Confidence interval to standard deviation converter

What this converts

Trials routinely report an effect as a point estimate with a confidence interval (a mean difference of 0.5 with a 95% CI of 0.12 to 0.88, say), but a meta-analysis needs the standard error (to weight the study) and often the standard deviation (to pool with studies that only reported SDs). This tool inverts a reported CI back to those two quantities.

The formula

A symmetric normal-based CI has a fixed width in standard-error units, so the SE falls straight out of the interval:

z    = 1.96 (95%), 1.645 (90%), 2.576 (99%)
SE   = (upper − lower) / (2 · z)

two groups:  SD = SE / √(1/n₁ + 1/n₂)
one group:   SD = SE · √n

For a difference in means, the SE of the difference equals the common SD times √(1/n₁ + 1/n₂), so dividing back out recovers the pooled within-group SD. For a single mean, SE = SD/√n, so SD = SE·√n. Both are the standard inversions in Cochrane Handbook §6.5.2.3.

When it is valid

The interval must be a large-sample normal (Wald) CI: the kind almost all trials report for means and mean differences. The SE step assumes the interval is symmetric about the point estimate; if the reported bounds are not equidistant from the estimate, the CI was not built this way and this inversion does not apply. Choosing the wrong confidence level silently rescales everything, so the tool warns you whenever the level is not 95%.

When NOT to use it

Do not use this on the CI of a ratio measure (odds ratio, risk ratio, hazard ratio): those intervals are symmetric on the log scale, not the raw scale, so (upper − lower) / 2z is meaningless. Convert on the log scale instead. For small samples where the trial used a t-based interval, the 1.96 / 1.645 / 2.576 multipliers understate the true critical value and the recovered SD will be a slight overestimate; with n below roughly 30, prefer the exact t quantile. And the two-group SD it returns is a single pooled within-arm SD. If you need per-arm SDs you cannot get them from the difference CI alone.