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Procedure: difference of two proportions, large sample

Short answer

State null p1=p2 and alternative. Compute sample proportions p̂1, p̂2, pooled p̂=(x1+x2)/(n1+n2), SE=√(p̂q̂(1/n1+1/n2)), z=(p̂1−p̂2)/SE. Compare |z| with critical normal value; reject H0 if it exceeds the critical value, else do not reject.

Describe the procedure of the test of significance for difference of two proportions for large sample.

The test checks whether two population proportions p₁ and p₂ are equal, using large independent samples of sizes n₁ and n₂ with x₁ and x₂ items having the attribute.

Procedure

  1. Hypotheses. H0: p₁ = p₂ (no difference) H1: p₁ ≠ p₂ (two-tailed); or p₁ > p₂ / p₁ < p₂ (one-tailed)
  2. Level of significance α, for example 0.05.
  3. Sample proportions and the pooled estimate (valid because H0 says p₁ = p₂):
p̂₁ = x₁/n₁,  p̂₂ = x₂/n₂
p̂  = (x₁ + x₂)/(n₁ + n₂),  q̂ = 1 − p̂
  1. Standard error and test statistic:
SE = √( p̂ q̂ (1/n₁ + 1/n₂) )
z  = (p̂₁ − p̂₂)/SE          (follows N(0,1) for large n)
  1. Critical value. Take it from the normal table for α: ±1.96 for a two-tailed test at 5%, 1.645 for a one-tailed test at 5%, ±2.576 and 2.326 at 1%.
  2. Decision. Reject H0 when |z| exceeds the critical value (two-tailed) or z lies in the stated tail (one-tailed). Otherwise do not reject H0.
  3. Conclusion. State in plain words whether the two proportions differ significantly.

Example layout

If 16 of 200 and 14 of 300 are defective: p̂₁ = 0.08, p̂₂ = 0.0467, p̂ = 0.06, SE = 0.0217, z = 1.54 < 1.96, so H0 is not rejected at 5%.

✓ Answer checked by Claude Sonnet 5.5

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