7.6 Sample Size for Hypothesis Tests

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2 Chapter 2: Organization-Wide Planning and Deployment

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6 Chapter 6: Measure Phase

6.1 Process Flow Metrics and the Hidden Factory 6.2 Process Analysis Tools and Value Stream Mapping 6.3 Types of Data and Measurement Scales 6.4 Sampling Concepts, Methods, and Sample Size for Estimation 6.5 Data Collection Plans and Common Pitfalls 6.6 Gage R&R for Variable Measurement Systems 6.7 Attribute Agreement Analysis and Audit MSA 6.8 Measurement Systems Across the Organization, and Metrology 6.9 Population Parameters, Sample Statistics, and Valid Statistical Conclusions 6.10 Descriptive Statistics: Central Tendency, Dispersion, and Shape 6.11 Graphical Methods for Exploratory Data Analysis 6.12 Probability Concepts, Common Distributions, and the Central Limit Theorem 6.13 Additional Distributions: Hypergeometric, Exponential, Lognormal, Weibull, and Bivariate 6.14 Process Capability and Performance Indices 6.15 Capability Studies, Attributes Capability, Yield Metrics, and Long-Term Capability

7 Chapter 7: Analyze Phase

7.1 The Correlation Coefficient, Its Confidence Interval, and Causation 7.2 Linear Regression and Residual Analysis 7.3 Multivariate Tools: Factor Analysis, Discriminant Analysis, and MANOVA 7.4 Hypothesis Testing Terminology, Errors, and Single-Sample Tests 7.5 Statistical Versus Practical Significance 7.6 Sample Size for Hypothesis Tests 7.7 Point Estimates, Confidence, Prediction, and Tolerance Intervals 7.8 Tests for Means, Variances, and Proportions 7.9 Analysis of Variance 7.10 Chi-Square Goodness-of-Fit and Tests of Independence 7.11 Types of Risk and Failure Mode and Effects Analysis 7.12 Gap Analysis and Root Cause Analysis 7.13 Waste Analysis: The Seven Classic Wastes and Resource Under-Utilization

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10 Chapter 10: Design for Six Sigma (DFSS)

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7.6 Sample Size for Hypothesis Tests

Key Takeaways Explain This Topic AI

Sample size depends on four quantities: alpha, power, the difference to be detected, and the process standard deviation.

For a two-sided two-sample comparison of means, n per group is approximately 2 times the square of (z alpha over 2 plus z beta) times sigma squared divided by delta squared.

Required sample size scales with the inverse square of the effect size, so halving the detectable difference quadruples the sample.

Attribute data requires far larger samples than continuous data for the same practical sensitivity.

Power is one minus beta, the probability of correctly detecting a real difference; 80% or 90% are the conventional targets.

Last updated: August 2026

• The four inputs
• Every sample size calculation needs the same four quantities, and three of them are business decisions

rather than statistical ones.

• Input Symbol Typical value Who decides
• Significance level α \alpha α 0.05 Team, from the cost of a false alarm
• Power 1 − β 1 - \beta 1 − β 0.80 or 0.90 Team, from the cost of missing a real effect
• Difference to detect δ \delta δ The practical threshold Process owner or customer
• Process standard deviation σ \sigma σ From baseline data or a pilot Measured
• The most common failure is treating δ \delta δ as a statistical parameter. It is not: it is the smallest
• difference worth detecting, and it comes from the economics of the process. If a 2% yield improvement

is the smallest change that would justify implementation, size the study to detect 2%.

Standard normal values needed:

• Quantity Value
• z α / 2 z_{\alpha/2} z α /2 ​ at α = 0.05 \alpha = 0.05 α = 0.05 two-sided 1.960
• z α z_{\alpha} z α ​ at α = 0.05 \alpha = 0.05 α = 0.05 one-sided 1.645
• z β z_{\beta} z β ​ at 80% power 0.842
• z β z_{\beta} z β ​ at 90% power 1.282
• Comparing two means

For a two-sided test of H 0 : μ 1 = μ 2 H_0: \mu_1 = \mu_2 H 0 ​ : μ 1 ​ = μ 2 ​ with equal group sizes: n = 2 ( z α / 2 + z β ) 2 σ 2 δ 2 per group n = \frac{2(z_{\alpha/2} + z_{\beta})^2 \sigma^2}{\delta^2} \quad \text{per group} n = δ 2 2 ( z α /2 ​ + z β ​ ) 2 σ 2 ​ per group

