I'm planning to combine several observational studies, some of which have produced odds ratios and some hazard ratios. How do I get the standard error from the odds ratio $1.57$ ($95\%$ CI: $1.08-2.27$) and the standard error from the hazard ratio $1.56$ ($95\%$ CI: $0.96-2.52$)?


You can back-calculate the standard errors from the CIs. Odds and hazard ratios are typically analyzed on the log scale. So, you will find that on the log scale, you get a symmetric confidence interval around the log of the estimate. For example, for $OR = 1.57$, you get $log(OR) = 0.451$ and the lower and upper CI bounds on the log scale are $0.077$ and $0.820$. Now these CI bounds are calculate with $log(OR) \pm 1.96 \times SE$, which implies that $SE = (UB - LB) / (2 \times 1.96)$. So, we get: $SE = (0.820 - 0.077) / (2 \times 1.96) = 0.189$. The same things works for the hazard ratio, where you will find that $SE = 0.246$. Note that these standard errors are for the log odds ratio and the log hazad ratio. But this is the scale we work on anyway when combining results from several studies with these outcome measures.

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    $\begingroup$ @Kate If you feel that my answer is acceptable, then please consider actually accepting it. $\endgroup$ – Wolfgang Aug 1 '11 at 15:53

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