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In a meta-analysis of the risk difference, does it make sense to apply the weights & back calculate the expected proportions in each group?

For example, if you had the following:

Study    RD     [95% Conf.  Interval]   % Weight
                        
1        -0.132 -0.336  0.071           34.28
2        -0.033 -0.220  0.153           40.74
3        -0.080 -0.318  0.159           24.98
                            
PooledRD -0.079 -0.198  0.040           100.00
                        

Study   n/N      prop1      n/N      prop0
1       23/35    .6571429   30/38    .7894737
2       32/50    .64        33/49    .6734694
3       12/32    .375       15/33    .4545455

Would it be correct to apply the weights the proportions? ie, to give 58.0% vs 65.9%? [note: calculated by weighting the proportions by the meta-analysis weights = .6571429*.3428 + .64*.4074 + .375*.2498 = 58.0% for prop1]

Or would you use the raw proportions (57.3% vs 65.0%) [note: calculated by summing events over total = (23+32+12)/(35+50+32)*100 = 57.3% for prop1]

Megan Moreton
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  • Can you edit to clarify this? Where do 58.0, 65.9, 57.3, and 65.0 come from? Is there any reason you want to do this or is it just curiosity? – mdewey Jul 07 '20 at 13:43
  • Sorry - I wasn't very clear. The 57.3% and 65.0% come from summing the total with the event over the total in arms 0 and 1 (ie, for arm 1 [23+32+12]/[35+50+32]*100 = 57.3%). The 58.0% and 65.9% come from weighting the proportions by the weights from the meta-analysis (ie, for arm 1 (23/35)*.3428 + (32/50)* .4074 + (12/32)*.2498 = 58.0%). Regarding the reason, I have been asked to present the proportions by arm rather than just the risk difference, & am unsure whether the best way to proceed is to use the raw proportions or the weighted proportions? – Megan Moreton Jul 07 '20 at 16:07
  • Would you not want to meta-analyse the proportions, which for arm 1 gives 56.3 using the logit transformation? Arm 2 would be 64.8. Admittedly this is incompatible with the risk difference as it yields -0.084. – mdewey Jul 07 '20 at 16:18
  • Thank you, I hadn't considered this - but it makes sense. I am concerned about the overall message to the "lay" public if the proportions/differences don't match up – Megan Moreton Jul 07 '20 at 17:17

1 Answers1

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Just to summarise the discussion from the comments.

In a comparative study the usual procedure is to compute some measure of discrepancy (a difference or a ratio usually) and then meta-analyse that using its standard error. An alternative approach might be to meta-analyse the values being compared and then compute the discrepancy measure from those summaries. In the specific case being considered here we have a risk difference and underlying it proportions. In general since in both cases the measures are analysed transformed and weighted the two procedures I outlined above will not give the same result.

I suppose this is an instance of Jensen's inequality.

mdewey
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