At α = 0.05 \alpha = 0.05 α = 0.05 two-sided and 80% power, ( 1.960 + 0.842 ) 2 = 7.85 (1.960 + 0.842)^2 = 7.85 ( 1.960 + 0.842 ) 2 = 7.85 , so a useful shortcut is n ≈ 15.7 σ 2 δ 2 per group n \approx \frac{15.7\,\sigma^2}{\delta^2} \quad \text{per group} n ≈ δ 2 15.7 σ 2 ​ per group

Worked example. A team wants to detect a 1.5-unit shift in mean cycle time. Baseline data gives σ = 3.0 \sigma = 3.0 σ = 3.0 units. At α = 0.05 \alpha = 0.05 α = 0.05 and 80% power: n = 2 ( 1.960 + 0.842 ) 2 ( 3.0 ) 2 ( 1.5 ) 2 = 2 × 7.851 × 9 2.25 = 141.3 2.25 = 62.8 → 63 per group n = \frac{2(1.960 + 0.842)^2 (3.0)^2}{(1.5)^2} = \frac{2 \times 7.851 \times 9}{2.25} = \frac{141.3}{2.25} = 62.8 \rightarrow 63 \text{ per group} n = ( 1.5 ) 2 2 ( 1.960 + 0.842 ) 2 ( 3.0 ) 2 ​ = 2.25 2 × 7.851 × 9 ​ = 2.25 141.3 ​ = 62.8 → 63 per group

Always round up . Raising power to 90% changes the multiplier to ( 1.960 + 1.282 ) 2 = 10.51 (1.960 + 1.282)^2 = 10.51 ( 1.960 + 1.282 ) 2 = 10.51 and the

requirement to 85 per group -- a 35% increase in data for a 10 percentage point gain in power.

One sample against a target n = ( z α / 2 + z β ) 2 σ 2 δ 2 n = \frac{(z_{\alpha/2} + z_{\beta})^2 \sigma^2}{\delta^2} n = δ 2 ( z α /2 ​ + z β ​ ) 2 σ 2 ​

Exactly half the two-sample requirement, because only one mean is being estimated.

• When sigma is unknown and n is small
• These formulas use z z z values, which assumes σ \sigma σ is known. When σ \sigma σ is estimated from the
• sample, the t t t distribution applies and the requirement is slightly larger. The standard practical fix
• is to compute with z z z , then add two to four observations per group, or iterate using t t t values at the

resulting degrees of freedom. For n n n above about 30 per group the difference is negligible.

Comparing two proportions n = ( z α / 2 + z β ) 2 [ p 1 ( 1 − p 1 ) + p 2 ( 1 − p 2 ) ] ( p 1 − p 2 ) 2 per group n = \frac{(z_{\alpha/2} + z_{\beta})^2 \left[p_1(1-p_1) + p_2(1-p_2)\right]}{(p_1 - p_2)^2} \quad \text{per group} n = ( p 1 ​ − p 2 ​ ) 2 ( z α /2 ​ + z β ​ ) 2 [ p 1 ​ ( 1 − p 1 ​ ) + p 2 ​ ( 1 − p 2 ​ ) ] ​ per group

Worked example. Current defect rate is 5%, and the team wants to detect a reduction to 3%. At α = 0.05 \alpha = 0.05 α = 0.05 two-sided and 80% power:

p 1 ( 1 − p 1 ) = 0.05 × 0.95 = 0.0475 p_1(1-p_1) = 0.05 \times 0.95 = 0.0475 p 1 ​ ( 1 − p 1 ​ ) = 0.05 × 0.95 = 0.0475

p 2 ( 1 − p 2 ) = 0.03 × 0.97 = 0.0291 p_2(1-p_2) = 0.03 \times 0.97 = 0.0291 p 2 ​ ( 1 − p 2 ​ ) = 0.03 × 0.97 = 0.0291

Sum = 0.0766 = 0.0766 = 0.0766 ; ( p 1 − p 2 ) 2 = ( 0.02 ) 2 = 0.0004 (p_1 - p_2)^2 = (0.02)^2 = 0.0004 ( p 1 ​ − p 2 ​ ) 2 = ( 0.02 ) 2 = 0.0004 n = 7.851 × 0.0766 0.0004 = 0.6014 0.0004 = 1,503 per group n = \frac{7.851 \times 0.0766}{0.0004} = \frac{0.6014}{0.0004} = 1{,}503 \text{ per group} n = 0.0004 7.851 × 0.0766 ​ = 0.0004 0.6014 ​ = 1 , 503 per group

• Just over 1,500 units per group -- 3,000 units in total -- to detect a 2 percentage point change. This
• single result explains why continuous data is preferred wherever it is available: a comparable

continuous study typically needs a few dozen observations per group.

