One-Way ANOVA

Summary

One-way analysis of variance compares the means of several groups defined by a single categorical factor. It partitions total variation into variation between groups and variation within groups, then tests whether the between-group variation is large relative to the within-group variation using an F-statistic. Significant results are usually followed by post-hoc comparisons to identify which means differ.^[1]

Prerequisites

Definition / Notation

The model for group $i = 1, \dots, k$ and observation $j = 1, \dots, n_{i}$ is

Y_{i j} = μ + α_{i} + ε_{i j},

with the constraint $\sum_{i = 1}^{k} n_{i} α_{i} = 0$ .

Symbol	Meaning
$Y_{i j}$	Observation $j$ in group $i$
$μ$	Overall mean
$α_{i}$	Effect of group $i$
${\bar{Y}}_{i .}$	Mean of group $i$
${\bar{Y}}_{. .}$	Grand mean
$S S T$	Total sum of squares
$S S B$	Between-group sum of squares
$S S E$	Within-group (error) sum of squares
$N$	Total sample size, $\sum_{i} n_{i}$

The sums of squares are

S S T = \sum_{i = 1}^{k} \sum_{j = 1}^{n_{i}} (Y_{i j} - {\bar{Y}}_{. .})^{2},

S S B = \sum_{i = 1}^{k} n_{i} ({\bar{Y}}_{i .} - {\bar{Y}}_{. .})^{2},

S S E = \sum_{i = 1}^{k} \sum_{j = 1}^{n_{i}} (Y_{i j} - {\bar{Y}}_{i .})^{2} .

They satisfy $S S T = S S B + S S E$ .

Parameters / Assumptions

Independent random samples from $k$ populations.
Each population is approximately normal.
The populations share a common variance $σ^{2}$ (homoscedasticity).
Observations are independent within and across groups.

Essential Result

The hypotheses are

H_{0} : μ_{1} = μ_{2} = \dots = μ_{k}

against

H_{a} : at least two means differ .

The test statistic is

F = \frac{M S B}{M S E} = \frac{S S B / (k - 1)}{S S E / (N - k)} \sim F_{k - 1, N - k} under H_{0} .

Reject $H_{0}$ if $F > F_{α, k - 1, N - k}$ . When $H_{0}$ is rejected, post-hoc procedures such as Tukey's HSD or Bonferroni corrections control the family-wise error rate for pairwise comparisons.^[2]

Worked Example

Three teaching methods produce scores:

Method A: 78, 82, 85
Method B: 88, 90, 92
Method C: 80, 84, 86

Group means are $81.67$ , $90$ , and $83.33$ ; the grand mean is $85$ . The sums of squares are

S S B \approx 116.67, S S E \approx 51.33, S S T \approx 168.00 .

With $k = 3$ and $N = 9$ , $M S B = 58.33$ and $M S E = 8.56$ , giving

F = \frac{58.33}{8.56} \approx 6.82 .

Since $F_{0.05, 2, 6} \approx 5.14$ , we reject $H_{0}$ and conclude that at least one method mean differs.

Common Mistakes

Running many pairwise t-tests instead of ANOVA, which inflates the Type I error rate.
Ignoring the equal-variance assumption.
Concluding that all groups differ when only some do.
Using ANOVA when the response is not continuous or the groups are not independent.

Connections

References

OpenStax, Introductory Statistics, "One-Way ANOVA", https://openstax.org/details/books/introductory-statistics ↩︎
NIST/SEMATECH, e-Handbook of Statistical Methods, "Analysis of Variance", https://www.itl.nist.gov/div898/handbook/ ↩︎