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What is my test’s statistical power?

How to calculate power in statistics: pick the test, then type the effect size (Cohen’s d), the sample size and α. The statistical power calculator gives the chance your test detects a real effect, with β and the noncentrality.

Your numbers

Alternative
About 0.2 is small, 0.5 medium and 0.8 large.
Power (1 − β)
75.40%

The power is 75.40%, so the chance of missing a real effect of d = 0.5 is 24.60%.

Type II error rate (β)
24.60%
Noncentrality (λ)
2.73861
Critical value
2.04523
Degrees of freedom (ν)
29
Total sample size
30

Power (1 − β): 75.40%. The power is 75.40%, so the chance of missing a real effect of d = 0.5 is 24.60%.

Power (1 − β) by effect size (Cohen’s d)

How to calculate

Finds the statistical power of a one-sample, paired or two-sample t test or z test from the effect size (Cohen’s d), the sample size and α, with β, the noncentrality and the critical value.

Example with the default inputs (Test t test: one sample or paired, Alternative Two-sided, Effect size (Cohen’s d) 0.5, Sample size (n per group) 30, Significance level (α) 0.05): The power is 75.40%, so the chance of missing a real effect of d = 0.5 is 24.60%.

Method: λ = d√n (one sample) or d√(n/2) (two groups of n). z test: power = Φ(λ − z(1 − α/2)) + Φ(−λ − z(1 − α/2)), or Φ(λ − z(1 − α)) one-sided. t test: power = P(T′ > t(1 − α/2; ν)) + P(T′ < −t(1 − α/2; ν)) for T′ noncentral t with ν = n − 1 or 2(n − 1) and noncentrality λ; one-sided P(T′ > t(1 − α; ν)).

  • The data are normal (or n is large), the observations independent, and two groups have the same size and standard deviation.
  • d is the true effect in the tested direction, so it is 0 or more.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. Test z test: one sample, Alternative One-sided, Effect size (Cohen’s d) 1, Sample size (n per group) 9, Significance level (α) 0.05 gives Power (1 − β) 91.23%, Noncentrality (λ) 3, Critical value 1.64485, Type II error rate (β) 8.77%.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §7.2.2.2 Sample sizes required (N = (z(1 − α/2) + z(1 − β))² (σ/δ)² for a two-sided test and (z(1 − α) + z(1 − β))² (σ/δ)² one-sided; α = 0.05, β = 0.10 and a shift of one standard deviation give N = (1.645 + 1.282)² = 8.567 ≈ 9; with s, (1.860 + 1.397)² = 10.6 ≈ 11), https://www.itl.nist.gov/div898/handbook/prc/section2/prc222.htm (retrieved 2026-10-19)
  2. Test t test: one sample or paired, Alternative One-sided, Effect size (Cohen’s d) 1, Sample size (n per group) 11, Significance level (α) 0.05 gives Power (1 − β) 92.45%, Degrees of freedom (ν) 10, Critical value 1.81246.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §7.2.2.2 Sample sizes required (N = (z(1 − α/2) + z(1 − β))² (σ/δ)² for a two-sided test and (z(1 − α) + z(1 − β))² (σ/δ)² one-sided; α = 0.05, β = 0.10 and a shift of one standard deviation give N = (1.645 + 1.282)² = 8.567 ≈ 9; with s, (1.860 + 1.397)² = 10.6 ≈ 11), https://www.itl.nist.gov/div898/handbook/prc/section2/prc222.htm (retrieved 2026-10-19); SAS/STAT, The Analyst Application, Sample Size and Power Calculations, Hypothesis Tests (Power = Prob(t > t_crit, ν, NC) one-sided, + Prob(t < t_critl, ν, NC) two-sided; ν = n − 1 and NC = δ√n for one-sample and paired t tests; ν = 2(n − 1) and NC = δ ÷ √(2/n) for two samples of n each; δ = (μ − μ₀) ÷ s), https://support.sas.com/documentation/cdl/en/anlystug/58352/HTML/default/chap12_sect7.htm (retrieved 2026-10-19)
  3. Test t test: two independent samples, Alternative Two-sided, Effect size (Cohen’s d) 0.5, Sample size (n per group) 64, Significance level (α) 0.05 gives Power (1 − β) 80.15%, Degrees of freedom (ν) 126, Total sample size 128, Noncentrality (λ) 2.82843.Source: SAS/STAT, The Analyst Application, Sample Size and Power Calculations, Hypothesis Tests (Power = Prob(t > t_crit, ν, NC) one-sided, + Prob(t < t_critl, ν, NC) two-sided; ν = n − 1 and NC = δ√n for one-sample and paired t tests; ν = 2(n − 1) and NC = δ ÷ √(2/n) for two samples of n each; δ = (μ − μ₀) ÷ s), https://support.sas.com/documentation/cdl/en/anlystug/58352/HTML/default/chap12_sect7.htm (retrieved 2026-10-19)
  4. Test t test: one sample or paired, Alternative Two-sided, Effect size (Cohen’s d) 0, Sample size (n per group) 20, Significance level (α) 0.05 gives Power (1 − β) 5.00%.Source: SAS/STAT, The Analyst Application, Sample Size and Power Calculations, Hypothesis Tests (Power = Prob(t > t_crit, ν, NC) one-sided, + Prob(t < t_critl, ν, NC) two-sided; ν = n − 1 and NC = δ√n for one-sample and paired t tests; ν = 2(n − 1) and NC = δ ÷ √(2/n) for two samples of n each; δ = (μ − μ₀) ÷ s), https://support.sas.com/documentation/cdl/en/anlystug/58352/HTML/default/chap12_sect7.htm (retrieved 2026-10-19)
  5. Test z test: two independent samples, Alternative Two-sided, Effect size (Cohen’s d) 0.5, Sample size (n per group) 64, Significance level (α) 0.05 gives Power (1 − β) 80.74%, Noncentrality (λ) 2.82843, Critical value 1.95996.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §7.2.2.2 Sample sizes required (N = (z(1 − α/2) + z(1 − β))² (σ/δ)² for a two-sided test and (z(1 − α) + z(1 − β))² (σ/δ)² one-sided; α = 0.05, β = 0.10 and a shift of one standard deviation give N = (1.645 + 1.282)² = 8.567 ≈ 9; with s, (1.860 + 1.397)² = 10.6 ≈ 11), https://www.itl.nist.gov/div898/handbook/prc/section2/prc222.htm (retrieved 2026-10-19)
  6. Test t test: one sample or paired, Alternative Two-sided, Effect size (Cohen’s d) 0.5, Sample size (n per group) 20, Significance level (α) 0.05 gives Power (1 − β) 56.45%.Source: SAS/STAT, The Analyst Application, Sample Size and Power Calculations, Hypothesis Tests (Power = Prob(t > t_crit, ν, NC) one-sided, + Prob(t < t_critl, ν, NC) two-sided; ν = n − 1 and NC = δ√n for one-sample and paired t tests; ν = 2(n − 1) and NC = δ ÷ √(2/n) for two samples of n each; δ = (μ − μ₀) ÷ s), https://support.sas.com/documentation/cdl/en/anlystug/58352/HTML/default/chap12_sect7.htm (retrieved 2026-10-19)

