One-way analysis of variance

Definition taught in Analysis of variance, part of Statistical inference

One-way analysis of variance

Corn yields by variety, three plots each, with the variety means in red and the grand mean dashed. From Analysis of variance.
Type Definition
Taught in Analysis of variance
Part Statistical inference
Properties 2
Worked examples 4
Questions using it 4

In a completely randomised one-factor design, responses at level j=1,\ldots,k follow

Y_{ij}=\mu_j+\varepsilon_{ij}, \qquad \varepsilon_{ij}\ \text{independent}\ \operatorname{N}(0,\sigma^2),

with n_j observations at level j and n=\sum_j n_j. The levels are the treatments. The hypothesis of equal means is H_0:\mu_1=\cdots=\mu_k.

Properties

ANOVA sum of squares decomposition

Main article: ANOVA sum of squares decomposition

With

\mathrm{SS}_{\mathrm{total}}=\sum_{j=1}^k\sum_{i=1}^{n_j}(Y_{ij}-\overline Y_{\cdot\cdot})^2, \qquad \mathrm{SS}_{\mathrm{treatment}}=\sum_{j=1}^k n_j(\overline Y_{\cdot j}-\overline Y_{\cdot\cdot})^2, \qquad \mathrm{SS}_{\mathrm{error}}=\sum_{j=1}^k\sum_{i=1}^{n_j}(Y_{ij}-\overline Y_{\cdot j})^2,

the total sum of squares splits as \mathrm{SS}_{\mathrm{total}}=\mathrm{SS}_{\mathrm{treatment}}+\mathrm{SS}_{\mathrm{error}}.

ANOVA F distribution

Main article: ANOVA F distribution

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.

Five treatments for cold sores

Worked example from Analysis of variance examples

Thirty patients with cold sores were assigned at random to five treatments, six to each; treatment 1 is a placebo. The days until the sore healed were

Placebo T2 T3 T4 T5
5 4 6 7 9
8 6 4 4 3
7 6 4 6 5
7 3 5 6 7
10 5 4 3 7
8 6 3 5 6

Question. Do the treatments differ in mean healing time?

Solution. The table is not in standard form: for an analysis the healing times must be one column and the treatment a factor in another. stack does that.

x <- data.frame(Placebo = c(5, 8, 7, 7, 10, 8), T2 = c(4, 6, 6, 3, 5, 6),
                T3 = c(6, 4, 4, 5, 4, 3), T4 = c(7, 4, 6, 6, 3, 5),
                T5 = c(9, 3, 5, 7, 7, 6))
coldsore <- stack(x)
names(coldsore) <- c("times", "treatment")
tapply(coldsore$times, coldsore$treatment, mean)
# Placebo  T2    T3    T4    T5
#   7.500 5.000 4.333 5.167 6.167
coldsore.lm <- lm(times ~ treatment, data = coldsore)
anova(coldsore.lm)
#           Df Sum Sq Mean Sq F value  Pr(>F)
# treatment  4 36.467  9.1167  3.896 0.01359
# Residuals 25 58.500  2.3400

Under equal means F=3.90 would come from an F_{4,25} distribution, and the p-value is 0.014, so the treatments do differ. The placebo has the longest mean healing time, 7.5 days. The residual mean square, 2.34, estimates the common variance, so the within-group standard deviation is about 1.5 days.

Heart rates in four age groups

Question page: Heart rates in four age groups

Four age groups each have 10 measurements of the increase in heart rate. The group totals are 309, 275, 295 and 282. The error sum of squares about the four group means is 935.500.

  1. Compute the treatment sum of squares and the F statistic for equal means.

  2. The 0.95 quantile of F_{3,36} is about 2.87. What does the test conclude at level 0.05?

The grand total is 309+275+295+282=1161, so the grand mean is 1161/40=29.025. The group means are 30.9, 27.5, 29.5 and 28.2. With 10 observations in each group,

\mathrm{SS}_{\mathrm{treatment}}=10\bigl[(30.9-29.025)^2+(27.5-29.025)^2+(29.5-29.025)^2+(28.2-29.025)^2\bigr]=67.475.

F=\frac{67.475/3}{935.500/36}=\frac{22.492}{25.986}\approx 0.866.

This is smaller than 2.87, so the test does not reject equal mean increases at level 0.05. The p-value is 0.47.

Two groups

Question page: Two groups

Show that in a one-way analysis of variance with k=2 groups, F=T^2, where T is the pooled two-sample t statistic for H_0:\mu_1=\mu_2.

With two groups, the grand mean is \overline Y_{\cdot\cdot}=(n_1\overline Y_1+n_2\overline Y_2)/n, so \overline Y_1-\overline Y_{\cdot\cdot}=n_2(\overline Y_1-\overline Y_2)/n and \overline Y_2-\overline Y_{\cdot\cdot}=-n_1(\overline Y_1-\overline Y_2)/n. Then

\mathrm{SS}_{\mathrm{treatment}}=\frac{n_1n_2^2+n_2n_1^2}{n^2}(\overline Y_1-\overline Y_2)^2=\frac{n_1n_2}{n}(\overline Y_1-\overline Y_2)^2=\frac{(\overline Y_1-\overline Y_2)^2}{1/n_1+1/n_2},

on 1 degree of freedom. The error mean square is \mathrm{SS}_{\mathrm{error}}/(n-2)=s_p^2, the pooled variance. So

F=\frac{(\overline Y_1-\overline Y_2)^2}{s_p^2(1/n_1+1/n_2)}=T^2.

Figures

One factor at three levels. Each observation’s gap from its level mean, Y_{ij}-\overline Y_{\cdot j}, goes into \mathrm{SS}_{\mathrm{error}}. Each level mean’s gap from the grand mean, \overline Y_{\cdot j}-\overline Y, counted once per observation, goes into \mathrm{SS}_{\mathrm{treatment}}. From Analysis of variance.

Healing times by treatment, one dot per patient, with the treatment means marked. The placebo group heals slowest; treatment 3 fastest. From Analysis of variance examples.

Used in

See also

ANOVA sum of squares decomposition · ANOVA F distribution · Contrast · Tukey intervals · Randomised block design

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