Assume we have a population of 100,000 where groups A and B are independent with \(p_A = .55\) and \(p_B = .6\) and \(n_A = 99,000\) (99% of the population) and \(n_B = 1,000\) (1% of the population). We can sample from the population (that includes groups A and B) and from group B of sample sizes of 1,000 and 100, respectively. We can also calculate \(\hat{p}\) for group A independent of B.

propA <- .55 # Proportion for group A
propB <- .6 # Proportion for group B
pop.n <- 100000 # Population size
sampleA.n <- 1000
sampleB.n <- 100
pop <- data.frame(
group = c(rep('A', pop.n * 0.99),
rep('B', pop.n * 0.01) ),
response = c(
sample(c(1,0), size = pop.n * 0.99, prob = c(propA, 1 - propA),
replace = TRUE),
sample(c(1,0), size = pop.n * 0.01, prob = c(propB, 1 - propB),
replace = TRUE) )
)
sampA <- pop[sample(nrow(pop), size = sampleA.n),]
sampB <- pop[sample(which(pop$group == 'B'), size = sampleB.n),]