---
title: "ANOVA: breaking strength of three cable brands"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:anova-three-brands-cables"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-three-brands-cables
site: BCT Engineering
---

# ANOVA: breaking strength of three cable brands

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2075 Chaitra, 5 marks; 2069 Chaitra, 6 marks (asked 2 times)
- Video: [MA25C11 | MA25C05 | MA3251 | Completely Randomized Design | One-way Classification](https://www.youtube.com/watch?v=4vfHNqxO8rc) — Mathematics Kala
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-three-brands-cables
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

The following are the breaking strength of three different brands of cables.
| Brand | Breaking Strength |
|---|---|
| A | 40 30 50 60 30 |
| B | 60 40 55 65 |
| C | 60 50 70 65 75 40 |
Construct ANOVA table and test for the equality of the average breaking strength of cables at α = 5%.

## Short answer

One‑way ANOVA gives F = 2.915 (df: 2, 12). At α = 0.05, F_crit = 3.885, so fail to reject H₀. No significant difference in mean breaking strengths.

## Answer

One-way ANOVA with three brands (k = 3).

**Hypotheses**

- H₀: μ_A = μ_B = μ_C (mean breaking strengths are equal)
- H₁: at least one mean is different

**Level of significance:** α = 0.05

Sample sizes: n_A = 5, n_B = 4, n_C = 6, N = 15.
Group totals: T_A = 210, T_B = 220, T_C = 360; grand total G = 790.

- Correction factor CF = G²/N = 790²/15 = 41606.667
- Σx² (all observations) = 44400
- Total SS = Σx² − CF = 2793.333
- SS between = Σ(Tᵢ²/nᵢ) − CF = 42520.000 − 41606.667 = 913.333
- SS within (error) = Total SS − SS between = 1880.000

#### ANOVA table

| Source | SS | df | MS | F |
|---|---|---|---|---|
| Between groups | 913.333 | 2 | 456.667 | 2.915 |
| Within groups (error) | 1880.000 | 12 | 156.667 | |
| Total | 2793.333 | 14 | | |

**Critical value:** F(0.05; 2, 12) = 3.885

**Decision:** calculated F = 2.915 < 3.885 (p-value ≈ 0.093), so we do not reject H₀.

**Conclusion:** there is not sufficient evidence of a difference; the average breaking strengths of the three brands of cable can be regarded as equal at the 5% level.

## Check yourself

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

1. What is H0 in one-way ANOVA?
   A. All brand means are equal
   B. At least one mean differs
   C. All variances differ
   D. Totals are equal
   Answer: A. H0: μA = μB = μC. (Key point: Hypotheses)

2. Between and within df here?
   A. 3 and 15
   B. 2 and 12
   C. 3 and 12
   D. 2 and 14
   Answer: B. k − 1 = 2 and N − k = 12. (Key point: Degrees of freedom)

3. How is F computed?
   A. MS within / MS between
   B. SS between / SS within
   C. MS between / MS within = 2.915
   D. Total SS / df
   Answer: C. F = 456.667/156.667 = 2.915. (Key point: F ratio)

4. Why not reject H0?
   A. F = 2.915 < 3.885
   B. F = 2.915 > 3.885
   C. p < 0.05
   D. SS between is zero
   Answer: A. F is less than F(0.05; 2, 12), so the brand means can be regarded as equal. (Key point: Decision)

## Related in Probability and Statistics

- [Breaking strength: average, median, SD](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Abreaking-strength-three-materials) — Probability and Statistics, chapter 1
- [Test: wear of two tyre brands](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Atyres-two-brands-wear-test) — Probability and Statistics, chapter 6
- [Test: two bulb brands differ in quality](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Alight-bulbs-two-brands-differ-quality) — Probability and Statistics, chapter 6
- [ANOVA: productivity of three operators](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aanova-three-operators) — Probability and Statistics, chapter 6, asked 2×
- [ANOVA table: wheat varieties on four plots](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aanova-wheat-varieties-plots) — Probability and Statistics, chapter 6
- [ANOVA: three oil spill clean-up systems](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aanova-oil-spill-systems) — Probability and Statistics, chapter 6

## Related in other subjects

- [Three reservoirs ZA 65, ZB 70, ZC 40](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Athree-reservoir-pipes-ad-bd-dc) — Hydraulics, chapter 3
- [Cable size MDB to SDB-1 (60 kW) and SDB-2 (75 kW)](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Acable-mdb-sdb-60kw-75kw) — Industrial Power Distribution and Illumination, chapter 5, asked 4×
- [Factory hall 55.75 m × 40.25 m LED lighting](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Afactory-hall-led-55x40) — Industrial Power Distribution and Illumination, chapter 7, asked 4×
- [Power/light sub-circuits and cable for 50 kW, 60 kW](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Asub-circuits-cable-50kw-60kw) — Industrial Power Distribution and Illumination, chapter 5, asked 4×