Discovering Statistics

A Group Contrast Meta-analysis

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  • #743
    Jon
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    I’m looking at a meta-analysis of randomized trials that consist of comparing two treatments for a disorder (i.e., behavioral treatment versus a comparison treatment).  My meta-analysis is thus based on whether behavioral treatments are more/less effective than their compared treatment (i.e., a group contrast) as opposed to just a meta-analysis solely looking at the overall effectiveness of a treatment.  

     Some studies are only giving me proportions in the forms of percentage of people who no longer have the disorder at the end of treatment. It is with these studies that I’m questioning my calculation of the inverse variance weight of the effect size.  

    I’m using the 1/(SE^2) formula recommended by Lipsey and Wilson, but it it’s giving me what seem to be huge w values.  I provided an example below:  

    Study 1 has n=59 in Treatment1 versus n=62 in Treatment2.  Treatment1 had 31% who were “cured” while Treatment2 had 22% who were cured.  

    Weighted mean = (59 * .31 + 62 * .22) / (59 + 62) = .2639

    Standard error of the effect size = Sqrt( .2639(1 – .2639) * (1/59 + 1/62)) = .0798

    Inverse variance weight of the effect size = 1 / .0798^2 = 157.0342

    157 seems huge…am I doing this right?   

      

    #744

    Why do you compute Weighted mean for data in the form of percentages? I think you should use  Odds Ratios or ϕ or Cramér’s V (if I am not wrong). 

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