I am conducting a meta-analysis and one of the studies doesn't have the standard deviation, only the CI and mean. The CI's are of 95%. The sample size is very large. I have tried to calculate by taking the square root of the sample size and multiplying it by the CI then dividing by 3.92 but the result does not seem right.

Here is the data:

mean = 28,839.4 95% CI= (24,909.3,32,769.4) N= 3,007

The standard deviation I got was 109,881 (I'm assuming because of the large sample size)


1 Answer 1


$CI = \bar{x}\pm t_{n-1, 0.975}s/\sqrt{n}$, at least for the usual $95\%$ confidence interval for the mean.

Start by determining the half-width, $ts/\sqrt{n}$, by subtracting the sample mean from the upper endpoint of the confidence interval. Then it’s some algebra to solve for $s$.

You can find the value of $t$ from software, such as qt(n-1, 0.975) in R, but with a large sample size, it will be very close to $1.96$. (Why?) It sounds like you forgot this step of including $t$ in your calculation, but the rest of your work sounds right.


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