni)) > pred pred se ci.lb ci.ub 0.8306 NA 0.7529 0.8977 In order to use the transf.ipft.hm() function, you need to supply a list with the individual n's via the targs argument, so that the harmonic mean can be computed. Okay, so that works as well. But if you now want to draw a forest plot, things get messy. For example, neither of these plots: > ### wrong transformation for the individual studies > forest(res, digits=3, atransf=transf.ipft.hm, targs=list(ni=ni), refline=NA, > xlim=c(-1.5,3.5)) > forest(res, digits=3, transf=transf.ipft.hm, targs=list(ni=ni), refline=NA, > xlim=c(-0.5,2), alim=c(0,1)) will work, since the individual values will also be back-transformed using the harmonic mean of the n's, which now yields the wrong back-transformed values. So, here is what you need to do: > ### calculate back-transformed CI bounds manually > ci.lb <- transf.ipft(dat$yi - 1.96*sqrt(dat$vi), dat$ni) > ci.ub <- transf.ipft(dat$yi + 1.96*sqrt(dat$vi), dat$ni) > > ### create forest plot with CI bounds supplied and then add the model > estimate using addpoly() > forest(dat$pi, ci.lb=ci.lb, ci.ub=ci.ub, ylim=c(-0.5,8), refline=NA, > xlim=c(-.5,1.8), alim=c(0,1), digits=3, xlab="Proportion") > addpoly(pred$pred, ci.lb=pred$ci.lb, ci.ub=pred$ci.ub, row=0, digits=3) > abline(h=0.5) And this plot has the correct back-transformed values for the individual proportions and the estimated average. Again, it will take a few days before the new version of the metafor package is available via CRAN. Then that will allow you to do this. Best, -- Wolfgang Viechtbauer Department of Psychiatry and Neuropsychology School for Mental Health and Neuroscience Maastricht University, P.O. Box 616 6200 MD Maastricht, The Netherlands Tel: +31 (43) 368-5248 Fax: +31 (43) 368-8689 Web: http://www.wvbauer.com > -----Original Message----- > From: [email protected] [mailto:[email protected]] > On Behalf Of Bruijning-Verhagen, P.C.J.L. > Sent: Wednesday, December 15, 2010 17:22 > To: [email protected] > Subject: [R] Using Metafor package: how to backtransform model > coefficients when Freeman Tukey double arcine transformation is used > > Hello, > > I am performing a meta-analysis using the metafor package. My data are > proportions and I used the Freeman Tukey double arcine (FT) > transformation to fit the random effects model. Now I want to create a > forest plot with my estimates backtransformed to the original scale of > proportions. Can this be done? > > Regards, > Patricia ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
