---
title: "ANOVA: three oil spill clean-up systems"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:anova-oil-spill-systems"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-oil-spill-systems
site: BCT Engineering
---

# ANOVA: three oil spill clean-up systems

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2076 Asoj, 5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-oil-spill-systems
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

A research company has designed three different systems to clean up oil spills. The following table contains the results, measured by how much surface area in square meters is cleared in one hour. The data were found by testing each method in several trials. Are the three systems equally effective? Use the 0.05 level of significance.
| System A | 55 | 60 | 63 | 56 | 59 | 55 |
|---|---|---|---|---|---|---|
| System B | 57 | 53 | 64 | 49 | 62 | - |
| System C | 66 | 52 | 61 | 57 | - | - |

## Short answer

One-way ANOVA yields F = 0.172 < F_crit = 3.885, so H₀ is not rejected. There is no significant difference among the three clean-up systems at the 5% level.

## Answer

One-way ANOVA with three systems (k = 3) and unequal sample sizes.

**Hypotheses**

- H₀: μ_A = μ_B = μ_C (the three systems are equally effective)
- H₁: at least one system has a different mean

**Level of significance:** α = 0.05

Sample sizes: n_A = 6, n_B = 5, n_C = 4, N = 15.
Group totals: T_A = 348, T_B = 285, T_C = 236; grand total G = 869.

- Correction factor CF = G²/N = 869²/15 = 50344.067
- Σx² (all observations) = 50665
- Total SS = Σx² − CF = 320.933
- SS between = Σ(Tᵢ²/nᵢ) − CF = 50353.000 − 50344.067 = 8.933
- SS within (error) = Total SS − SS between = 312.000

#### ANOVA table

| Source | SS | df | MS | F |
|---|---|---|---|---|
| Between groups | 8.933 | 2 | 4.467 | 0.172 |
| Within groups (error) | 312.000 | 12 | 26.000 | |
| Total | 320.933 | 14 | | |

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

**Decision:** calculated F = 0.172 < 3.885 (p-value ≈ 0.844), so we do not reject H₀.

**Conclusion:** there is no significant difference between the systems at the 5% level; the three clean-up systems can be regarded as equally effective.

## Check yourself

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

1. H0 here?
   A. Three systems equally effective
   B. At least one differs
   C. Variances equal only
   D. Sizes equal
   Answer: A. H0: μA = μB = μC. (Key point: Hypotheses)

2. Within-group df with N = 15, k = 3?
   A. 2
   B. 14
   C. 12
   D. 3
   Answer: C. N − k = 12. (Key point: df)

3. F is:
   A. MS between/MS within = 0.172
   B. SS between/SS within
   C. 26/4.467
   D. 3.885
   Answer: A. 4.467/26 = 0.172. (Key point: F)

4. Conclusion?
   A. Systems differ
   B. No significant difference at 5%
   C. Reject H0
   D. Insufficient data
   Answer: B. 0.172 < 3.885, so fail to reject H0. (Key point: Decision)

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