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Test of difference of two means: large sample

Short answer

For large independent samples, test H₀: μ₁ = μ₂ with Z = (x̄₁ − x̄₂) / √(s₁²/n₁ + s₂²/n₂). Compare |Z| to the standard normal critical value Z_α/₂ (two‑tailed) or Z_α (one‑tailed); reject H₀ if the statistic exceeds the critical value.

Describe the procedure of the test of significance for difference of two population means for large samples.

When two independent large samples (n₁, n₂ ≥ 30) are drawn from populations with means μ₁ and μ₂, we test whether the difference of the sample means x̄₁ − x̄₂ is significant, using the Z-test (by the central limit theorem, x̄₁ − x̄₂ is approximately normal).

Procedure

  1. Hypotheses: H₀: μ₁ = μ₂ (or μ₁ − μ₂ = 0). H₁: μ₁ ≠ μ₂ (two-tailed), or μ₁ > μ₂ / μ₁ < μ₂ (one-tailed).
  2. Level of significance: choose α (usually 0.05 or 0.01).
  3. Test statistic: Z = (x̄₁ − x̄₂) / √(σ₁²/n₁ + σ₂²/n₂) If σ₁², σ₂² are unknown, replace them by the sample variances s₁², s₂². If σ₁² = σ₂² = σ², the standard error is σ√(1/n₁ + 1/n₂).
  4. Critical value: from the normal table.
αTwo-tailedOne-tailed
0.05±1.961.645
0.01±2.5762.326
  1. Decision: reject H₀ if |Z| > Z_α/2 (two-tailed) [or Z > Z_α, Z < −Z_α for one-tailed]; otherwise accept (fail to reject) H₀.
  2. Conclusion: state in words whether the two population means differ significantly.

Example

Sample A: n₁ = 100, x̄₁ = 1120, s₁ = 75; sample B: n₂ = 100, x̄₂ = 1062, s₂ = 82. Test H₀: μ₁ = μ₂ vs H₁: μ₁ ≠ μ₂ at α = 0.05.

Z = (1120 − 1062)/√(75²/100 + 82²/100) = 58/11.11 = 5.22 > 1.96, so H₀ is rejected: the means differ significantly.

✓ Answer checked by Claude Sonnet 5.5

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