The standard error appears in almost no psychology paper, and it is the number a meta-analysis really needs in order to weight. It is hidden in the confidence interval: <strong>SE = (upper − lower) / 3.92</strong> at 95%. The subtraction is exact as long as the interval is symmetric and you know which multiplier built it, and this page also tells you <em>when it is not</em>.
SE = (upper limit − lower limit) / (2 · z)
95% → z = 1.96 → SE = width / 3.92
90% → z = 1.645 → SE = width / 3.29
99% → z = 2.576 → SE = width / 5.15
If the interval came from a t: SE = width / (2 · critical t with its df) df = 10 → 4.46 · SE
Almost every published interval is built the same way: estimate ± multiplier · SE. Subtract the lower limit from the upper and the estimate cancels itself out; what is left is 2 · multiplier · SE. Solving for the standard error is one line. The only thing you have to get right is which multiplier they used: 1.96 is the default in meta-analysis (metafor::rma), in logistic and Cox regression, in GEE and in confint.default; for means and coefficients of a linear regression — SPSS in Explore, in the T-Test and in the CIs for B; R in t.test and in confint on an lm — the default is the t with its degrees of freedom. That is why the conversion does not force you to accept anything new: the assumption that the estimator’s sampling distribution is normal, and therefore the interval symmetric, you already accepted when you read the interval as the authors published it.
A concrete case, deliberately chosen with a large sample so the multiplier does not get in the way. A mean difference of 4.20 with 95% CI [1.10; 7.30] in a trial with 200 per group: there the t is 1.966 and the standard error comes out the same, to two decimals, with the t as with the z. The width is 6.20, and 6.20 / 3.92 = 1.58. With that you recover the rest: z = 4.20 / 1.58 = 2.66, and from there p = 0.008. That is the p that should appear in the paper. If the paper says p = 0.04, one of the two numbers is misprinted and you have just found it without asking for the data.
The center of the interval, (1.10 + 7.30) / 2 = 4.20, matches the published estimate. That match is the license to use the formula, and checking it costs less than the conversion itself. When it does not hold — and for an odds ratio it never does — the interval was built on another scale, and the number that comes out of the subtraction is not the standard error of anything.
The first is the multiplier. A 95% interval for a mean with a small sample is built with the t, not the z, and the critical value is larger: 2.228 with ten degrees of freedom against 1.96. Dividing by 3.92 when you should have divided by 4.46 returns a standard error 14% larger than the real one. With df = 30 the excess drops to 4%, with df = 60 it is 2% and from there it keeps falling: with df = 120 it is 1% and tends to zero. From sixty onwards it no longer moves any result.
The second is ratios: odds ratio, risk ratio, hazard ratio. Their intervals are computed in logarithms and then exponentiated, so on the original scale they come out asymmetric. An OR of 2.50 with CI [1.20; 5.20] has its center at 3.20, not 2.50, and right there you know the shortcut does not hold. The right conversion goes through logs: (ln 5.20 − ln 1.20) / 3.92 = 0.374. The direct shortcut gives 1.020, almost three times as much and in units that are not the model’s.
The third is intervals that are not Wald: Wilson and Clopper-Pearson for proportions, the percentile of a bootstrap, Fisher’s z for correlations. They are asymmetric on purpose, because Wald works badly there. The fourth is multiplicity: a Tukey or Bonferroni interval is wider than an ordinary 95%, and converting it with 1.96 inflates the standard error without any warning. A Bayesian credible interval does not count either, unless the posterior is approximately normal.
The main use is weighting. In a fixed-effect meta-analysis each study’s weight is 1/SE²: with SE = 1.58, that study weighs 0.40. In random effects the weight is 1/(SE² + τ²). Both need each study’s standard error, and that is exactly the figure that is not in the paper: there are whole tables with means, standard deviations, p values and not a single SE.
The second use is control. With the standard error you recover z, and with z the p. Comparing that p with the published one uncovers misprints, intervals computed at a level other than the one the table footnote declares, and results described as significant whose interval crosses zero. It is the same check the reviewer runs on the whole manuscript, applied by hand to a single row.
And a precision warning that is rarely given. The limits arrive rounded, almost always to two decimals, so the width carries up to 0.01 of uncertainty. In the example above that moves the standard error by less than 0.2%. In a narrow interval — an OR of [0.98; 1.02], width 0.04 — that same 0.01 is 25% of the width, and the SE you get can be off by that same proportion: of its two significant figures, the second one is fake.
All thirteen effect size conversions, grouped by what each one is for, are in the index.
If you use it in your thesis or in a paper, cite it so that readers of your Methods section know how the calculation was done.
Ciudad-Fernández, V. (2026). Confidence interval to standard error converter [Web application]. Ciudad Science. Retrieved October 14, 2026, from https://analisisdedatospsicologia.com/en/convert/confidence-interval-to-standard-error
For your Methods section: The standard error was recovered from the reported confidence interval, using the Ciudad Science confidence interval to standard error converter (Ciudad-Fernández, 2026).
@misc{ciudadfernandez2026convintervaloaerrorestandar,
author = {Ciudad-Fern{\'a}ndez, V{\'i}ctor},
title = {{Confidence interval to standard error converter}},
year = {2026},
publisher = {Ciudad Science},
howpublished = {Web application},
url = {https://analisisdedatospsicologia.com/en/convert/confidence-interval-to-standard-error},
urldate = {2026-10-14},
note = {Accessed 2026-10-14},
}It is almost always 95%, and 1.96 is right in the vast majority of cases. Check it rather than assume it: recover the p with that multiplier and compare it with the published one. If yours comes out 0.008 and the paper says 0.03, either the interval was not 95% or it was not built with the z.
Yes. Convert the two-tailed p into its z and divide the estimate by it: SE = estimate / z. It works just as well when the paper reports the exact p. What does not work is a p written as "< 0.05" or "n.s.": that only yields a bound, and a bound weights nothing in a meta-analysis.
Not on the original scale. Take the natural log of both limits and apply the same formula: SE(ln OR) = (ln upper − ln lower) / 3.92. With OR = 2.50 and CI [1.20; 5.20], the right route gives 0.374 and the shortcut 1.020, almost three times as much. The same goes for risk ratios and hazard ratios.
No, and confusing them is expensive. For a mean, SD = SE · √n: with SE = 1.58 and n = 40, the standard deviation is 10.0. If your extraction asks for SD per group and you give it the SE, that study enters looking about six times more precise than it is, and drags the pooled result with it.