---
title: "Test: steel rod diameters on two machines"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:steel-rods-two-machines-mean-diameter"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/steel-rods-two-machines-mean-diameter
site: BCT Engineering
---

# Test: steel rod diameters on two machines

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2072 Kartik, 5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/steel-rods-two-machines-mean-diameter
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

The diameter of steel rods manufactured on two different extrusion machines is being investigated. Two random samples of sizes n1 = 15 and n2 = 17 are selected, and the sample means and sample variances are x̄1 = 8.73, s1² = 0.35, x̄2 = 8.68, and s2² = 0.40, respectively. Assume that σ1² = σ2² and that the data are drawn from a normal distribution. Is there evidence to support the claim that the two machines produce rods with different mean diameters? Use α = 0.05 in arriving at this conclusion.

## Short answer

Pooled t-test (equal variances): Sp = 0.6137, t = 0.230, df = 30. At α = 0.05, critical value = ±2.042; |t| < 2.042, fail to reject H0. There is no evidence that the two machines produce rods with different mean diameters.

## Answer

Variances are assumed equal and populations normal, with small samples, so the pooled t-test is used.

1. H0: μ₁ = μ₂ (both machines give the same mean diameter)
   H1: μ₁ ≠ μ₂ (two-tailed)
2. Level of significance: α = 0.05
3. Test statistic (small samples, equal variances assumed, pooled t-test):

```text
Sp² = [(n₁−1)s₁² + (n₂−1)s₂²] / (n₁+n₂−2)
    = [14×0.3500 + 16×0.4000] / 30 = 0.3767
Sp  = 0.6137
t   = (x̄₁ − x̄₂) / [Sp √(1/n₁ + 1/n₂)]
    = (8.73 − 8.68) / [0.6137 × √(1/15 + 1/17)]
    = 0.0500 / 0.2174 = 0.230
```

4. Critical value: t at α = 0.05, df = n₁+n₂−2 = 30: ±2.042. Reject H0 if |t| > 2.042.
5. Decision: the calculated value falls in the acceptance region (p-value = 0.8197), so **fail to reject H0**.

**Conclusion:** there is no evidence that the two machines produce rods of different mean diameter.

## Check yourself

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

1. Why pooled t?
   A. Equal variances assumed, normal data, small samples
   B. Large samples
   C. Paired data
   D. Unknown distribution
   Answer: A. The question assumes sigma1^2 = sigma2^2 and normality with small n. (Key point: Choice of test)

2. What is Sp^2?
   A. 0.35 x 0.40
   B. (0.35 + 0.40)/2 = 0.375
   C. [14 x 0.35 + 16 x 0.40]/30 = 0.3767
   D. [15 x 0.35 + 17 x 0.40]/32
   Answer: C. Weighted by n-1: (4.9 + 6.4)/30 = 0.3767. (Key point: Pooled variance)

3. What is df and critical value?
   A. 15; +-2.131
   B. 30; +-2.042
   C. 32; +-2.037
   D. 29; +-2.045
   Answer: B. df = 15 + 17 - 2 = 30. (Key point: Degrees of freedom)

4. What is the conclusion?
   A. Machines differ
   B. t = 0.23 so reject H0
   C. t = 2.042 so reject
   D. t = 0.23; fail to reject H0; no evidence of difference
   Answer: D. |t| is far below 2.042. (Key point: Conclusion)

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