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ANOVA: three oil spill clean-up systems

Short answer

One-way ANOVA yields F = 0.172 < F_crit = 3.885, so H₀ is not rejected. There is no significant difference among the three clean-up systems at the 5% level.

A research company has designed three different systems to clean up oil spills. The following table contains the results, measured by how much surface area in square meters is cleared in one hour. The data were found by testing each method in several trials. Are the three systems equally effective? Use the 0.05 level of significance.

System A556063565955
System B5753644962-
System C66526157--

One-way ANOVA with three systems (k = 3) and unequal sample sizes.

Hypotheses

  • H₀: μ_A = μ_B = μ_C (the three systems are equally effective)
  • H₁: at least one system has a different mean

Level of significance: α = 0.05

Sample sizes: n_A = 6, n_B = 5, n_C = 4, N = 15. Group totals: T_A = 348, T_B = 285, T_C = 236; grand total G = 869.

  • Correction factor CF = G²/N = 869²/15 = 50344.067
  • Σx² (all observations) = 50665
  • Total SS = Σx² − CF = 320.933
  • SS between = Σ(Tᵢ²/nᵢ) − CF = 50353.000 − 50344.067 = 8.933
  • SS within (error) = Total SS − SS between = 312.000

ANOVA table

SourceSSdfMSF
Between groups8.93324.4670.172
Within groups (error)312.0001226.000
Total320.93314

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

Decision: calculated F = 0.172 < 3.885 (p-value ≈ 0.844), so we do not reject H₀.

Conclusion: there is no significant difference between the systems at the 5% level; the three clean-up systems can be regarded as equally effective.

✓ Answer checked by Claude Sonnet 5.5

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