---
title: "ANOVA: chicken weight gain on three diets"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:anova-chickens-three-diets"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-chickens-three-diets
site: BCT Engineering
---

# ANOVA: chicken weight gain on three diets

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2074 Asoj, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-chickens-three-diets
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

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.

## 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.

## Answer

One-way ANOVA is used to compare three means (equal variances and normality assumed).

1. 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)
2. Level of significance: α = 0.05
3. Computations:

```text
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
```

4. 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 | | |

5. Critical value: F(0.05; 2, 12) = 3.885. Reject H0 if F > 3.885.
6. 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.

## Check yourself

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

1. Why is one-way ANOVA suited to this problem?
   A. It compares two paired means
   B. It compares three group means with one factor
   C. It tests one proportion
   D. It tests a single variance
   Answer: B. Three diets are three levels of one factor, so one-way ANOVA is used. (Key point: Method)

2. How is SS between computed?
   A. Sum of x^2 minus sum of Ti
   B. Total SS + SS within
   C. Sum of Ti^2/ni minus the correction factor
   D. Sum of (x - 6)^2
   Answer: C. SS between = 550 - 540 = 10 using CF = T^2/N. (Key point: SS between)

3. What are the d.f. in the F-test?
   A. (2, 12)
   B. (3, 12)
   C. (3, 15)
   D. (2, 14)
   Answer: A. Between: k-1 = 2; within: N-k = 15-3 = 12. (Key point: Degrees of freedom)

4. What is the decision?
   A. F = 2.308 < 3.885, fail to reject H0
   B. F = 5 > 3.885, reject H0
   C. F = 2.308 > 3.885, reject H0
   D. F = 2.308 < 3.885, reject H0
   Answer: A. Calculated F is below the critical value, so the diets do not differ significantly. (Key point: Decision)

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