# [R-meta] question

Viechtbauer, Wolfgang (SP) wolfg@ng@viechtb@uer @ending from m@@@trichtuniver@ity@nl
Thu Nov 8 17:09:17 CET 2018

```Hi Frank,

I don't think treating the values as proportions makes sense. You do not have 26 'successes' out of 234 independent 'trials' in study A. You have a mean of 1 on a scale that ranges from 0 (?) to 9 based on 26 patients. You could also say that you have a mean of 1/9 on a scale that ranges from 0 to 1 based on 26 patients. The variance of the re-scaled mean is SD^2 / (9^2 * 26), where SD is the standard deviation of the scores (on the 0-9 scale) of the 26 patients. You can rescale the other studies in the same manner and then proceed to meta-analyze those rescaled means.

Best,
Wolfgang

-----Original Message-----
From: R-sig-meta-analysis [mailto:r-sig-meta-analysis-bounces using r-project.org] On Behalf Of Frank van Boven
Sent: Monday, 05 November, 2018 19:51
To: r-sig-meta-analysis using r-project.org
Subject: [R-meta] question

I have a question on the choice of an outcome measure in Metafor.
Given are data on asthma symptom scores, measured by a questionnaire.
The question is for a meta-analysis of the mean scores.
For instance.
Trial A: mean score 1, maximum score 9, totally 26 patients;
Trial B:  mean score 6, maximum score 60, totally 157 patients.
I suppose the data per trial are binomially distributed, in a meta-analysis
this would be multinomial as the maximum score differs due to different
questionnaires.

My problem is the choice of the outcome measure.
Would it be a solution to multiply both the mean score and the maximum score
by the number of patients for each trial, eventually followed by re-scaling.
Then the choice for measure = proportion.

For instance:

x <- c(26.0, 942.0)

n <- c(234.0, 9420.0)

dat <- escalc(measure="PR", xi= x, ni= n, data=dat, append=TRUE)

Or is there may be a better solution?