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Test: wear of two tyre brands

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.

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.

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:
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
  1. Degrees of freedom = 12 + 12 − 2 = 22. Critical value: t(0.025; 22) = ±2.074. Reject H0 if |t| > 2.074.
  2. 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.

✓ Answer checked by Claude Sonnet 5.5

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