---
title: "ANOVA table: wheat varieties on four plots"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:anova-wheat-varieties-plots"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-wheat-varieties-plots
site: BCT Engineering
---

# ANOVA table: wheat varieties on four plots

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2070 Asar, 6 marks
- Video: [BCS301 Module 5 | One-Way ANOVA | Wheat Variety Production | New Model Question Paper-I](https://www.youtube.com/watch?v=Coy1WkfoFW0) — 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-wheat-varieties-plots
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

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 |

## 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).

## Answer

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.

1. 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.)
2. Level of significance: α = 0.05
3. Row totals, column totals and sums of squares:

```text
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
```

4. 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 | | |

5. Critical values: varieties F(0.05; 2, 6) = 5.14; plots F(0.05; 3, 6) = 4.76.
6. 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why two-way ANOVA?
   A. Samples are unequal
   B. Each variety is grown on every plot, so plots act as blocks
   C. Only one factor exists
   D. Variance is known
   Answer: B. Plots and varieties are crossed factors. (Key point: Method)

2. How is SS error obtained?
   A. SS plots - SS varieties
   B. 8 - 18
   C. Total + SS plots
   D. Total - SS plots - SS varieties = 6
   Answer: D. 32 - 18 - 8 = 6, with 6 d.f. (Key point: Error SS)

3. What is the result for varieties?
   A. F = 8 significant
   B. F = 6 < 5.14
   C. F = 4.0 > 5.14, significant
   D. F = 4.0 < 5.14, not significant
   Answer: D. MS varieties/MS error = 4/1 = 4 against F(0.05; 2, 6) = 5.14. (Key point: Varieties test)

4. What is the result for plots?
   A. F = 6.0 > 4.76, significant
   B. F = 18, not significant
   C. F = 6.0 < 4.76
   D. Same as varieties
   Answer: A. MS plots/MS error = 6 exceeds F(0.05; 3, 6) = 4.76. (Key point: Plots test)

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