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Does anybody know whether it's possible to conduct a meta-analysis, outcome of interest being difference in standard deviations between two medical interventions, instead of the difference in mean? We are assessing continuous variables from RCTs.

kjetil b halvorsen
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P. Jansen
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    Variances should be quite doable. The variance is the actual parameter of a normal distribution, not the SD. You could always take the square root of the final values for interpretability. – gung - Reinstate Monica Oct 16 '15 at 18:38
  • Thank you. Any suggestions how I could do this using software packages as R or Stata? Standard meta-analyses use means and inverse of variances for weighting. How would that work for a meta-analysis where the variance is the outcome of interest. – P. Jansen Oct 21 '15 at 15:18

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One possibility you can look into is comparing the variance between your two medical interventions in each study using Levene's Test or Brown-Forsythe test. Then you can combine the p-values from each study using Stoufer's test (allows weighting) or Fisher's test (harder to weight).

For more on comparing variances look here (Levene vs Brown-Forsythe) and for more on combining p-values look here (Fisher's and Stouffer's Methods).

Community
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JustGettinStarted
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    If the OP has extracted the variances from the primary articles I do not see how s/he can do either Levene or Brown-Forsythe as they require the raw data which s/he will not have. – mdewey May 10 '16 at 14:07
  • The indicated reference contains methodology on how to use summary measures of means and variances to produce weighted Levene statistics. Have not come across Brown-Forsythe methods. See the following: Deng, W. Q., et al. (2014). Meta-analysis of SNPs involved in variance heterogeneity using Levene's test for equal variances. European Journal of Human Genetics : EJHG, 22(3), 427–430) – JustGettinStarted May 14 '16 at 04:33