I'm trying to calculate the weight of studies for inclusion in a meta analysis as part of homework.

I've been given a number of studies and their final results represented as a risk ratio with 95% confidence instervals.

To calculate their weight I want to use the inverse of the square root of the standard error but I'm not sure if the 95% CIs are conventionally calculated by SE or SD.

If they are SE I was going to halve the difference between the CIs and then divide by 1.96 to get the SE. If 95% CIs use SD then I would use the same method to calculate SD and then divide this by the square root of the sample size to get SE.

Does anyone know if scienctific papers (specifically in the medical field in this instance) calculate 95% CIs with SD or SE?

  • $\begingroup$ what MDs should do and what they actually do with statistics are not always the same thing. $\endgroup$
    – pjs
    Commented Feb 6 at 15:49

3 Answers 3


A standard deviation (SD) quantifies variability among a set of values.

A standard error (SE) quantifies the precision of a parameter you estimated by a statistical calculation. The parameter could be a mean, a proportion, or a risk ratio.

A confidence interval (CI) is another way to quantify the precision of an estimated parameter, the risk ratio in your example. So a CI is calculated from the SE, not the SD.

  • $\begingroup$ Not sure CIs have anything to do with a 'computed parameter'. $\endgroup$ Commented Feb 5 at 18:52
  • $\begingroup$ @GrahamBornholt. The confidence interval of a mean, tells you about the precision of the mean you calculated. The confidence interval of a proportion tells you about the precision of the proportion you calculated. Yes, that is simplified and a full explanation is a lot more complicated, but the idea is right. SD assesses variation and SEM assesses precision. $\endgroup$ Commented Feb 5 at 19:11
  • 2
    $\begingroup$ Ok, by computed parameter you mean the parameter estimate. $\endgroup$ Commented Feb 5 at 21:06
  • $\begingroup$ @GrahamBornholt. I agree and edited my answer. Thanks. $\endgroup$ Commented Feb 6 at 0:54

The formula for a 95% CI based on the a normal distribution is: $\mu \pm 1.96 \times se$ = $\mu \pm 1.96 \times \frac{sd}{\sqrt{n}}$, so you can use the standard error (se) or calculate the standard error using standard deviation (sd) and the number of observations in your sample (n). But in the end, it is the standard error itself that you want.

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    $\begingroup$ That is the equation for a CI of a mean. CIs can be calculated for other parameters too.. $\endgroup$ Commented Feb 6 at 0:53
  • $\begingroup$ Indeed, this was just by means of giving a simple, well-known example to answer the initial question as the structure of the interval stays the same for his case. $\endgroup$ Commented Feb 6 at 10:22

Others have already answered which you need, standard deviation or standard error, but to form the appropriate weights you need the inverse of the sampling variance not its square root, still less the square root of the standard error.


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