One-way ANOVA tests \(H_0: \mu_1 = \mu_2 = \dots = \mu_k\) by comparing the variance between group means with the variance within groups through \(F = MSB/MSW\).
Steps
- State \(H_0\): all group means equal; \(H_1\): at least one differs. Choose \(\alpha\) (say 0.05).
- Find each group mean \(\bar x_i\) and the grand mean \(\bar x\).
- Between-groups sum of squares: \(SSB = \sum n_i(\bar x_i - \bar x)^2\), df \(= k-1\).
- Within-groups sum of squares: \(SSW = \sum\sum (x_{ij} - \bar x_i)^2\), df \(= N-k\).
- Mean squares: \(MSB = SSB/(k-1)\), \(MSW = SSW/(N-k)\); then \(F = MSB/MSW\).
- Compare with table value \(F_\alpha(k-1, N-k)\). If calculated \(F\) > table \(F\), reject \(H_0\).
Example. Scores under three teaching methods:
- A: 2, 3, 4 (mean 3)
- B: 4, 5, 6 (mean 5)
- C: 6, 7, 8 (mean 7)
Grand mean \(= 45/9 = 5\), \(k = 3\), \(N = 9\).
\[SSB = 3[(3-5)^2 + 0^2 + (7-5)^2] = 24\]
\[SSW = 2 + 2 + 2 = 6\]
(each group: \(1 + 0 + 1 = 2\)).
| Source | SS | df | MS | F |
|---|---|---|---|---|
| Between | 24 | 2 | 12 | 12 |
| Within | 6 | 6 | 1 | |
| Total | 30 | 8 |
Table value \(F_{0.05}(2, 6) = 5.14\). Since \(12 > 5.14\), reject \(H_0\): the teaching methods give significantly different mean scores. (A post-hoc test such as Tukey's HSD then shows which pairs differ.)
Assumptions: independent samples, normality in each group, and equal variances.