One-Factor Analysis of Variance

A one-factor analysis of variance (one-way ANOVA) tests whether $k$ population means are all equal, using independent random samples from each population. The hypotheses are

$$H_0:\ \mu_1 = \mu_2 = \cdots = \mu_k$$ $$H_a:\ \text{at least one } \mu_i \text{ differs from the others.}$$

We assume each population is approximately normal with a common variance $\sigma^2$.

The test statistic is

$$F = \dfrac{\text{MSB}}{\text{MSW}}, \qquad \text{MSB} = \dfrac{\text{SSB}}{k-1}, \qquad \text{MSW} = \dfrac{\text{SSW}}{N-k},$$

where $k$ is the number of groups and $N = n_1 + n_2 + \cdots + n_k$ is the total sample size. Under $H_0$, the statistic follows an F-distribution with $k-1$ numerator and $N-k$ denominator degrees of freedom.

The numerator MSB measures variation between group means; the denominator MSW measures variation within groups. A large value of $F$ indicates that the between-group variation is too large to be explained by within-group variation alone.

For example, suppose $k = 3$ groups with $n_1 = n_2 = n_3 = 5$, so $N = 15$. If SSB $= 24$ and SSW $= 36$, then

$$\begin{align*} \text{MSB} &= \dfrac{24}{3-1} = 12, \\ \text{MSW} &= \dfrac{36}{15-3} = 3, \\ F &= \dfrac{12}{3} = 4, \end{align*}$$

with $(2,\ 12)$ degrees of freedom.