- 2072 Kartik · 5 marks
ANOVA: distance to fire stations in four areas
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.
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? |
One-way ANOVA with four areas (5 residences each).
- 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)
- Level of significance: α = 0.05
- Computations:
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
- 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 |
- Critical value: F(0.05; 3, 16) = 3.239. Reject H0 if F > 3.239.
- 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.
✓ Answer checked by Claude Sonnet 5.5
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