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
...

 A: 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  

