SS(par1.1, par1.2, par2.1, par2.2=NULL, rho, width,
costs=matrix(c(0, 0, 1, (1-rho)/rho), 2, 2, byrow=TRUE),
var.equal=FALSE, alpha=0.05)var.equal=TRUE) and in this situation the common variance is assumed to be equal to par1.2.FALSE.SS which is a list with eight components:par1.1 <- 0
par1.2 <- 1
par2.1 <- 2
par2.2 <- 1
rho <- 0.3
width <- 0.5
SS(par1.1, par1.2, par2.1, par2.2, rho, width, var.equal=TRUE)Run the code above in your browser using DataLab