# How to work out 95% confidence interval from odds ratio and standard error

$lnOR = -0.909 - (-1.707) = 0.798$ and its standard error is $SE(lnOR) = \sqrt{0.2182 + 0.3182} = 0.386$. The results of the adjusted indirect comparison suggest that x was less effective than y for (odds ratio $2.22$; 95% CI: $1.04, 4.72$).

How was the the confidence interval worked out in that example?

• Note: $\sqrt{0.2182 + 0.3182} \ne 0.386$. The $0.386$ appears to be correct (since it yields the correct CI bounds), but something is off with the square-root part. Apr 16 '16 at 14:57
• Can you explain what the numbers in your post refer to? Apr 17 '16 at 8:25
• @wolfgang I think each one should have been squared first Apr 17 '16 at 13:56

The CI was computed with $$\exp[0.798 \pm 1.96 \times 0.386] = (1.04, 4.73)$$ (the discrepancy between the 4.72 and the 4.73 is due to rounding error).