---
title: "ANOVA: distance to fire stations in four areas"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:anova-fire-station-distance"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-fire-station-distance
site: BCT Engineering
---

# ANOVA: distance to fire stations in four areas

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2072 Kartik, 5 marks
- Video: [What is ANOVA (Analysis of Variance) in Statistics ? | Explained with Examples (ANOVA F - test)](https://www.youtube.com/watch?v=x_gHAly3mJo) — Digital E-Learning
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-fire-station-distance
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

In the investigation of a citizens' committee complaint about the availability of fire protection within the country, the distance in miles to the nearest fire station was measured for each of five randomly selected residences in each of four areas.
| Area 1 | 7 | 5 | 5 | 6 | 8 |
|---|---|---|---|---|---|
| Area 2 | 1 | 4 | 3 | 4 | 5 |
| Area 3 | 7 | 9 | 8 | 7 | 8 |
| Area 4 | 4 | 6 | 3 | 7 | 5 |
Do these data provide sufficient evidence to indicate a difference in mean distance for the four areas at the α = 0.05 level of significance?

## Short answer

One-way ANOVA gives F = 9.630. At α = 0.05, critical F(3,16) = 3.239, so reject H0. The mean distance to the nearest fire station differs significantly among the four areas.

## Answer

One-way ANOVA with four areas (5 residences each).

1. H0: μ₁ = μ₂ = μ₃ = μ₄ (mean distance is the same in all four areas)
   H1: at least one area has a different mean distance (F-test, right-tailed)
2. Level of significance: α = 0.05
3. Computations:

```text
Area 1       n=5  total=31  mean=6.2000
Area 2       n=5  total=17  mean=3.4000
Area 3       n=5  total=39  mean=7.8000
Area 4       n=5  total=25  mean=5.0000

Grand total T = 112,  N = 20
Correction factor CF = T²/N = 112²/20 = 627.2000
Total SS  = Σx² − CF = 708.0000 − 627.2000 = 80.8000
SS between = Σ(Ti²/ni) − CF = 679.2000 − 627.2000 = 52.0000
SS within  = Total SS − SS between = 28.8000
```

4. ANOVA table:

| Source | SS | d.f. | MS | F |
|---|---|---|---|---|
| Between areas | 52.0000 | 3 | 17.3333 | 9.630 |
| Within (error) | 28.8000 | 16 | 1.8000 | |
| Total | 80.8000 | 19 | | |

5. Critical value: F(0.05; 3, 16) = 3.239. Reject H0 if F > 3.239.
6. Decision: the calculated value falls in the rejection region (p-value = 0.0007), so **reject H0**.

**Conclusion:** the mean distance to the nearest fire station differs significantly among the four areas.

## Check yourself

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

1. Which analysis applies?
   A. z-test
   B. One-way ANOVA with four areas
   C. Two-way ANOVA
   D. Paired t-test
   Answer: B. Four independent groups of one factor: one-way ANOVA. (Key point: Method)

2. How is SS within found?
   A. SS between - Total SS
   B. 52/16
   C. Total SS - SS between = 80.8 - 52 = 28.8
   D. Total SS + 52
   Answer: C. SS within = 28.8, d.f. 20-4 = 16, MS = 1.8. (Key point: Within SS)

3. What is F?
   A. 52/28.8
   B. 28.8/80.8
   C. 1.8/17.333
   D. MS between / MS within = 17.333/1.8 = 9.63
   Answer: D. MS between = 52/3 = 17.333, so F = 9.63. (Key point: F value)

4. What is the decision?
   A. Cannot decide
   B. F = 9.63 > 3.239, reject H0; areas differ
   C. Fail to reject
   D. F = 9.63 < 3.239
   Answer: B. F exceeds the critical value F(0.05; 3, 16) = 3.239. (Key point: Decision)

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