How it works

Power is the chance that the test rejects H₀ when the true effect is d standard deviations: power = 1 − β.

Noncentrality. λ = d × √n for one sample (or n pairs), and λ = d × √(n/2) for two independent groups of n each. Under the alternative, the z statistic is normal with mean λ and standard deviation 1; the t statistic follows the noncentral t distribution T′ with ν degrees of freedom and noncentrality λ.

z test (σ known). With z = z(1 − α/2), the upper α/2 point of the standard normal:

  • Two-sided: power = Φ(λ − z) + Φ(−λ − z).
  • One-sided (with z = z(1 − α)): power = Φ(λ − z).

This is the relation behind the NIST e-Handbook's sample size formula: power 1 − β is reached when λ = z(1 − α) + z(1 − β) (one-sided), so N = (z(1 − α) + z(1 − β))² (σ/δ)². NIST's two-sided formula leaves out the small second term Φ(−λ − z); the page keeps it.

t test (σ estimated). ν = n − 1 for one sample or pairs, ν = 2(n − 1) for two groups, and t_c = t(1 − α/2; ν) (two-sided) or t(1 − α; ν) (one-sided), the upper points of Student's t:

  • Two-sided: power = P(T′ > t_c) + P(T′ < −t_c).
  • One-sided: power = P(T′ > t_c).

The noncentral t. T′ = (Z + λ) ÷ √(V/ν), with Z standard normal and V chi-square with ν degrees of freedom, independent. Its upper tail is P(T′ > t) = ∫₀^∞ Φ(λ − t·s) g(s) ds, where g is the density of √(V/ν). The page sums the equivalent Poisson series of Lenth's algorithm AS 243. For t ≥ 0, with x = t² ÷ (t² + ν), y = ν ÷ (t² + ν), pⱼ = e^(−λ²/2)(λ²/2)ʲ ÷ j! and qⱼ = λ e^(−λ²/2)(λ²/2)ʲ ÷ (√2 Γ(j + 3/2)):

  • P(T′ > t) = ½ Σⱼ [pⱼ I_y(ν/2, j + ½) + qⱼ I_y(ν/2, j + 1)], used when λ ≤ t;
  • P(T′ > t) = 1 − P(T′ ≤ t), with P(T′ ≤ t) = Φ(−λ) + ½ Σⱼ [pⱼ I_x(j + ½, ν/2) + qⱼ I_x(j + 1, ν/2)], used when λ > t;

where I is the regularized incomplete beta function. The lower tail P(T′ < −t) is the upper tail at t with noncentrality −λ. At λ = 0 the series is Student's t, so the power is exactly α.

