- 2070 Asar · 6 marks
ANOVA table: wheat varieties on four plots
Short answer
Two-way ANOVA with plots as blocks: variety F=4.000, critical F(0.05;2,6)=5.14, so variety differences not significant at 5%. Plot differences are significant (F=6.000 > 4.76).
Set up an ANOVA table for the following acre production of data for three varieties of wheat, each growth on 4 plots and state if the variety differences are significant. Use α = 0.05
| Plot of land | Variety of wheat |
|---|---|
| 1 | 6 5 5 |
| 2 | 7 5 4 |
| 3 | 3 3 3 |
| 4 | 8 7 4 |
Each variety is grown on every plot, so plots act as blocks. A two-way ANOVA (plots = rows, varieties = columns) separates plot differences from variety differences.
- H0 (varieties): μ_V1 = μ_V2 = μ_V3 (no difference between varieties) H1: at least one variety mean differs (For plots: H0: all plot means are equal.)
- Level of significance: α = 0.05
- Row totals, column totals and sums of squares:
plots totals: 1=16, 2=16, 3=9, 4=19
varieties totals: V1=24, V2=20, V3=16
T = 60, N = 12, CF = T²/N = 300.0000
Total SS = Σx² − CF = 332 − 300.0000 = 32.0000
SS plots = Σ(row total²)/3 − CF = 18.0000
SS varieties = Σ(col total²)/4 − CF = 8.0000
SS error = Total − SSplots − SSvarieties = 6.0000
- Two-way ANOVA table:
| Source | SS | d.f. | MS | F |
|---|---|---|---|---|
| Between plots | 18.000 | 3 | 6.000 | 6.000 |
| Between varieties | 8.000 | 2 | 4.000 | 4.000 |
| Error | 6.000 | 6 | 1.000 | |
| Total | 32.000 | 11 |
- Critical values: varieties F(0.05; 2, 6) = 5.14; plots F(0.05; 3, 6) = 4.76.
- Decision:
- Varieties: F = 4.000 < 5.14, so H0 is not rejected.
- Plots: F = 6.000 > 4.76, so H0 is rejected.
Conclusion: The differences between the wheat varieties are not significant at the 5% level; the differences between plots are significant.
✓ Answer checked by Claude Sonnet 5.5
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