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Test: mean weights of two processes

Short answer

Two-sample z-test: SE = √(12²/250+14²/400)=1.0325, z = (120−124)/1.0325 = −3.874. Since |z|>1.96, reject H0; the two mean weights differ significantly.

In a certain factory, there are two independent processes manufacturing the same item. The average weight in a sample of 250 items produced from one process is found to be 120 gram with a standard deviation of 12 gram, while the corresponding figures in a sample of 400 items from the other process are 124 and 14 respectively. Test whether the two mean weights differ significantly or not at 5 percent level of significance.

Both samples are large, so the two-sample z-test is applied.

  1. H0: μ₁ = μ₂ (mean weights are equal) H1: μ₁ ≠ μ₂ (two-tailed)
  2. Level of significance: α = 0.05
  3. Test statistic (large samples, z-test):
z = (x̄₁ − x̄₂) / √(s₁²/n₁ + s₂²/n₂)
SE = √(12²/250 + 14²/400) = √(0.5760 + 0.4900) = 1.0325
z  = (120 − 124) / 1.0325 = -3.874
  1. Critical value: z at α = 0.05 (two-tail) = ±1.960. Reject H0 if |z| > 1.960.
  2. Decision: the calculated value falls in the rejection region (p-value = 0.0001), so reject H0.

Conclusion: the two mean weights differ significantly; the second process gives heavier items.

✓ Answer checked by Claude Sonnet 5.5

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