---
title: "Test: wear of two tyre brands"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:tyres-two-brands-wear-test"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/tyres-two-brands-wear-test
site: BCT Engineering
---

# Test: wear of two tyre brands

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2068 Jestha (old course), 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/tyres-two-brands-wear-test
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

A large automobile manufacturing company is trying to decide whether to purchase brand A or brand B tyres for its new models. To help arrive at a decision, an experiment is conducted using 12 of each brand. The tyres are run until they wear out. The results are:
Brand A: x̄1 = 37,900 kilometers, S1 = 5,100 kilometers.
Brand B: x̄2 = 39,800 kilometers, S2 = 5,900 kilometers.
Test the hypothesis that there is no difference in the average wear of 2 brands of tyres. Assume the population to be approximately normally distributed with equal variances.

## Short answer

Pooled t-test: t=(37900−39800)/2251.3=−0.844, df=22, critical t=±2.074. |t|<2.074, fail to reject H0. No significant difference in average tyre wear between brands.

## Answer

Both samples are small (12 each), the populations are normal with equal variances, so a pooled two-sample t-test is used.

1. H0: μ_A = μ_B (no difference in average wear)
   H1: μ_A ≠ μ_B (two-tailed)
2. Level of significance: α = 0.05 (not stated; the usual 5% is taken)
3. Test statistic:

```text
Sp² = [(n₁−1)S₁² + (n₂−1)S₂²]/(n₁+n₂−2)
    = [11×5100² + 11×5900²]/22 = (26,010,000 + 34,810,000)/2 = 30,410,000
Sp  = 5514.5
SE  = Sp·√(1/12 + 1/12) = 5514.5 × 0.40825 = 2251.3
t   = (37,900 − 39,800)/2251.3 = -0.844
```

4. Degrees of freedom = 12 + 12 − 2 = 22. Critical value: t(0.025; 22) = ±2.074. Reject H0 if |t| > 2.074.
5. Decision: |t| = 0.844 < 2.074 (p-value = 0.41), so **fail to reject H0**.

**Conclusion:** there is no significant difference in the average wear of the two brands (brand B lasts 1,900 km longer in the samples, but this is within chance variation). The company may buy either brand; price and other factors can decide.

## Check yourself

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

1. Why is a pooled t-test used for the tyres?
   A. Samples are small (12 each) with normal populations and equal variances
   B. sigma is known
   C. Samples are paired
   D. The data are counts
   Answer: A. Small samples from normal populations with equal variances call for the pooled t. (Key point: Choice of test)

2. With equal sample sizes of 12, Sp² equals which value?
   A. Average of S1 and S2 squared
   B. Average of S1² and S2²
   C. S1² × S2²
   D. (S1 + S2)²
   Answer: B. With n1 = n2 the pooled variance is the simple mean of the two variances: (5100² + 5900²)/2 = 30,410,000. (Key point: Pooled variance)

3. What are the degrees of freedom and critical value?
   A. 24 and ±2.064
   B. 11 and ±2.201
   C. 22 and ±2.074
   D. 23 and ±2.069
   Answer: C. d.f. = 12 + 12 − 2 = 22; t(0.025; 22) = ±2.074. (Key point: Degrees of freedom)

4. |t| = 0.844 < 2.074. What is concluded?
   A. Brand B is significantly better
   B. Brand A is significantly better
   C. Reject H0
   D. No significant difference in average wear; fail to reject H0
   Answer: D. The 1,900 km sample difference is within chance variation. (Key point: Decision)

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