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One-way ANOVA

One-way ANOVA (analysis of variance) tests whether the means of three or more groups are equal. It partitions the total variation in the data into two sources: between-group variation (due to the factor) and within-group variation (random noise). The F-statistic is the ratio of these two estimates of variance.

The procedure

Null hypothesis: H₀: μ₁ = μ₂ = μ₃ (all group means equal).
Alternative: H₁: at least one mean differs.

SS_between = Σ nᵢ × (x̄ᵢ − x̄_grand)² (df = k − 1) SS_within = Σᵢ Σⱼ (xᵢⱼ − x̄ᵢ)² (df = N − k) MS_between = SS_between / (k − 1) MS_within = SS_within / (N − k) F = MS_between / MS_within

Reject H₀ if F > F_critical (from F-table at df₁ = k−1, df₂ = N−k, α).

Worked example

Three machines (k = 3), five measurements each (n = 5, N = 15).

Machine A	Machine B	Machine C
10, 12, 11, 13, 9	16, 18, 17, 19, 15	13, 14, 13, 15, 15
mean = 11	mean = 17	mean = 14

Grand mean x̄_grand = (55 + 85 + 70) / 15 = 14 SS_between = 5 × 9 + 5 × 9 + 5 × 0 = 90 SS_within(A) = (10−11)²+(12−11)²+(11−11)²+(13−11)²+(9−11)² = 10 SS_within(B) = (16−17)²+(18−17)²+(17−17)²+(19−17)²+(15−17)² = 10 SS_within(C) = (13−14)²+(14−14)²+(13−14)²+(15−14)²+(15−14)² = 4 SS_within = 10 + 10 + 4 = 24 MS_between = 90 / 2 = 45 MS_within = 24 / 12 = 2 F = 45 / 2 = 22.5

F_critical(2, 12, 0.05) ≈ 3.89. Since 22.5 >> 3.89, reject H₀. The machines produce significantly different means.

Practice problem

SS_between = 60 (df = 2), SS_within = 90 (df = 18). Compute F.

Show answer

MS_between = 60/2 = 30. MS_within = 90/18 = 5. F = 30/5 = 6.0.

Common exam traps

ANOVA assumes equal variances and normality within groups. Bartlett’s or Levene’s test checks the equal-variance assumption. Severe violations require a Welch ANOVA or Kruskal–Wallis test.
A significant F tells you some mean differs, not which ones. Post-hoc tests (Tukey, Bonferroni) identify the specific pairs.
SS_between + SS_within = SS_total. Use this to check your arithmetic.

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