Bonferroni correction

Definition taught in Analysis of variance examples, part of Data analysis

Bonferroni correction

Healing times by treatment, one dot per patient, with the treatment means marked. From Analysis of variance examples.
Type Definition
Taught in Analysis of variance examples
Part Data analysis
Worked examples 1
Questions using it 1

To make m comparisons with a familywise error rate of at most \alpha, run each at level \alpha/m. Equivalently, multiply each p-value by m, capping at 1.

Examples

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.

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