- 2074 Asoj · 6 marks
ANOVA: chicken weight gain on three diets
Short answer
One-way ANOVA: SS between = 10, MS between = 5, F = 2.308 with df (2,12). F < critical 3.885; fail to reject H0. Mean weight gains do not differ significantly across diets.
Three randomly selected groups of chickens are fed on three different diets. Each group consists of five chickens. Their weight gains during a specified period of time are as follows:
| Diet I | 4 | 4 | 7 | 7 | 8 |
|---|---|---|---|---|---|
| Diet II | 3 | 4 | 5 | 6 | 7 |
| Diet III | 6 | 7 | 7 | 7 | 8 |
| Test the hypothesis that mean gains of weight due to the three diets are equal. |
One-way ANOVA is used to compare three means (equal variances and normality assumed).
- H0: μ₁ = μ₂ = μ₃ (mean weight gains due to the three diets are equal) H1: at least one diet mean differs (F-test, right-tailed)
- Level of significance: α = 0.05
- Computations:
Diet I n=5 total=30 mean=6.0000
Diet II n=5 total=25 mean=5.0000
Diet III n=5 total=35 mean=7.0000
Grand total T = 90, N = 15
Correction factor CF = T²/N = 90²/15 = 540.0000
Total SS = Σx² − CF = 576.0000 − 540.0000 = 36.0000
SS between = Σ(Ti²/ni) − CF = 550.0000 − 540.0000 = 10.0000
SS within = Total SS − SS between = 26.0000
- ANOVA table:
| Source | SS | d.f. | MS | F |
|---|---|---|---|---|
| Between diets | 10.0000 | 2 | 5.0000 | 2.308 |
| Within (error) | 26.0000 | 12 | 2.1667 | |
| Total | 36.0000 | 14 |
- Critical value: F(0.05; 2, 12) = 3.885. Reject H0 if F > 3.885.
- Decision: the calculated value falls in the acceptance region (p-value = 0.1419), so fail to reject H0.
Conclusion: the mean weight gains due to the three diets do not differ significantly; they may be taken as equal.
✓ Answer checked by Claude Sonnet 5.5
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