One sample against a target proportion n = ( z α / 2 p 0 ( 1 − p 0 ) + z β p 1 ( 1 − p 1 ) ) 2 ( p 1 − p 0 ) 2 n = \frac{(z_{\alpha/2}\sqrt{p_0(1-p_0)} + z_{\beta}\sqrt{p_1(1-p_1)})^2}{(p_1 - p_0)^2} n = ( p 1 ​ − p 0 ​ ) 2 ( z α /2 ​ p 0 ​ ( 1 − p 0 ​ ) ​ + z β ​ p 1 ​ ( 1 − p 1 ​ ) ​ ) 2 ​

The inverse-square relationship

Because δ \delta δ appears squared in the denominator, sensitivity is expensive:

• Detectable difference Relative sample size
• δ \delta δ 1x
• δ / 2 \delta / 2 δ /2 4x
• δ / 4 \delta / 4 δ /4 16x
• The same relationship works in your favour on the other side: halving σ \sigma σ quarters the required
• sample . Improving the measurement system is therefore often cheaper than collecting more data. If gage
• R&R contributes a large share of the observed variance, fixing the gauge reduces σ \sigma σ and shrinks

every subsequent sample size requirement. σ o b s e r v e d 2 = σ p r o c e s s 2 + σ m e a s u r e m e n t 2 \sigma^2_{observed} = \sigma^2_{process} + \sigma^2_{measurement} σ o b ser v e d 2 ​ = σ p rocess 2 ​ + σ m e a s u re m e n t 2 ​

• Power analysis after the fact
• If a study has already been run and returned a non-significant result, compute the difference it was

actually able to detect: δ d e t e c t a b l e = ( z α / 2 + z β ) σ 2 n \delta_{detectable} = (z_{\alpha/2} + z_{\beta})\,\sigma\sqrt{\frac{2}{n}} δ d e t ec t ab l e ​ = ( z α /2 ​ + z β ​ ) σ n 2 ​ ​

With n = 12 n = 12 n = 12 per group and σ = 3.0 \sigma = 3.0 σ = 3.0 : δ = 2.802 × 3.0 × 2 / 12 = 2.802 × 3.0 × 0.408 = 3.43 \delta = 2.802 \times 3.0 \times \sqrt{2/12} = 2.802 \times 3.0 \times 0.408 = 3.43 δ = 2.802 × 3.0 × 2/12 ​ = 2.802 × 3.0 × 0.408 = 3.43

• The study could only detect a shift of 3.4 units. Reporting "no significant difference" when the
• practically important difference was 1.5 units is misleading; the honest statement is that the study

was inconclusive.

Practical adjustments

• Situation Adjustment
• Expected dropout, invalid records, or destructive test loss Inflate n n n by the expected loss fraction

Unequal group sizes with ratio k k k n 1 = ( 1 + 1 / k ) 2 × n e q u a l n_1 = \frac{(1 + 1/k)}{2} \times n_{equal} n 1 ​ = 2 ( 1 + 1/ k ) ​ × n e q u a l ​

• Multiple comparisons Use a Bonferroni-adjusted α \alpha α in the calculation
• Finite population Apply the finite population correction when n / N > 0.05 n/N > 0.05 n / N > 0.05
• Clustered or subgrouped data Inflate by the design effect; observations within a cluster are not independent
• Sizing checklist

Is δ \delta δ the smallest practically important difference, agreed with the process owner?

Is σ \sigma σ from real baseline data on this process, not a textbook value?

Does σ \sigma σ include measurement variation, and would improving the gauge be cheaper than more data?

Are α \alpha α and power chosen from the relative cost of the two errors?

Has n n n been rounded up and inflated for expected loss?

Is collecting that many observations feasible in the project timeline? If not, renegotiate δ \delta δ or power before collecting, not after.