---
title: "Two proportions: injection-molding machines"
subject: "Probability and Statistics"
chapter: 7
key: "semester_5/probability_and_statistics:injection-molding-defective-fraction"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/injection-molding-defective-fraction
site: BCT Engineering
---

# Two proportions: injection-molding machines

Probability and Statistics · Chapter 7: Inference concerning proportion

- Asked in: 2070 Asar, 5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/injection-molding-defective-fraction
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-7

## Question

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?

## 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.

## Answer

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

```text
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:

```text
SE = √( p̂q̂ (1/300 + 1/300) ) = √(0.03833×0.96167×0.006667) = 0.01568
z  = (0.0500 − 0.0267)/0.01568 = 1.488
```

4. Critical value: z(0.05, two-tail) = ±1.960. Reject H0 if |z| > 1.960.
5. 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why and how is the pooled proportion computed in this test?
   A. Because H0 says p1 = p2; p-hat = (15 + 8)/600
   B. Because the samples are small; p-hat = (0.05 + 0.0267)/2 with no weighting needed
   C. To raise power; p-hat = 15/300
   D. It is not needed here
   Answer: A. Under H0 both machines share one common p, estimated by total defectives over total sample, 23/600 = 0.0383. (Key point: Pooled proportion)

2. Which critical value rule applies?
   A. Right-tailed, z > 1.645
   B. Left-tailed, z < -1.645
   C. Two-tailed, |z| > 1.960
   D. Two-tailed, |z| > 2.576
   Answer: C. The alternative p1 not equal p2 gives a two-tailed test at 5%, critical +/-1.96. (Key point: Type of test)

3. z = 1.488 is compared with 1.960. What is concluded?
   A. Reject H0; machines differ
   B. Fail to reject H0; no significant difference in defective fraction
   C. Machine 1 is definitely worse
   D. Reject H0 because 0.05 > 0.0267
   Answer: B. |1.488| < 1.96, so the observed difference is within sampling variation. (Key point: Decision)

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