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Statistical Power Calculator

Work out the statistical power of a two-group study — a t-test on means or a test of two proportions — from your sample size and effect size. It also gives the sample size you need and the minimum effect you can detect, with every formula and z-value shown. Free, no signup.

By Induwara AshinsanaUpdated Jul 13, 2026
Power & sample sizetwo independent groups
z-values verified two ways

Standardised mean difference. 0.2 small · 0.5 medium · 0.8 large.

Effect size presets

Participants in each arm (equal groups assumed).

Significance level (α)

False-positive risk you accept. 0.05 is the usual default.

Tails

Two-tailed tests for a difference in either direction.

Target power (for required n)

Power you want the required-n panel to reach. 80% is conventional.

Statistical power (1 − β)
80.74%
Adequate
0.807 · Adequate
Effect size used (d)
0.5
Cohen's d, medium
Required n/group for 80%
63
Total N = 126
Min. detectable d at n = 64
0.5
Smallest d detectable at 80% power

Work shown

Effect size (ES)
Cohen's d0.5
Noncentrality δ = ES·√(n/2)
2.83
Critical z (z₁₋α/₂)
Φ⁻¹ check: 1.961.96
z for target power (z₁₋β)
80% power0.84
Power = Φ(δ − z) + Φ(−δ − z)
0.8074

Normal-approximation z-test, equal group sizes. Post-hoc power uses your stated a priori effect size — not one computed from an observed p-value (see the FAQ on why observed power is a fallacy).

Method: normal (large-sample) approximation for equal group sizes — the model behind G*Power's z-tests. Sources cited: Cohen (1988, 1992) for the power and sample-size formulas and effect-size bands; Faul et al. (2007) for the reference implementation. z-constants cross-checked against the inverse-normal CDF (Acklam 2003). Full references are listed below the calculator.

How it works

This calculator uses the normal (large-sample) approximation for a two-group test with equal group sizes — the same z-test model used by G*Power's z-test family and by most online power calculators. It is quick, transparent, and accurate to the third decimal of power. Write Φ for the standard-normal cumulative distribution function and Φ⁻¹ for its inverse. The steps are:

  1. Fix the effect size (ES).For two means, ES is Cohen's d— the difference in means divided by the pooled standard deviation. For two proportions, ES is the absolute value of Cohen's h = 2·arcsin(√p₁) − 2·arcsin(√p₂), the arcsine transform that stabilises the variance so a given h means the same detectability across the 0–1 range.
  2. Scale it by sample size. The noncentrality parameter for n per group is δ = ES · √(n / 2). This is ES divided by the standard error of the difference when the two groups are equal.
  3. Compare against the critical value. The two-tailed critical value is z₁₋α/₂ (e.g. 1.959964 for α = 0.05); one-tailed uses z₁₋α. Power is then Φ(δ − z) + Φ(−δ − z) for a two-tailed test — the second term covers detecting the effect in the wrong direction and is negligible unless power is very low — or Φ(δ − z) for one-tailed.
  4. Invert for planning. The sample size needed for a target power (1 − β) is n = 2·(z + z₁₋β)² / ES², rounded up. The minimum detectable effect at your current n is MDE = (z + z₁₋β)·√(2 / n).

Φ is evaluated with the Abramowitz & Stegun 7.1.26 error-function approximation (absolute error below 1.5×10⁻⁷), and the critical z-constants are cross-checked against the inverse-normal CDF (Acklam 2003) so the values are verified two independent ways. The formulas and the small-/medium-/large-effect conventions (d and h = 0.2 / 0.5 / 0.8) come from Cohen (1988) and the "Power Primer" (Cohen 1992); the required-n figures reconcile with that paper's tables to within the one-participant noncentral-t correction described in the FAQ.

Worked examples

Two means — the power you actually have

d = 0.5, n = 64/group, α = 0.05, two-tailed

  1. Noncentrality: δ = 0.5 × √(64/2) = 0.5 × √32 = 2.828427
  2. Critical z (two-tailed, α = 0.05): z₀.₉₇₅ = 1.959964
  3. Power = Φ(2.828427 − 1.959964) + Φ(−2.828427 − 1.959964)
  4. = Φ(0.868463) + Φ(−4.788391) ≈ 0.8074 + 0.0000 = 0.807
  5. Read: 80.7% power — adequate. Cohen (1992) tables 64/group here. ✓

Two proportions — 60% vs 40%

p₁ = 0.60, p₂ = 0.40, n = 100/group, α = 0.05, two-tailed

  1. Cohen's h = 2·arcsin(√0.60) − 2·arcsin(√0.40)
  2. = 1.772154 − 1.369438 = 0.402716
  3. Noncentrality: δ = 0.402716 × √(100/2) = 0.402716 × 7.071068 = 2.847653
  4. Power = Φ(2.847653 − 1.959964) = Φ(0.887689) ≈ 0.813
  5. Read: 81.3% power — adequate to detect a 60/40 split. ✓

Planning the sample size (edge case at the boundary)

d = 0.5, target power 0.80, α = 0.05, two-tailed

  1. n = 2 × (1.959964 + 0.841621)² / 0.5²
  2. = 2 × (2.801585)² / 0.25 = 2 × 7.848879 / 0.25 = 62.79
  3. Round up: 63 per group (total N = 126). Power at 63 = 0.801.
  4. Cohen's table lists 64 — the extra participant is the noncentral-t
  5. correction the normal approximation leaves out (see FAQ).

Frequently asked questions

Sources & references

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