Solved – Converting odds ratios to Cohen’s d for meta analysis

cohens-deffect-sizemeta-analysisodds-ratio

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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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 $$ \mathrm{SE}_{d}=\mathrm{SE}_{\mathrm{LogOR}}\times\frac{\sqrt{3}}{\pi} $$

Related Solutions

Solved – How to calculate confidence intervals for Cohen’s d

According to p238 of standard text on meta-analysis in social science The Handbook of Research Synthesis, the variance of Cohen's $d$ is $$\left( \frac{n_1 + n_2}{n_1 n_2} + \frac{d^2}{2(n_1+n_2-2)}\right) \left(\frac{n_1 + n_2}{n_1+n_2-2} \right), $$ where $n_1$ and $n_2$ are the sample sizes of the two groups being compared and $d$ is Cohen's $d$.

Taking the square-root of this variance gives the standard error of $d$, needed as input by several of the user-written meta-analysis packages for Stata. (Some of them also accept confidence intervals as input, but they simply convert them to standard errors internally anyway.)

Solved – How to calculate Standard Error of Odds Ratios

You can calculate/approximate the standard errors via the p-values. First, convert the two-sided p-values into one-sided p-values by dividing them by 2. So you get $p = .0115$ and $p = .007$. Then convert these p-values to the corresponding z-values. For $p = .0115$, this is $z = -2.273$ and for $p = .007$, this is $z = -2.457$ (they are negative, since the odds ratios are below 1). These z-values are actually the test statistics calculated by taking the log of the odds ratios divided by the corresponding standard errors (i.e., $z = log(OR) / SE$). So, it follows that $SE = log(OR) / z$, which yields $SE = 0.071$ for the first and $SE = .038$ for the second study.

Now you have everything to do a meta-analysis. I'll illustrate how you can do the computations with R, using the metafor package:

library(metafor)
yi  <- log(c(.85, .91))     ### the log odds ratios
sei <- c(0.071, .038)       ### the corresponding standard errors
res <- rma(yi=yi, sei=sei)  ### fit a random-effects model to these data
res

Random-Effects Model (k = 2; tau^2 estimator: REML)

tau^2 (estimate of total amount of heterogeneity): 0 (SE = 0.0046)
tau (sqrt of the estimate of total heterogeneity): 0
I^2 (% of total variability due to heterogeneity): 0.00%
H^2 (total variability / within-study variance):   1.00

Test for Heterogeneity: 
Q(df = 1) = 0.7174, p-val = 0.3970

Model Results:

estimate       se     zval     pval    ci.lb    ci.ub          
 -0.1095   0.0335  -3.2683   0.0011  -0.1752  -0.0438       **

Note that the meta-analysis is done using the log odds ratios. So, $-0.1095$ is the estimated pooled log odds ratio based on these two studies. Let's convert this back to an odds ratio:

predict(res, transf=exp, digits=2)

 pred  se ci.lb ci.ub cr.lb cr.ub
 0.90  NA  0.84  0.96  0.84  0.96

So, the pooled odds ratio is .90 with 95% CI: .84 to .96.

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Solved – How to calculate confidence intervals for Cohen’s d

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