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I need to calculate the pooled odds ratio and associated 95% confidence interval for meta-analysis of 2 studies about the risk of bleeding. The only information I have is the odds ratios and 95% confidence intervals. They are 2.7 (1.8 – 4.0) in the first, and 1.3 (0.5 – 3.4) in the second study.

I computed the standard errors, weights, and pooled SE, OR and CI from the available ORs and CIs. According to my calculations, the standard errors are 0.204 in the first, and 0.489 in the second study, which gives the pooled OR and CI of 2.49 (1.72 – 3.60).

I’m not sure about this and I would appreciate if someone checks this up.

Thanks in advance!

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I did the following in Stata, the first is fixed effect and the second is random effect. I got different answers than you did.

           Study     |     ES    [95% Conf. Interval]     % Weight
---------------------+---------------------------------------------------
1                    |  2.700       1.800     4.000         63.47
2                    |  1.300       0.500     3.400         36.53
---------------------+---------------------------------------------------
I-V pooled ES        |  2.189       1.312     3.065        100.00
---------------------+---------------------------------------------------
 Heterogeneity calculated by formula
  Q = SIGMA_i{ (1/variance_i)*(effect_i - effect_pooled)^2 } 
where variance_i = ((upper limit - lower limit)/(2*z))^2 



 Heterogeneity chi-squared =   2.27 (d.f. = 1) p = 0.132
  I-squared (variation in ES attributable to heterogeneity) =  56.0%

  Test of ES=0 : z=   4.89 p = 0.000

. metan or ll ul, effect(Odds Ratio) null(1) lcols(trialname) texts(200) random

           Study     |     ES    [95% Conf. Interval]     % Weight
---------------------+---------------------------------------------------
1                    |  2.700       1.800     4.000         55.93
2                    |  1.300       0.500     3.400         44.07
---------------------+---------------------------------------------------
D+L pooled ES        |  2.083       0.721     3.445        100.00
---------------------+---------------------------------------------------
 Heterogeneity calculated by formula
  Q = SIGMA_i{ (1/variance_i)*(effect_i - effect_pooled)^2 } 
where variance_i = ((upper limit - lower limit)/(2*z))^2 

  Heterogeneity chi-squared =   2.27 (d.f. = 1) p = 0.132
  I-squared (variation in ES attributable to heterogeneity) =  56.0%
  Estimate of between-study variance Tau-squared =  0.5488

  Test of ES=0 : z=   3.00 p = 0.003
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  • $\begingroup$ Thank you very much. However, I’m not sure what to do now. As I calculated the numbers by myself, probably one of your results is completely accurate. Still, I’m not sure whether fixed effect or random effect is more appropriate in this case. Thanks again. $\endgroup$ – Jozo Apr 18 '12 at 18:43
  • $\begingroup$ I figured out that random effect should be used. Thanks again. $\endgroup$ – Jozo Apr 21 '12 at 14:51
  • $\begingroup$ So, did you ever get the same numbers? $\endgroup$ – pmgjones Apr 22 '12 at 17:52
  • $\begingroup$ Unfortunately, I have not. $\endgroup$ – Jozo Apr 30 '12 at 17:53

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