- 2073 Chaitra · 5 marks
ANOVA: thiamin content of four cereals
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.
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. |
One-way ANOVA with four cereal types (6 samples each).
- 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)
- Level of significance: α = 0.05
- Computations:
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
- 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 |
- Critical value: F(0.05; 3, 20) = 3.098. Reject H0 if F > 3.098.
- 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.
✓ Answer checked by Claude Sonnet 5.5
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