Skip to main content

Type I error; difference of two means procedure

Short answer

Type I error is rejecting a true null hypothesis, with probability α. For large samples, test μ₁=μ₂ using Z = (x̄₁−x̄₂)/√(s₁²/n₁+s₂²/n₂); reject H₀ if |Z| > Zα/2.

What is type I error? Describe the procedure of the test for difference of two mean for large sample.

Type I error

Type I error is the error of rejecting a true null hypothesis H₀. Its probability is the level of significance:

α = P(reject H₀ | H₀ is true)

(Type II error, with probability β, is accepting H₀ when it is false.) Example: concluding that a new machine gives a different mean output when in fact it does not.

Test for the difference of two means (large samples)

For independent large samples (n₁, n₂ ≥ 30) with means x̄₁, x̄₂ and standard deviations s₁, s₂:

  1. Hypotheses: H₀: μ₁ = μ₂ against H₁: μ₁ ≠ μ₂ (or μ₁ > μ₂, μ₁ < μ₂ for one-tailed).
  2. Level of significance: fix α, e.g. 0.05.
  3. Test statistic: Z = (x̄₁ − x̄₂) / √(σ₁²/n₁ + σ₂²/n₂) with s₁², s₂² used in place of σ₁², σ₂² when these are unknown.
  4. Critical value: two-tailed ±Z_α/2 (±1.96 at 5%, ±2.576 at 1%); one-tailed Z_α (1.645 at 5%, 2.326 at 1%).
  5. Decision: if |Z| > critical value, reject H₀; otherwise do not reject H₀.
  6. Conclusion: state in words whether the population means differ significantly.

Example: x̄₁ = 52, s₁ = 6, n₁ = 50; x̄₂ = 49, s₂ = 5, n₂ = 60 gives Z = 3/√(36/50 + 25/60) = 3/1.0661 = 2.81 > 1.96, so H₀ is rejected at 5%.

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Probability and Statistics chapter 6 questions →