rsu.sssep.rbmrg(pstar, rr, ppr, spr, spr.rg, se.p, se.u)

## EXAMPLE 1:
## You are working with a disease of cattle where the prevalence is believed 
## to vary according to herd type. The risk of disease is 5 times greater 
## in dairy herds and 3 times greater in mixed herds compared with the 
## reference category, beef herds. The distribution of dairy, mixed and beef 
## herds in the population of interest is 0.10, 0.10 and 0.80, respectively. 
## You intend to distribute your sampling effort 0.4, 0.4 and 0.2 across dairy, 
## mixed and beef herds, respectively.

## Within each of the three risk groups a single test with a diagnostic 
## sensitivity of 0.95 will be used. How many herds need to be sampled if 
## you want to be 95% certain of detecting disease if it is present in the 
## population at a prevalence of 1% or greater?

## Generate a matrix listing the proportions of samples for each test in 
## each risk group (the number of rows equal the number of risk groups, 
## the number of columns equal the number of tests):

m <- rbind(1,1,1)

rsu.sssep.rbmrg(pstar = 0.01, rr = c(5,3,1), ppr = c(0.1,0.1,0.8),
   spr = c(0.4,0.4,0.2), spr.rg = m, se.p = 0.95, se.u = 0.95)

## A total of 147 herds need to be sampled: 59 dairy, 59 mixed and 29 
## beef herds.

## EXAMPLE 2:
## Now assume that one of two tests will be used for each herd. The first 
## test has a diagnostic sensitivity of 0.92. The second test has a diagnostic 
## sensitivity of 0.80. The proportion of dairy, mixed and beef herds receiving 
## the first test is 0.80, 0.50 and 0.70, respectively (which means that 0.20,
## 0.50 and 0.30 receive the second test, respectively).

## Recalculate the sample size.

m <- rbind(c(0.8,0.2), c(0.5,0.5), c(0.7,0.3))

rsu.sssep.rbmrg(pstar = 0.01, rr = c(5,3,1), ppr = c(0.1,0.1,0.8),
   spr = c(0.4,0.4,0.2), spr.rg = m, se.p = 0.95, se.u = c(0.92,0.80))

## A total of 159 herds need to be sampled: 64 dairy, 64 mixed and 31 
## beef herds.