# A question about the trimfill function in the "meta" package in R

I was doing a meta-analysis of single proportion using the meta package in R.

I performed a double arcsine transformation to my data. I also wanted to do a trim and fill procedure. However, when backtransf= in the trimfill function is TRUE, the result came back as NA. I was not able to get a transformed back proportion. Interestingly, when I applied other transformations, including PRAW, PLOGIT, and PAS, everything worked fine. So, my question is how to get the original proportion after you perform a PFT in this situation?

Below is my code:

proportion = read.table("D:\\...\\Example.csv", header=T, sep=",")
metaproportion = metaprop(cases, total, author, data=proportion,
sm="PFT", method.tau="DL", method.ci="CP",
taf = trimfill(metaproportion, backtransf=TRUE, comb.fixed=FALSE,
comb.random=TRUE, ma.fixed=FALSE, type="R")
taf


The results:

          proportion     95%-CI            z  p-value
Random effects model         NA        16.10 < 0.0001


• Despite the appearance there is a statistical issue here independent of what software the OP is using. The link which I provide in my answer sets it out in some detail. Jun 4, 2017 at 12:38

$$y_i = \sin^{-1} \sqrt{\frac{r_i}{n_i+1}} + \sin^{-1} \sqrt{\frac{r_i+1}{n_i+1}}$$ Where the $i$th study has $r_i$ successes out of $n_i$ trials. We can weight each $y_i$ by the inverse of its variance. $$V(y_i) = (n_i + 0.5)^{-1}$$ The weights $w_i=(n_i+0.5)$ are then applied in the usual way. Having formed the weighted average $y$ it is usual to back--transform it onto the original scale by noting that for large $r_i, n_i$ reduces to $$y = 2 \sin^{-1} \sqrt{\frac{r}{n}}$$ and using $$p = \left(\sin \frac{y}{2}\right)^2$$ Incidentally Miller gives a more accurate formula for small $n$. $$p = 0.5\left(1-\mathrm{sign}(\cos y) \sqrt{1 - \left(\sin y + \frac{\sin y - \frac{1}{\sin y}}{\tilde{n}}\right)^2}\right)$$ where $\tilde{n}$ is the harmonic mean of the study sizes, $$\tilde{n} = \frac{k}{\sum{n_i^{-1}}}$$
Note that the Freeman--Tukey transformation although often referred to as variance stabilising only stabilises the variance for equal $n_i$ hence the need for the weighting step.
• @NaikeWang, questions about how to use software (eg, how to do this in meta) are off topic here. Your question is staying open because of the statistical issues mdewey addresses. You might have to code this up yourself, or ask on the r-help listserv. Jun 4, 2017 at 16:49