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ANOVA: breaking strength of three cable brands

Short answer

One‑way ANOVA gives F = 2.915 (df: 2, 12). At α = 0.05, F_crit = 3.885, so fail to reject H₀. No significant difference in mean breaking strengths.

The following are the breaking strength of three different brands of cables.

BrandBreaking Strength
A40 30 50 60 30
B60 40 55 65
C60 50 70 65 75 40
Construct ANOVA table and test for the equality of the average breaking strength of cables at α = 5%.

One-way ANOVA with three brands (k = 3).

Hypotheses

  • H₀: μ_A = μ_B = μ_C (mean breaking strengths are equal)
  • H₁: at least one mean is different

Level of significance: α = 0.05

Sample sizes: n_A = 5, n_B = 4, n_C = 6, N = 15. Group totals: T_A = 210, T_B = 220, T_C = 360; grand total G = 790.

  • Correction factor CF = G²/N = 790²/15 = 41606.667
  • Σx² (all observations) = 44400
  • Total SS = Σx² − CF = 2793.333
  • SS between = Σ(Tᵢ²/nᵢ) − CF = 42520.000 − 41606.667 = 913.333
  • SS within (error) = Total SS − SS between = 1880.000

ANOVA table

SourceSSdfMSF
Between groups913.3332456.6672.915
Within groups (error)1880.00012156.667
Total2793.33314

Critical value: F(0.05; 2, 12) = 3.885

Decision: calculated F = 2.915 < 3.885 (p-value ≈ 0.093), so we do not reject H₀.

Conclusion: there is not sufficient evidence of a difference; the average breaking strengths of the three brands of cable can be regarded as equal at the 5% level.

✓ Answer checked by Claude Sonnet 5.5

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