530 Chapter 12:Nonparametric Hypothesis Tests
From this it follows that choosing thenrankings of the first sample is probabilis-
tically equivalent to randomly choosingnof the (possible rank) values 1, 2,...,
n+m. Using this, it can be shown thatThas a mean and variance given by
EH 0 [T]=
n(n+m+1)
2
VarH 0 (T)=
nm(n+m+1)
12
In addition, it can be shown that when bothnandmare of moderate size (both
being greater than 7 should suffice)Thas, underH 0 , approximately a normal
distribution. Hence, whenH 0 is true
T−
n(n+m+1)
√^2
nm(n+m+1)
12
∼ ̇N(0, 1) (12.4.6)
If we letddenote the absolute value of the difference between the observed
value ofTand its mean value given above, then based on Equation 12.4.6 the
approximatep-value is
p-value=PH 0 {|T−EH 0 [T]|>d}
≈P
{
|Z|>d/
√
nm(n+m+1)
12
}
whereZ∼N(0, 1)
= 2 P
{
Z>d/
√
nm(n+m+1)
12
}
EXAMPLE 12.4e In Example 12.4a,n=5,m=6, and the test statistic’s value is 21. Since
n(n+m+1)
2
= 30
nm(n+m+1)
12
= 30
we have thatd=9 and so
p-value≈ 2 P
{
Z>
9
√
30
}
= 2 P{Z>1.643108}