Meta-Analysis – How to Calculate Standard Error of Odds Ratios

geneticsmeta-analysis

I have two datasets from genome-wide association studies. The only
information available is the odds ratio and the p-value for the first
data set. For the second data set I have the Odds Ratio, p-value and allele frequencies (AFD= disease, AFC= controls) (e.g: 0.321). I'm trying to do a meta-analysis of these data but I
don't have the effect size parameter to perform this. Is there a
possibility to calculate the SE and OR confidence intervals for each of
these data only using the info that is provided??
Thank you in advance

example:
Data available: 

    Study     SNP ID      P        OR    Allele   AFD    AFC
    1         rs12345    0.023    0.85
    2         rs12345    0.014    0.91     C      0.32   0.25

With these data can I calculate the SE and CI95% OR ?
Thanks

Best Answer

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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