Brown Mood Median Test

Brown-Mood Median Test

对于两独立样本尺度中的位置参数（中位数）检验问题：

\(H_0: med_x = med_y\)   \(H_1=med_x > med_y\)

在\(H_0\)假设下，两组数据具有相同的中位数

对于X,Y两组数据组成的两个数组，进行中位数比较。

核心是比较\(med_xy\) 和\(med_x\) \(med-y\)的关系。

>\(M_{xy}\)	A	B	t
<\(M_{xy}\)	C	D	(m+n)-(A+B)
\(\sum\)	m	n	m=n=A+B+C+D

when m,n,t are fixed.

则A的密度函数如下：

\(P(A=k)=\frac{\binom{m}{k}\binom{n}{n-k}}{\binom{m+n}{t}}\), k\(\leqslant\) min{m,t}

以下是Brown-Mood R代码

#Brown-Mood median test

BM.test <- function(x,y,alt)

{

  xy <- c(x,y)

  md.xy <-median(xy)

  t <- sum(xy > md.xy)

  lx <- length(x)

  ly <- length(y)

  lxy <- lx + ly

  A <- sum(x > md.xy)

  if(alt == "greater")

    {w <- 1 - phyper(A,lx,ly,t)}

  else if (alt == "less")

    {w <- phyper(A,lx,ly,t)}

  conting.table = matrix(c(A,lx-A,lx,t-A,ly-(t-A),ly,t,lxy-t,lxy),3,3)

  col.name <- c("X","Y","X+Y")

  row.name <- c(">MXY","<MXY","TOTAL")

  dimnames(conting.table)<-list(row.name,col.name)

  list(contingencu.table=conting.table,p.value = w)

}

\[phyper(q,m,n,k) = \frac{\binom{m}{k}\binom{n}{n-k}}{\binom{m+n}{t}}, k\leqslant min(m,t)
\]

大样本计算时，趋于正态分布:

binom 分布的 $$E[X] = μ = t\cdot\frac{m}{m+n}$$

\[D[x]
\]

example:

x <- c(698,688,675,656,655,648,640,639,620)

y <- c(780,754,740,712,693,680,621)

BM.test(x,y,"less")

$contingencu.table

      X Y X+Y

>MXY  2 6   8

<MXY  7 1   8

TOTAL 9 7  16