I am conducting a meta-analysis and am trying to produce a forest plot displaying the mean weighted effect size with and without the outliers. I looked on so many websites and tried a lot of syntaxes, however, didn't really find anything about what I am looking for.
Assume there are 10 studies in the meta-analysis. Outlier analyses showed that Study 3 is an outlier leading to an overestimation of the mean weighted ES. Accordingly, I calculate a mean ES with (k = 10) and without the outlier (k = 9). I would like to display the studies sorted according to their ES and report both Mean ESs in the forest plot. I would appreciate any help and/or advice.
Study3 Study9 Study5 ... Study7
Mean ES (k = 10) Mean ES (k = 9)
Thank you again for your response. I have been able to produce the graphs I wanted, there were only two issues, and I was wondering if there is an easier way to do it than I did.
APA style requires to write statistical terms italic, e.g., k = 10, and at sometimes also subscripted, e.g., u in Hedges g[u]. My apologies, the subscript command here seems not to work. I tried this syntax to finish the forest plot but without success as the program would draw the ** and , as well as the command "italic()" in the graphs.
Further, the command digits=2 was successful for the single ES and CIs, however not for the tables left hand side where I have mean age and the mean scores for each of the experimental groups (Note: the DV is continuous).
I created a syntax as follows, however, as I said, I was wondering if there is an easier was to do this. The meta-analyses for each of the subgroups have been already calculated.
### decrease margins so the full space is used par(mar=c(4,4,1,2)) ### draw a plot par(font=1) with(SMetaHits, forest(rma.allHits, xlim=c(-12, 6), ylim=c(-1, 41), xlab="Hedges gu", slab=paste(author), ilab=cbind(SMetaHits$N, SMetaHits$MYou, SMetaHits$MOld), ilab.xpos=c(-6, -4.5, -3.5), ilab.pos=c(2, 4, 4), order=order(SMetaHits$FacAgeNum), rows=c(36:30, 24:15, 9:4))) text(-12, 40, "Author(s), Year [Exp.]", pos=4) ### add text for the subgroups par(font=3) text(-12.1, c(37.5,25.5,10.5), pos=4, c("Young", "Mixed-age", "Old"))
This is where it gets complicated and I was hoping that there is an easier way than this
### add column headings to the plot and draw Y and O subscripted and italic par(font=3) text(c(-6), 40, c("N"), pos=2) text(c(-4.5, -3.5), 40, c("M", "M"), pos=4) text(c(-4.3, -3.3), 39.7, c("Y", "O"), pos=4, cex=.75) ##Y and O subscripted ###Command to draw 0 to the mean scores; otherwise e.g., .6 instead of .60. par(font=1) text(-3.75, c(36,16,4), "0", pos=2) ###add 0 to MY text(-2.75, c(7,4), "0", pos=2) ###add 0 to MO
At this stage, there were 2 problems: 1. I chose pos=2 to have all values left handed, which is a good solution only if the values are all in the same direction. As soon as there is one value negative, the values are not exactly underneath each other anymore. There are a lot of trial and errors to get the 0 at the right place, a similar problem I had below with gu
###Command to write Hedges *gu* as required, i.e., gu italic and u subscript. text(6, 40, "Hedges [95% CI]", pos=2) par(font=3) text(4.93, 40, "g", pos=2) text(5.02, 39.7, "u", pos=2, cex=.75)
The last command, I noticed only now that k is not italic yet
### add summary polygons for the three subgroups par(font=2) addpoly(res.YoF, row=28.5, cex=1, mlab="FE Model (k = 7)") addpoly(res.YoFNoOutl, row=27.3, cex=1, mlab="FE Model (k = 6)") addpoly(res.MxF, row=13.5, cex=1, mlab="FE Model (k = 10)") addpoly(res.MxFNoOutl, row=12.3, cex=1, mlab="FE Model (k = 9)") addpoly(res.OlF, row=2.5, cex=1, mlab="FE Model (k = 6)") addpoly(res.OlFNoOutl, row=1.3, cex=1, mlab="FE Model (k = 5)")}'