[R-meta] Proportional meta-analysis

Ahmed Tarek tbedair.ahmed at gmail.com
Thu Feb 15 19:50:12 CET 2018

Dear Wolfgang, 

Thank you very much for your reply. I did the subgroup meta-analysis, and it did indeed show me the relationship between the material (intercept) and the other materials. I then simply stated that the pooled estimates previously calculated where signifiant or not when compared to the intercept. 

Should I also report the numbers in the first column? “estimate”. Are they of releveance? because I don’t seem to know how to interpret them. 

Ahmed Bedir

> On Feb 12, 2018, at 6:49 PM, Viechtbauer Wolfgang (SP) <wolfgang.viechtbauer at maastrichtuniversity.nl> wrote:
> Dear Ahmed,
> See my comment below.
>> -----Original Message-----
>> From: R-sig-meta-analysis [mailto:r-sig-meta-analysis-bounces at r-
>> project.org] On Behalf Of Ahmed Tarek
>> Sent: Friday, 09 February, 2018 20:46
>> To: r-sig-meta-analysis at r-project.org
>> Subject: [R-meta] Proportional meta-analysis
>> Dear All,
>> In a previous email I have been explaining about a proportional meta
>> analysis I have been performing. I am working on comparing the retention
>> rates of different materials. I first took the approach of grouping
>> different materials with similar follow up times , and calculating a
>> summary proportion of all materials, however I later found out that
>> performing a moderator analysis ( in this case “materials”) would only
>> compare the materials against the summary proportion (?) , which is not
>> my aim. I aim to see how each material performed, and compare directly
>> between them. For example, I have materials A,B,C,D and I wish to compare
>> the retention between A and B, to see if they are significantly
>> different.
> But this is exactly what a moderator analysis will do. Fit a single model to all outcomes, using 'materials' as a moderator. One level of the 'materials' factor will become the reference level to which all other levels are compared. Using contrasts, you can then also compare all other levels against each other.
> Best,
> Wolfgang

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