--- title: "ANOVA: thiamin content of four cereals" subject: "Probability and Statistics" chapter: 6 key: "semester_5/probability_and_statistics:anova-cereal-thiamin" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-cereal-thiamin site: BCT Engineering --- # ANOVA: thiamin content of four cereals Probability and Statistics · Chapter 6: Inference concerning mean - Asked in: 2073 Chaitra, 5 marks - Video: [How To Calculate and Understand Analysis of Variance (ANOVA) F Test.](https://www.youtube.com/watch?v=-yQb_ZJnFXw) — statisticsfun - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-cereal-thiamin - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6 ## Question Six sample of each of four types of cereal grain grown in a certain region were analyzed to determine thiamin content, resulting in the following data (mg/g): | Wheat | 5.2 | 4.5 | 6.0 | 6.1 | 6.7 | 5.8 | |---|---|---|---|---|---|---| | Barley | 6.5 | 8.0 | 6.1 | 7.5 | 5.9 | 5.6 | | Maize | 5.8 | 4.7 | 6.4 | 4.9 | 6.0 | 5.2 | | Oats | 8.3 | 6.1 | 7.8 | 7.0 | 5.5 | 7.2 | Does this data suggest that at least one of the grains differ with respect to true average thiamin content? Use 0.05 level of significance. ## Short answer One-way ANOVA yields F = 3.957. At α = 0.05, critical F(3,20) = 3.098, so we reject H0. At least one cereal grain type has a significantly different true average thiamin content. ## Answer One-way ANOVA with four cereal types (6 samples each). 1. H0: μ_wheat = μ_barley = μ_maize = μ_oats H1: at least one cereal has a different true mean thiamin content (F-test, right-tailed) 2. Level of significance: α = 0.05 3. Computations: ```text Wheat n=6 total=34.3 mean=5.7167 Barley n=6 total=39.6 mean=6.6000 Maize n=6 total=33 mean=5.5000 Oats n=6 total=41.9 mean=6.9833 Grand total T = 148.8, N = 24 Correction factor CF = T²/N = 148.8²/24 = 922.5600 Total SS = Σx² − CF = 946.6800 − 922.5600 = 24.1200 SS between = Σ(Ti²/ni) − CF = 931.5433 − 922.5600 = 8.9833 SS within = Total SS − SS between = 15.1367 ``` 4. ANOVA table: | Source | SS | d.f. | MS | F | |---|---|---|---|---| | Between cereals | 8.9833 | 3 | 2.9944 | 3.957 | | Within (error) | 15.1367 | 20 | 0.7568 | | | Total | 24.1200 | 23 | | | 5. Critical value: F(0.05; 3, 20) = 3.098. Reject H0 if F > 3.098. 6. Decision: the calculated value falls in the rejection region (p-value = 0.0229), so **reject H0**. **Conclusion:** at least one cereal differs in true average thiamin content. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does H1 say? A. At least one cereal has a different true mean B. All four means differ C. The variances differ D. Wheat is lowest Answer: A. ANOVA's alternative is that at least one mean differs. (Key point: Alternative) 2. What are the d.f.? A. (4, 24) B. (3, 20) C. (4, 20) D. (3, 23) Answer: B. Between: 4-1 = 3; within: 24-4 = 20. (Key point: Degrees of freedom) 3. How is F obtained? A. MS within / MS between B. SS between / SS within C. MS between / MS within = 2.9944/0.7568 = 3.957 D. SS between / 3.098 Answer: C. F = 3.957 using MS between over MS within. (Key point: F ratio) 4. What is the decision? A. F = 3.957 < 3.098, fail to reject B. F = 3.957 > 3.098, reject H0 C. F = 3.957 > 4.5, reject D. F = 0.25, reject Answer: B. F exceeds 3.098, so at least one cereal differs. 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