ANOVA F distribution
Theorem taught in Analysis of variance, part of Statistical inference
ANOVA F distribution
| Type | Theorem |
| Taught in | Analysis of variance |
| Part | Statistical inference |
| Worked examples | 1 |
In the one-way model, \mathrm{SS}_{\mathrm{error}}/\sigma^2 is \chi^2_{n-k}, and it is independent of \mathrm{SS}_{\mathrm{treatment}}. Under H_0:\mu_1=\cdots=\mu_k, \mathrm{SS}_{\mathrm{treatment}}/\sigma^2 is \chi^2_{k-1}, and therefore
F=\frac{\mathrm{SS}_{\mathrm{treatment}}/(k-1)}{\mathrm{SS}_{\mathrm{error}}/(n-k)}=\frac{\mathrm{MS}_{\mathrm{treatment}}}{\mathrm{MS}_{\mathrm{error}}}
has the F_{k-1,\,n-k} distribution.
Examples
Five varieties of corn
Worked example from Analysis of variance
Five varieties of corn are each grown on three plots.
| Variety | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Yields | 46.2, 51.9, 48.7 | 49.2, 58.6, 57.4 | 60.3, 58.7, 60.4 | 48.9, 51.4, 44.6 | 52.5, 54.0, 49.3 |
| Mean | 48.93 | 55.07 | 59.80 | 48.30 | 51.93 |
Question. Do the varieties differ in mean yield?
Solution. The grand mean is 52.81. Then \mathrm{SS}_{\mathrm{treatment}}=3\sum_j(\overline y_{\cdot j}-52.81)^2=270.3 on 4 degrees of freedom, and \mathrm{SS}_{\mathrm{error}}=105.7 on 10.
| Source | Df | SS | MS | F | p |
|---|---|---|---|---|---|
| Variety | 4 | 270.3 | 67.57 | 6.39 | 0.008 |
| Error | 10 | 105.7 | 10.57 | ||
| Total | 14 | 375.9 |
The p-value is 0.008, so the varieties differ. The pooled standard deviation is \sqrt{10.57}=3.25. The F test says only that the five means are not all equal; which varieties differ is the next question.
Used in
- Taught in Analysis of variance, where the derivation and examples are.
- All 6 questions on Analysis of variance
See also
One-way analysis of variance · ANOVA sum of squares decomposition · Contrast · Tukey intervals · Randomised block design