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I am giving the number of trials and number of successes for 4 Bernoulli samples.

I want to test the null hypothesis that p1 = p2 = p3 = p4 versus the alternative hypothesis that pi != pj for any i and j between 1 and 4. (px is the probability of success for population x)

How do I go about this? I only understand how to do this with two populations.

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You can use Pearson's chi-squared test to do this. See

A test of homogeneity compares the distribution of counts for two or more groups using the same categorical variable (e.g. choice of activity—college, military, employment, travel—of graduates of a high school reported a year after graduation, sorted by graduation year, to see if number of graduates choosing a given activity has changed from class to class, or from decade to decade).[3]

Code example:

 #fake data

 nl <- c(20,25,30,20) # sample size
 pl1 <- c(0.2,0.3,0.4,0.5) # diffrent p
 pl2 <- c(0.3,0.3,0.3,0.3) # same p

 tab1 <- sapply(1:4,function(i){
   s <- sum(rbinom(nl[i],1,pl1[i]))
   return(c(nl[i]-s,s))
 })
 tab2 <- sapply(1:4,function(i){
   s <- sum(rbinom(nl[i],1,pl2[i]))
   return(c(nl[i]-s,s))
 })

 tab1
 #     [,1] [,2] [,3] [,4]
 #[1,]   19   19   24    8
 #[2,]    1    6    6   12


 # testing
 chisq.test(tab1) # p-value = 0.0006819
 chisq.test(tab2) # p-value = 0.8818
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