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I'm conducting a meta-analysis largely on mean differences data, however I have several articles that report only an odds ratio, N, p value, and confidence interval, and I need to convert this information to Cohen's d. I am having particular trouble calculating the variance for d and the confidence intervals for d. Here's an example from the data: the odds ratio for depression among partner violent men is 3.37, p<.0001, CI=2.08-5.47, and N=128. I calculated a d of .2909, converted the p-value to a z-value and obtained a z value of 3.291, and then calculated the SE at .1603. From here, I'm stuck on calculating the variance and confidence intervals of d. Any help would be very much appreciated! I am using Wilson's effect size calculator (not any additional meta-analysis programs).

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  • $\begingroup$ Je suis française.. thank a lot, it is the first explanation that I can understand with my poor english ! $\endgroup$ – user52829 Jul 25 '14 at 23:56
  • $\begingroup$ We have a problem. The source below and the formulas above give identical results only, if you use ln in the formula above, not LOG. Stat Med. 2000 Nov 30;19(22):3127-31. A simple method for converting an odds ratio to effect size for use in meta-analysis. Chinn S1. Abstract A systematic review may encompass both odds ratios and mean differences in continuous outcomes. A separate meta-analysis of each type of outcome results in loss of information and may be misleading. It is shown that a ln(odds ratio) can be converted to effect size by dividing by 1.81. The validity of effect size, the estim $\endgroup$ – user206446 Apr 30 '18 at 11:43
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Hi Erica and welcome to the site. Have a look at this (page 3) document and this paper. The basic formula for the conversion is $$ d=\mathrm{LogOR}\times \frac{\sqrt{3}}{\pi} $$ Applying the delta-method, we get the following expression for the the variance of $d$ (the standard error of $d$ is just the square root of its variance): $$ \mathrm{Var}_{d}=\mathrm{Var}_{\mathrm{LogOR}}\times \frac{3}{\pi^{2}} $$

Where $\mathrm{LogOR}$ denotes the log of the odds ratio and $\mathrm{Var}_{\mathrm{LogOR}}$ denotes the variance of the log odds ratio.

To get the variance of the log odds ratio, you can use the information given by the confidence interval. To get the standard error of the log odds ratio, use the following formula: $$ \mathrm{SE}_{\mathrm{LogOR}}=\frac{\log(\mathrm{CI}_{upper}) - \log(\mathrm{CI}_{lower})}{2\times z_{1-\alpha/2}} $$

Where $\mathrm{CI}_{upper}$ denotes the upper and $\mathrm{CI}_{lower}$ the lower bound of the confidence interval for the odds ratio (as given in the papers) and $z_{1-\alpha/2}$ is the $1-\alpha/2$ quantile of the standard normal distribution. For a 95%-CI, $\alpha = 0.05$ and $z_{1-\alpha/2}\approx 1.96$. To get the variance of the log odds ratio, just square the standard error. In your example, the standard error of the log odds ratio is about $0.247$. Hence, the variance of the log odds ratio is $0.247^{2}\approx0.061$. To calculate the confidence interval of $d$, you need the standard error of $d$, which is simply $$ \mathrm{SE}_{d}=\mathrm{SE}_{\mathrm{LogOR}}\times\frac{\sqrt{3}}{\pi} $$

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