600 Nonparametric and Robust StatisticsThe variance is displayed below (10.4.10) and a derivation of a more general case is
given in Section 10.5. It also can be shown thatWis asymptotically normal. We
summarize these items in the theorem below.Theorem 10.4.1.SupposeX 1 ,X 2 ,...,Xn 1 is a random sample from a distribution
with a continuous cdfF(x)andY 1 ,Y 2 ,...,Yn 2 is a random sample from a distribu-
tion with a continuous cdfG(x).SupposeH 0 : F(x)=G(x), for allx.IfH 0 is
true, then
Wis distribution free with a symmetric pmf (10.4.8)EH 0 [W]=n 2 (n+1)
2(10.4.9)VarH 0 (W)=n 1 n 2 (n+1)
12(10.4.10)
Wq−n 2 (n+1)/ 2
VarH 0 (W)has an asymptoticallyN(0,1)distribution. (10.4.11)The only item of the theorem not discussed above is the symmetry of the null
distribution, which we show later. First, consider this example:
Example 10.4.1(Water Wheel Data Set). In an experiment discussed in Abebe
et al. (2001), mice were placed in a wheel that is partially submerged in water. If
they keep the wheel moving, they avoid the water. The response is the number of
wheel revolutions per minute. Group 1 is a placebo group, while Group 2 consists
of mice that are under the influence of a drug. The data are
Group 1X 2.3 0.3 5.2 3.1 1.1 0.9 2.0 0.7 1.4 0.3
Group 2Y 0.8 2.8 4.0 2.4 1.2 0.0 6.2 1.5 28.8 0.7The data are in the filewaterwheel.rda. Comparison boxplots of the data (asked
for in Exercise 10.4.9) show that the two data sets are similar except for the large
outlier in the treatment group. A two-sided hypothesis seems appropriate in this
case. Notice that a few of the data points in the data set have the same value
(are tied). This happens in real data sets. We follow the usual practice and use
the average of the ranks involved to break ties. For example, the observations
x 2 =x 10 =0.3 are tied and the ranks involved for the combined data are 2 and- Hence we use 2.5 for the ranks of each of these observations. Continuing in
this way, the Wilcoxon test statistic isw=
∑ 10
j=1R(yj) = 116.50. The null mean
and variance ofW are 105 and 175, respectively. The asymptotic test statistic
isz= (116. 5 −105)/√
175 = 0.869 withp-value2*(1-pnorm(0.869))=0.3848.
HenceH 0 would not be rejected. The test confirms the comparison boxplots of the
data. Thet-test based on the difference in means is discussed in Exercise 10.4.9. In
Example 10.4.2, we discuss the R computation.
We next want to derive some properties of the test statistic and then use these
properties to discuss point estimation and confidence intervals for Δ. As in the
last section, another way of writingW proves helpful in these regards. Without
loss of generality, assume that theYjs are in order. Recall that the distributions