Outputs. Power and β = 100% − power in percent; λ; the critical value (z or t_c); ν for t tests; the total sample size, n or 2n.

Rules

  • d from 0 to 10, the true effect in the tested direction; n a whole number from 2 to 100,000 (per group for two samples); α from 0.0001 to 0.5. Limits include their ends.
  • The data are normal, or n is large; two groups have equal sizes and standard deviations.

Output format

  • Power shows in percent with exactly 2 decimal places, rounded half up from its 64-bit value. β is the text b% with b = 100 × (1 − power) to exactly 2 decimal places, with thousands commas. λ and the critical value show to 6 significant figures.
  • λ = d√n is computed in 64-bit floats. Φ and the normal quantile are 64-bit floats accurate to about 15 significant figures, the t quantile to about 13 (below 1,000 degrees of freedom it takes one Newton step on Student's t tail, ½ I_y(ν/2, ½) with y = ν ÷ (ν + t²)), and the noncentral t tails to about 12.
  • The answer sentence reads The power is P, so the chance of missing a real effect of d = d is B., with P as the power output shows it and d as typed.

Worked examples by hand

NIST's one-sided z plan. d = 1, n = 9, α = 0.05: λ = 1 × √9 = 3, z(0.95) = 1.644854, power = Φ(3 − 1.644854) = Φ(1.355146) = 91.23%, β = 8.77%. NIST's (1.645 + 1.282)² = 8.567 is the n for exactly 90%; rounding up to 9 gives a little more.

NIST's t version. d = 1, n = 11, α = 0.05 one-sided: ν = 10, λ = √11 = 3.316625, t(0.95; 10) = 1.812461. P(T′ > 1.812461) for T′ with 10 degrees of freedom and noncentrality 3.316625 is 92.45%, above the 90% NIST's iteration aimed for.

Two groups of 64, d = 0.5, two-sided α = 0.05. λ = 0.5 × √32 = 2.828427. z test: Φ(2.828427 − 1.959964) + Φ(−2.828427 − 1.959964) = Φ(0.868463) + Φ(−4.788391) = 80.74%. t test: ν = 126, t(0.975; 126) = 1.978971, and P(T′ > 1.978971) + P(T′ < −1.978971) = 80.15%.

No effect. d = 0, n = 20, two-sided α = 0.05: λ = 0, so T′ is Student's t and the power is 5.00%, α.

One sample of 20, d = 0.5, two-sided α = 0.05. ν = 19, λ = √5 = 2.236068, t(0.975; 19) = 2.093024: power = 56.45%.

Other questions people ask

What is statistical power?

The chance that a test rejects the null hypothesis when the effect you expect is real: power = 1 − β, where β is the chance of a type II error (missing the effect). Studies usually aim for 80% or 90%.

How do you calculate the power of a z test?

Find λ = d√n and the critical value z. One-sided: power = Φ(λ − z(1 − α)). With d = 1, n = 9 and α = 0.05: Φ(3 − 1.645) = Φ(1.355) = 91.2%, which is why the NIST e-Handbook rounds its 8.57 up to 9 for 90% power.

How is the power of a t test calculated?

From the noncentral t distribution: power = P(T′ > t_c) (+ P(T′ < −t_c) two-sided), where T′ has ν = n − 1 degrees of freedom (2(n − 1) for two groups) and noncentrality λ = d√n (d√(n/2) for two groups). Two groups of 64 with d = 0.5 and α = 0.05 give 80.1%.

What effect size should I use?

The smallest difference that would matter, divided by the standard deviation: d = (μ₁ − μ₂) ÷ σ. A common rule of thumb calls 0.2 small, 0.5 medium and 0.8 large. A smaller d needs a larger sample for the same power.

Why is a t test’s power lower than a z test’s?

The t test also estimates σ from the data, which adds uncertainty: its critical value is larger and its statistic more spread out. For two groups of 64 at d = 0.5, the z test gives 80.7% and the t test 80.1%; with small samples the gap grows.

What happens to power when the effect is 0?

It equals α: with no real effect, the test rejects only by chance. A two-sided t test at α = 0.05 with d = 0 has power exactly 5%.