Two-way ANOVA: coal varieties and chemists

Two-way ANOVA without replication (varieties = rows, chemists = columns) is used because each variety was analysed by every chemist.

1. H0 (varieties): μ_A = μ_B = μ_C (no difference in mean ash content between varieties) H1: at least one variety mean differs (Similarly for chemists: H0: all chemist means are equal.)
2. Level of significance: α = 0.05
3. Row totals, column totals and sums of squares:

varieties totals: A=25, B=21, C=18
chemists totals: Chem 1=18, Chem 2=17, Chem 3=14, Chem 4=15
T = 64, N = 12, CF = T²/N = 341.3333
Total SS   = Σx² − CF = 366 − 341.3333 = 24.6667
SS varieties = Σ(row total²)/4 − CF = 6.1667
SS chemists = Σ(col total²)/3 − CF = 3.3333
SS error   = Total − SSvarieties − SSchemists = 15.1667

4. Two-way ANOVA table:

5. Critical value for varieties: F(0.05; 2, 6) = 5.14. Reject H0 if F > 5.14. (The table values F(2,9) = 19.4 and F(3,9) = 8.81 quoted in the question do not match the correct degrees of freedom here, (2, 6) for varieties and (3, 6) for chemists; the correct F-table values at 5% are 5.14 and 4.76.)
6. Decision:
   • Varieties: F = 1.220 < 5.14, so H0 is not rejected.
   • Chemists: F = 0.440 < 4.76, so H0 is not rejected.

Conclusion: At the 5% level the varieties of coal do not differ significantly in ash content (the calculated F is smaller than the table value). Chemists also do not differ significantly.