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[R-meta] Difference in Proportions Meta-Analysis

10 messages · Wolfgang Viechtbauer, Michael Dewey, ne gic

ne gic
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Dear List,

I aim to perform a meta-analysis and present a forest plot of the
difference in proportions between two groups.

For each group I have proportion and standard error (SE).

It's straightforward to get the difference in the proportions, but am not
sure how to handle the SEs, I presume I can't just subtract them. Is there
a formula on how to handle SEs?

Sincerely,
nelly
Wolfgang Viechtbauer
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Dear Nelly,

For two independent estimates, y1 and y2, with standard errors SE1 and SE2, the SE of their difference is:

SE[y1 - y2] = sqrt(SE1^2 + SE2^2)

Best,
Wolfgang
ne gic
# 
Dear Wolfgang,

Thank you!

Sincerely,
nelly

On Mon, Jun 8, 2020 at 12:42 PM Viechtbauer, Wolfgang (SP) <
wolfgang.viechtbauer at maastrichtuniversity.nl> wrote:

            

      
      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Michael Dewey
    

    

      
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Dear Nelly

I am not sure what software you use but both meta and metafor provide 
analysis of risk differences (which is what differences in proportions 
are) so you may get what you want directly there.

Michael

        
On 08/06/2020 11:36, ne gic wrote:

            

      
      
      
    

        

      
      
    

        

  
    
  


  


      

    

    

      
      

        


  

        

      ne gic
    

    

      
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Dear Michael,

Thanks. I am using metafor (or rather planning to for this).

Initially I had performed two separate single arm meta-analysis of
proportion as follows to get an estimate for each of the arms:

rma(yi = arm1_prop_plogis, sei = arm1_se, slab=author_year ,
data=gastric_data, method="REML")

rma(yi = arm2_prop_plogis, sei = arm2_se, slab=author_year ,
data=gastric_data, method="REML")

But then it was pointed out that it would be more interesting to
meta-analyze the difference in proportions from both arms, and hence my
question.

So what I have done is:


   1. Calculate the raw proportion differences i.e. before using the R
   function "qlogis"
   2. Calculate a single SE from the SE of both arms using the equation
   provided by Wolfgang.

Then the two outputs are what I hope to provide as inputs to rma exactly as
I had done before for the single arm analysis.

Is there a more direct way to do this? or am I missing something?

Thanks for your help,

Sincerely,
nelly







On Mon, Jun 8, 2020 at 2:48 PM Michael Dewey <lists at dewey.myzen.co.uk>
wrote:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Wolfgang Viechtbauer
    

    

      
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I am not sure I really understand what you did. arm1_prop_plogis and arm2_prop_plogis actually sound like they are plogis()-transformed proportions, which doesn't make sense (plogis() is the inverse-logit transformation). The logit transformation is qlogis(). Not sure how you computed arm1_se and arm2_se. Why not use escalc(measure="PLO", ...), which will do things correctly for you?

But if you want to compare the two groups within studies directly, then you need to use measures such as the risk difference (measure="RD"), the log transformed risk ratio ("RR"), or the log transformed odds ratio ("OR"). In fact, the difference between two logit-transformed proportions *IS* the log transformed odds ratio.

So, just use escalc(measure="OR", ...) and then pass the 2x2 table counts via arguments 'ai', 'bi', 'ci', 'di' (or the 'event' counts via arguments 'ai' and 'ci' and the group sizes via 'n1i' and 'n2i'). See help(escalc) and search for "OR".

Best,
Wolfgang

            

      
      
      
    

  
  


  


      

    

    

      
      

        


  

        

      ne gic
    

    

      
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Sorry, that was a typing mistake:
I transformed the proportions using qlogis, then back transformed them
using the plogis.

Here is literally what I have done:

On excel:

1. Calculated the difference in proportions between arm1 and arm2 (arm1 -
arm2) to get arm12_prop_diff
2. Used the formula you provided to calculate the "joint" standard error
(se_diff)

In R:

# Import the excel data.

gastric_data$arm12_prop_diff <- as.numeric(gastric_data$arm12_prop_diff)
gastric_data$se_diff <- as.numeric(gastric_data$se_diff)

# Log transformation
gastric_data$arm12_prop_qlogs_diff <- qlogis(gastric_data$arm12_prop_diff)

# fit random-effects model
pes.da=rma(yi = arm12_prop_qlogs_diff, sei = se_diff, slab=author ,
data=gastric_data, method="REML")
forest(pes.da, xlab = "2-year survival difference (%)",
order=order(gastric_data$arm12_prop_diff), atransf=plogis)

Is this still not a reasonable way to go about this?


On Mon, Jun 8, 2020 at 5:27 PM Viechtbauer, Wolfgang (SP) <

        
wolfgang.viechtbauer at maastrichtuniversity.nl> wrote:

        

            

      
      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Wolfgang Viechtbauer
    

    

      
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No, this is not correct. You should not apply the logit-transformation to the difference and the SE is not correct either. You should apply the logit-transformation to the two proportions individually, compute the correct SE of the logit-transformed proportions, take the difference between the two logit-transformed proportions, and then you can use the equation to get the SE of this difference. But all of this can just be done in a single step with escalc(measure="OR", ...). As I mentioned, the difference between two logit-transformed proportions is the log odds ratio.

Best,
Wolfgang

            

      
      
      
    

  
  


  


      

    

    
1 day later

    

      
      

        


  

        

      ne gic
    

    

      
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Dear Wolfgang,

I have taken time to read the entire escalc documentation (
https://wviechtb.github.io/metafor/reference/escalc.html ) and I am sorry
to say I don't understand how I can do that in a single step with
escalc(measure="OR", ...) as you previously mentioned. I would like to do
this correctly following your suggestion so as to minimize any errors as I
redo steps that can be done in fewer steps.

For each of my two groups I have proportion and standard error. Could you
kindly show me how I can do this using one of the available datasets please?

Sincerely,
nelly

On Mon, Jun 8, 2020 at 6:48 PM Viechtbauer, Wolfgang (SP) <

        
wolfgang.viechtbauer at maastrichtuniversity.nl> wrote:

        

            

      
      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Wolfgang Viechtbauer
    

    

      
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Hi Nelly,

Don't you have the sample size on which the proportions are based? If not, the standard error of a proportion is SE[p] = sqrt[p(1-p)/n], so given p and SE[p], one can easily recover the sample size with n = p(1-p) / SE^2. Do this for both groups. Call the resulting sample sizes n1i and n2i. Also compute the number events with p*n. Do this for both groups. Call the resulting event counts ai and ci. Now you have everything to use escalc(measure="OR", ai, n1i, ci, n2i).

Best,
Wolfgang