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Cohen's d Effect Size Calculator (with Hedges' g)

Turn a mean difference into a standardised effect size. Enter means and SDs, paste raw data, or work from a t value — and get Cohen's d, the bias-corrected Hedges' g, a 95% confidence interval, the overlap, and a small/medium/large verdict, with every step shown.

By Induwara AshinsanaUpdated Jun 12, 2026
Cohen's d & Hedges' g
Cohen 1988 · cross-checked
Input mode

Standard deviation, > 0

Whole number ≥ 2

Standard deviation, > 0

Whole number ≥ 2

Try an example
Method: Independent two-sample d (pooled SD), df = 58
Cohen's d
0.5708
Hedges' g
0.5634
bias-corrected
95% CI for d
[0.055, 1.087]
normal approximation
Cohen's U₃
71.6%
overlap 77.5%
MagnitudeMedium|d| = 0.571
0.20.50.8

Cohen's 0.2 / 0.5 / 0.8 cut-offs are rules of thumb, not field-specific truths — interpret alongside your domain's typical effects.

Cohen's U₃

About 71.6% of the second group scores below the first group's mean.

Distribution overlap

The two distributions overlap by about 77.5% (assuming equal-variance normals).

Cohen's d = 0.57, 95% CI [0.05, 1.09], a medium effect (Hedges' g = 0.56).

Show working

  1. Group 1: M₁ = 78, s₁ = 10, n₁ = 30
  2. Group 2: M₂ = 72, s₂ = 11, n₂ = 30
  3. s_pooled = √[((n₁−1)s₁² + (n₂−1)s₂²)/(n₁+n₂−2)] = 10.511898
  4. d = (mean difference) / pooled SD = 6 / 10.511898 = 0.570782
  5. df = 58; J = 1 − 3/(4·df − 1) = 0.987013
  6. g = J · d = 0.987013 · 0.570782 = 0.563369
  7. SE(d) ≈ 0.263404; 95% CI = d ± 1.96·SE(d)

Cross-check: d computed directly and d recovered from the t statistic (d = t·√(1/n₁ + 1/n₂), Lakens 2013) agree to machine precision.

Sources cited

How it works

An effect size answers the question a p-value cannot: how big is the difference? Cohen's d expresses the gap between two means in standard-deviation units, so a result is comparable across studies, scales, and sample sizes. Journals and supervisors now expect it reported alongside the t-test p-value.

  1. Pooled SD (independent design), per Cohen (1988). s_pooled = √[((n₁−1)s₁² + (n₂−1)s₂²) / (n₁+n₂−2)] weights each group's variance by its degrees of freedom.
  2. Cohen's d. Divide the mean difference by the standardiser: d = (M₁ − M₂) / s_pooled. A one-sample design uses the sample SD against a reference μ₀; a paired design uses the SD of the difference scores (the d_z of Lakens 2013).
  3. Hedges' g (bias correction), per Hedges (1981). Small samples make d run high, so g multiplies it by J = 1 − 3/(4·df − 1), with df = n₁+n₂−2 (independent) or n−1 (one-sample and paired).
  4. Confidence interval (Lakens 2013). The standard error uses the large-sample normal approximation SE(d) ≈ √((n₁+n₂)/(n₁n₂) + d²/(2(n₁+n₂))), and the interval is d ± 1.96·SE. This is an approximation, labelled as such — it is close for moderate samples and slightly narrow for tiny ones.
  5. Overlap (Cohen 1988; McGraw & Wong 1992).Cohen's U₃ = Φ(d) is the share of one group below the other's mean, and the overlap coefficient 2·Φ(−|d|/2) is how much the two equal-variance normal curves share. Both use the standard-normal CDF.
  6. Magnitude.The verdict reads |d| against Cohen's 0.2 / 0.5 / 0.8 cut-offs. These are deliberately blunt benchmarks; the page flags them as rules of thumb rather than field-specific truth.

As a credibility check, the tool also recovers d from the t statistic — d = t·√(1/n₁ + 1/n₂) (Lakens 2013) — and confirms it equals the value computed directly from the means and SDs, to machine precision. If you are sizing a future study instead, the sample size calculator turns a target d back into a required n.

Worked examples

Independent groups (the default)

Intervention M₁=78, s₁=10, n₁=30 vs control M₂=72, s₂=11, n₂=30

  1. Pooled variance: ((29·100) + (29·121)) / 58 = 6409 / 58 = 110.5
  2. s_pooled = √110.5 = 10.512
  3. d = (78 − 72) / 10.512 = 0.571 → Medium
  4. df = 58, J = 1 − 3/231 = 0.9870, g = 0.571 · 0.9870 = 0.563
  5. SE = √(60/900 + 0.571²/120) = 0.263; 95% CI = 0.571 ± 0.516 = [0.055, 1.087]
  6. U₃ = Φ(0.571) ≈ 71.6% of control scores fall below the treatment mean

One-sample vs a reference

Class M=65, s=12, n=25 against a national benchmark μ₀=60

  1. d = (65 − 60) / 12 = 0.417 → Small
  2. df = 24, J = 1 − 3/95 = 0.9684, g = 0.417 · 0.9684 = 0.404
  3. SE = √(1/25 + 0.417²/50) = 0.209; 95% CI = [0.008, 0.825]
  4. The class outperforms the benchmark by about 0.42 SD

Edge case — a negative effect

Group 1 M₁=50, s₁=10, n₁=20 vs Group 2 M₂=55, s₂=10, n₂=20

  1. s_pooled = √100 = 10 (equal variances)
  2. d = (50 − 55) / 10 = −0.5 → Medium (magnitude), group 1 below group 2
  3. df = 38, J = 0.9801, g = −0.490
  4. 95% CI = [−1.129, 0.129] — it includes 0, so the effect is not conclusive
  5. U₃ = Φ(−0.5) ≈ 30.9%: most of group 2 scores above group 1's mean

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