# Three binomial random variables: test null that no two have equal mean

I have a dummy variable for health (0 or 1), and for each observation a treatment (one of three possible treatments for each observation). I want to test the null hypothesis that the proportion (mean) healthy within each treatment group is not significantly different than any other treatment group's proportion. \begin{align} H_0\!: P_j = P_{j'}\quad &\text{for some } j \ne j' \\ H_A\!\!: P_j\ne P_{j'}\quad &\text{for all } j \ne j' \end{align} That is, my alternative hypothesis is that every proportion (mean) is different than every other proportion. My data looks something like this:

treatment, health
1,         1
1,         0
2,         0
2,         0
3,         1
3,         1
...

• I think the simplest way to achieve this would be to test each pair. Your composite null is kind of tricky. – Glen_b Apr 20 '14 at 22:18

You can use a multiple-degrees of freedom test and logistic regression. I will show this with some R code. First simulating some data:

p <- c(0.5, 0.5, 0.25)
set.seed(7*11*13)# public seed
m <- 40
N <- 3*m
T <- as.factor(rep(1:3, rep(m, 3)))
Y <- rbinom(N, 1, rep(p, rep(m, 3)))


The null hypothesis is that $$p_2=p_3$$ and $$p_1=p_3$$ (and $$p_1=p_2$$, but that is already implied by the first two). We can represent that with a contrast hypothesis with two rows. We fit a logistic regression model without intercept, to get one parameter for each group. Note that since this is a saturated model, some other link function would give the same estimated probabilities.

library(car)# for car::linearHypothesis
mod <- glm(Y ~ 0+T, family=binomial() )
summary(mod) # output not shown

car::linearHypothesis(mod, rbind(c(0, 1, -1),
c(1, 0, -1)), test="Chisq", verbose=TRUE)

Hypothesis matrix:
[,1] [,2] [,3]
[1,]    0    1   -1
[2,]    1    0   -1

Right-hand-side vector:
[1] 0 0

Estimated linear function (hypothesis.matrix %*% coef - rhs)
[1] 1.2367626 0.9344818

Estimated variance/covariance matrix for linear function
[,1]      [,2]
[1,] 0.243369 0.1433690
[2,] 0.143369 0.2456708

Linear hypothesis test

Hypothesis:
T2 - T3 = 0
T1 - T3 = 0

Model 1: restricted model
Model 2: Y ~ 0 + T

Res.Df Df Chisq Pr(>Chisq)
1    119
2    117  2 6.548    0.03785 *
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

• "The null hypothesis is that $p_2 = p_3$ and $p_1 = p_3$" I believe that there is a composite null hypothesis rather than a single point hypothesis $p_1=p_2=p_3$. E.g the null hypothesis includes any of the four situations $$\begin {array}{}p_1=p_2 \neq p_3\\p_2=p_3 \neq p_1\\p_1=p_3 \neq p_2\\p_1 = p_2 = p_3\end {array}$$ – Sextus Empiricus Jan 23 '19 at 14:35
• What I have done is to formulate a composite hypothesis? so I cannot see what is wrong. This then must be tested simultaneouls, using a contrast matrix. – kjetil b halvorsen Jan 23 '19 at 14:49
• You state that the null hypothesis/effect is $p_1=p_2$ and $p_1=p_3$, but it should be or. – Sextus Empiricus Jan 23 '19 at 14:53