Tukey intervals

Definition taught in Analysis of variance, part of Statistical inference

Tukey intervals

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
Worked examples	3
Questions using it	1

In a one-way design with k groups of equal size r, and n=kr, the probability is 1-\alpha that every one of the \binom k2 differences \mu_i-\mu_j simultaneously satisfies

\overline Y_{\cdot i}-\overline Y_{\cdot j}-D\sqrt{\mathrm{MS}_{\mathrm{error}}}<\mu_i-\mu_j<\overline Y_{\cdot i}-\overline Y_{\cdot j}+D\sqrt{\mathrm{MS}_{\mathrm{error}}}, \qquad D=\frac{Q_{k,\,n-k;\,1-\alpha}}{\sqrt r},

where Q_{k,\,n-k;\,1-\alpha} is the quantile of the studentised range for k groups and n-k error degrees of freedom.

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.

A Tukey interval for two varieties

Question page: A Tukey interval for two varieties

Five corn varieties are grown in r=3 plots each. The variety means include \overline y_3=59.800 and \overline y_4=48.300. The residual mean square on 10 degrees of freedom is 10.57. The studentised-range quantile Q_{5,10;\,0.95} is 4.654.

Compute the Tukey interval for \mu_3-\mu_4 at familywise level 0.05, and say whether the test declares the two means different.

Worked answer (click to show)

The estimated difference is 59.800-48.300=11.5. The half-width is

4.654\sqrt{\frac{10.57}{3}}=4.654\times1.877\approx 8.74.

The interval is 11.5\pm 8.74, that is (2.76,\,20.24). It excludes zero, so the two variety means are declared different at familywise level 0.05.

Used in

Taught in Analysis of variance, where the derivation and examples are.
Also used in Analysis of variance examples
Questions that use it (1) — A Tukey interval for two varieties
All 6 questions on Analysis of variance