--- title: "Two-way ANOVA: coal varieties and chemists" subject: "Probability and Statistics" chapter: 6 key: "semester_5/probability_and_statistics:anova-two-way-coal-chemists" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-two-way-coal-chemists site: BCT Engineering --- # Two-way ANOVA: coal varieties and chemists Probability and Statistics · Chapter 6: Inference concerning mean - Asked in: 2075 Asoj, 6 marks - Video: [BCS301 Module 5 | Two-Way ANOVA | Wheat Seed & Fertilizer Study | New Model Question Paper-I](https://www.youtube.com/watch?v=1Im_HJYTYYs) — Mathematics Tutor - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-two-way-coal-chemists - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6 ## Question Three varieties of coal were analyzed by four chemists and the ash-content in the varieties were found as follows: | Varieties | Chemist 1 | Chemist 2 | Chemist 3 | Chemist 4 | |---|---|---|---|---| | A | 8 | 5 | 5 | 7 | | B | 7 | 6 | 4 | 4 | | C | 3 | 6 | 5 | 4 | Test whether the varieties differ significantly in their ash-content? Test at 5% level of significance. [F(2,9) = 19.4, F(3,9) = 8.81]. ## Short answer Two-way ANOVA without replication yields F(2,6) = 1.220 for varieties. Since 1.220 < critical value 5.14, H0 is not rejected; ash content does not differ significantly among coal varieties at the 5% level. ## Answer 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: ```text 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: | Source | SS | d.f. | MS | F | |---|---|---|---|---| | Between varieties | 6.167 | 2 | 3.083 | 1.220 | | Between chemists | 3.333 | 3 | 1.111 | 0.440 | | Error | 15.167 | 6 | 2.528 | | | Total | 24.667 | 11 | | | 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is two-way ANOVA used here? A. The population variance is known B. There are only two samples C. Each variety was analysed by every chemist, so there are two factors D. Chemists have unequal sample sizes Answer: C. Varieties and chemists form rows and columns, so both factors are separated in the analysis. (Key point: Choice of method) 2. What are the d.f. for varieties and error? A. 2 and 9 B. 3 and 6 C. 2 and 6 D. 3 and 9 Answer: C. Varieties: 3-1 = 2; error: (3-1)(4-1) = 6. The question's F(2,9) is not appropriate. (Key point: Degrees of freedom) 3. How is SS error obtained? A. Total SS + SS varieties + SS chemists B. SS varieties - SS chemists C. Total SS - SS varieties - SS chemists D. Total SS / N Answer: C. SS error = 24.667 - 6.167 - 3.333 = 15.167. (Key point: Error SS) 4. What is the conclusion for varieties? A. F = 1.22 > 5.14, so they differ B. F = 1.22 < 5.14, so no significant difference in ash content C. F = 3.083 > 5.14, so they differ D. F = 0.44 < 5.14, so they differ Answer: B. Calculated F = 3.083/2.528 = 1.220, below the table value 5.14, so H0 is not rejected. 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