> #Problem 4.8: data with ties
> #Using Charlie Geyer's Computer Code
> exdata <- read.table("http://www.rohan.sdsu.edu/~babailey/stat672/4.8.data", header=T)
> attach(exdata)
> exdata
    x   y
1 2.1 1.9
2 1.9 2.6
3 2.6 3.7
4 3.3  NA
> 
> #Wilcoxon Rank Sum Test
> mu <- 0 # hypothesized value of median difference
> x <- x[!is.na(x)]
> y <- y[!is.na(y)]
> print(nx <- length(x))
[1] 4
> print(ny <- length(y))
[1] 3
> y <- y - mu
> data <- c(x, y)
> names(data) <- c(rep("x", nx), rep("y", ny))
> data <- sort(data)
> r <- rank(data)
> rbind(data, r)
       x   y   x   x   y   x   y
data 1.9 1.9 2.1 2.6 2.6 3.3 3.7
r    1.5 1.5 3.0 4.5 4.5 6.0 7.0
> print(w <- sum(r[names(data) == "y"]))
[1] 13
> print(u <- w - ny * (ny + 1) / 2)
[1] 7
> pwilcox(u, nx, ny)
[1] 0.6857143
> 
> #Hodges-Lehmann estimator: median of the pairwise difference 
> x <- x[!is.na(x)]
> y <- y[!is.na(y)]
> print(diffs <- sort(as.vector(outer(y, x, "-"))))
 [1] -1.4 -0.7 -0.7 -0.2  0.0  0.0  0.4  0.5  0.7  1.1  1.6  1.8
> median(diffs)
[1] 0.2
> 
> #Associated CI: counts in k from each end of the list of 
> #sorted pairwise differences
> conf.level <- 0.95
> x <- x[!is.na(x)]
> y <- y[!is.na(y)]
> print(nx <- length(x))
[1] 4
> print(ny <- length(y))
[1] 3
> print(diffs <- sort(as.vector(outer(y, x, "-"))))
 [1] -1.4 -0.7 -0.7 -0.2  0.0  0.0  0.4  0.5  0.7  1.1  1.6  1.8
> print(m <- length(diffs))
[1] 12
> alpha <- 1 - conf.level
> k <- qwilcox(alpha / 2, nx, ny)
> if (k == 0) k <- k + 1
> print(k)
[1] 1
> cat("achieved confidence level:",
+     1 - 2 * pwilcox(k - 1, nx, ny), "\n")
achieved confidence level: 0.9428571 
> c(diffs[k], diffs[m + 1 - k])
[1] -1.4  1.8
> 
> #Or use the wilcox.exact test
> library("exactRankTests")
  Package 'exactRankTests' is no longer under development.
  Please consider using package 'coin' instead.
> wilcox.exact(y, x, alternative="less")

	Exact Wilcoxon rank sum test

data:  y and x 
W = 7, p-value = 0.6857
alternative hypothesis: true mu is less than 0 

> wilcox.exact(y, x, conf.int=T)

	Exact Wilcoxon rank sum test

data:  y and x 
W = 7, p-value = 0.8
alternative hypothesis: true mu is not equal to 0 
95 percent confidence interval:
 -1.4  1.8 
sample estimates:
difference in location 
                   0.2