x<-c(10,80,209,273,279,324,391,415,566,85,852,881,895,954,
1101,1133,1337,1393,1408,1444,1513,1585,1669,1823,1941)
dx<-c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,0,0,0,0,1,1,0)
y<-c(21,38,39,51,77,185,240,289,524,610,612,677,798,881,899,
946,1010,1074,1147,1154,1199,1269,1329,1484,1493,1559,1602,1684,1900,1952)
dy<-c(1,1,1,1,1,1,2,2,1,1,1,1,1,2,1,1,1,1,1,1,0,0,1,1,0,0,1,0,0,0)
nx<-length(x)
ny<-length(y)
xc<-1:nx
yc<-1:ny
wx<-rep(1,nx)
wy<-rep(1,ny)
maxit<-10
mean=c(0.5,0.5)
p<-2
H1<-matrix(NA,nrow=nx,ncol=ny)
H2<-matrix(NA,nrow=nx,ncol=ny)
for (i in 1:nx) {
for (j in 1:ny) {
H1[i,j]<-(x[i]>y[j])
H2[i,j]<-(x[i]>1060) } }
H=matrix(c(H1,H2),nrow=nx,ncol=p*ny)
# Ho1: X is stochastically equal to Y
# Ho2: mean of X equals mean of Y
el2.cen.EMm(x, dx, y, dy, p, H, xc=1:length(x), yc=1:length(y),
mean, maxit=10)
# Result: Pval is 0.6310234, so we cannot with 95 percent confidence reject the two
# simultaneous hypotheses Ho1 and Ho2Run the code above in your browser using DataLab