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Two proportions: injection-molding machines

Short answer

Using a two-proportion z test with pooled p̂=0.0383, z=1.488 is below the ±1.960 critical value; fail to reject H0. There is no significant difference, so both machines can reasonably produce the same fraction of defectives.

Two different types of injection-molding machines are used to form plastic parts. A part is considered defective if it has excessive shrinkage or is discolored. Two random samples, each of size 300, are selected, and 15 defective parts are found in the sample from machine 1 while 8 defective parts are found in the sample from machine 2. Is it reasonable to conclude that both machines produce the same fraction of defective parts, using α = 0.05?

Test for the difference of two proportions (large samples, two-tailed).

Machine 1: n₁ = 300, x₁ = 15, p̂₁ = 0.0500
Machine 2: n₂ = 300, x₂ = 8,  p̂₂ = 0.0267
Pooled p̂ = (15 + 8)/600 = 0.03833,  q̂ = 0.96167
  1. H0: p₁ = p₂ (both machines give the same fraction defective) H1: p₁ ≠ p₂
  2. Level of significance: α = 0.05
  3. Test statistic:
SE = √( p̂q̂ (1/300 + 1/300) ) = √(0.03833×0.96167×0.006667) = 0.01568
z  = (0.0500 − 0.0267)/0.01568 = 1.488
  1. Critical value: z(0.05, two-tail) = ±1.960. Reject H0 if |z| > 1.960.
  2. Decision: |z| = 1.488 < 1.960 (p-value = 0.137), so fail to reject H0.

Conclusion: there is no significant difference between the two machines; it is reasonable to conclude that both produce the same fraction of defective parts.

✓ Answer checked by Claude Sonnet 5.5

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