00000000800091ees00堂0++零00000*+格.1.21.41.61.82.01.0region
1.0 1.2 1.4 1.6 1.8 2.0 10000 12000 14000 16000 18000 20000 region salary
秩秩混合样本地区(1/2)混合样本地区(1/2)1113193431239814130978312552152279956126421822612675102582912312749102761212221037413199221319105331368322161810633140491117151082714060222014108371406111211110940159511224101120916079227251139316079126281186416441113291203217498212321204019723
ex.ST=function(O(xx<-read.table("D:llNonparallsalary.txt"):xx[(xx[,2]==1),1]<-xx[(xx[,2]==1),1]-median(outer(xx[(xx[,2]==1),1],xx[(xx[,2]==2),1],"_"))r1<-sum(xx[,2]==1);r2<-sum(xx[,2]==2)xx<-cbind(xx,rv1=rank(xx[,1)n<-nrow(xx);m<-floor(n/2);ind1<-ind2<-rep(0,n)for (i in 1:m)(ind1[(2*i):(2*i+1)]<-c(4*i,4*i+1)ind2[(2*i-1):(2*i)]<-c(4*i-2,4*i-1))ind1[1]<-1;ind1<-ind1[1:n];ind2<-rev(ind2[1:n])ind<-apply(cbind(ind1,ind2),1,min)yy<-cbind(xx[order(xx[, 1]),],ind)Wx<-sum(yy[(yy[,2]==1),4]);Wy<-sum(yy[(yy[,2]==2),4])Wxy<-Wy-r2*(r2+1)/2;Wyx<-Wx-r1*(r1+1)/2Wmin<-min(Wxy,Wyx)plot(xx[,2],xx[,1],pch=16,lwd=1,col=2,xlab="region",ylab="salary")list(yy,cbind(r1,r2,Wx,Wy,Wxy,Wyx),pwilcox(Wmin,r1,r2)))
ex.ST=function(){ xx<-read.table("D:\\Nonpara\\salary.txt"); xx[(xx[,2]==1),1]<-xx[(xx[,2]==1),1]- median(outer(xx[(xx[,2]==1),1],xx[(xx[,2]==2),1],"-")) r1<-sum(xx[,2]==1);r2<-sum(xx[,2]==2) xx<-cbind(xx,rv1=rank(xx[,1])) n<-nrow(xx);m<-floor(n/2);ind1<-ind2<-rep(0,n) for (i in 1:m){ind1[(2*i):(2*i+1)]<-c(4*i,4*i+1) ind2[(2*i-1):(2*i)]<-c(4*i-2,4*i-1)} ind1[1]<-1;ind1<-ind1[1:n];ind2<-rev(ind2[1:n]) ind<-apply(cbind(ind1,ind2),1,min) yy<-cbind(xx[order(xx[,1]),],ind) Wx<-sum(yy[(yy[,2]==1),4]);Wy<-sum(yy[(yy[,2]==2),4]) Wxy<-Wy-r2*(r2+1)/2;Wyx<-Wx-r1*(r1+1)/2 Wmin<-min(Wxy,Wyx) plot(xx[,2],xx[,1],pch=16,lwd=1,col=2,xlab="region",ylab="salary") list(yy,cbind(r1,r2,Wx,Wy,Wxy,Wyx),pwilcox(Wmin,r1,r2))}
假设检验：Ho: 1 = 02; H1 : 01 > 02r1r2WxWyWxyWyx177515180228300pwilcox(min(Wxy,Wyx),r1,r2)=0.02428558Pvalue为0.02428558对水平大于0.025拒绝零假设
假设检验： r1 r2 Wx Wy Wxy Wyx 17 15 228 300 180 75 pwilcox(min(Wxy,Wyx),r1,r2)=0.02428558 Pvalue 为0.02428558 对水平大于0.025， 拒绝零假设
5.2两样本尺度参数的Mood检验假定有两个独立样本X1,...,Xm ~ F二01yYi,..., Yn ~ FF()为连续分布函数02而且实际中，如01 = 02F(0) = 果两样本中位数不等，可估计中位数的差并平移使它们相等．检验:Ho : 1 = 02 对H1 : 01 >02
5.2 两样本尺度参数的Mood检验 假定有两个独立样本 . 为连续分布函数 . 而且 实际中, 如 果两样本中位数不等, 可估计中位数的差, 并平移使它们相等. 检验：