I have the results of a meta-analysis of 10 studies that reports a combined random effects odds ratio (computed using Woolf's method) and 95% confidence interval of an event happening in one group relative to another:

$OR = 7.1\ (95\%\ CI\ 4.4-11.7)$

I'm now building a model that needs to sample around this odds ratio (for the purposes of a probabilistic sensitivity analysis). Given that it's an odds ratio, I'm assuming that it's log-normally distributed and that 7.1 is the mean, but what's the best way to convert the confidence interval to a standard deviation so I can sample the distribution using Excel's LOGNORMDIST function?

(I've found similar questions for the normal and gamma distributions (From confidence interval to standard deviation - what am I missing? and How to calculate mean and standard deviation in R given confidence interval and a normal or gamma distribution?) and also questions calculating the confidence interval for a log-normal distribution (How do I calculate a confidence interval for the mean of a log-normal data set?), but I can't seem to find how to go the other way round.)


1 Answer 1


I've solved this as follows:

$$SD = \frac{\left(\frac{\ln(\text{OR})-\ln(\text{Lower CI bound})}{1.96}\right)}{\sqrt n}$$

This represents the difference in the $\ln$ of the mean and lower confidence interval bound (which gives the error), divided by 1.96 (which gives the standard error), divided by $\sqrt{n}$ (which gives the standard deviation).

Since the meta-analysis didn't make use of patient-level and just combined studies using assumptions of random effects, $n$ was simply the number of studies (10 in this case).

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    $\begingroup$ Are you sure about dividing by the square root of n? I don't think that will be helpful here. What he wants is a SD to use (with the log(OR) as the mean) to simulate odd's ratios. I think the top part of your equation (without the division by sqrt[n]) answers that. $\endgroup$ Aug 26, 2013 at 14:37
  • $\begingroup$ I would think that you would multiply by $\sqrt{n}$ as the standard error of the mean is $\frac{SD}{\sqrt{n}}$ which equals the numerator you've given. $\endgroup$ Sep 25, 2013 at 9:39
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    $\begingroup$ @probabilityislogic: What is the odds ratio of a single observation? To get the standard deviation of the log-odds, we neither need to divide nor multiply by root $n$. $\endgroup$
    – Michael M
    Nov 24, 2013 at 12:14
  • 2
    $\begingroup$ @HarveyMotulsky's suggestion is how I would calculate this. Usually we'd call this (in biostats) the standard error of the ln(OR). $\endgroup$ Jan 23, 2014 at 20:25
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    $\begingroup$ The above provides you with the standard error of the logged odds ratio. Meta-analysis of odds ratios is performed on the logged values. Under the fixed effects model, this standard error is a function of the standard errors of the individual logged odds ratios. Under the random effects model, this standard error is a function of both the standard errors of the individual logged odds ratios and the variability across these individual logged odds ratios. As such, you cannot recover the standard deviation in the individual logged odds ratios from the meta-analytic standard error. $\endgroup$
    – dbwilson
    Jun 9, 2018 at 11:53

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