Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

10.3. Signed-Rank Wilcoxon 587

other hand, ifH 1 is true, then we expect more than half of theXis to be positive
and further, the positive observations are more likely to receive the higher ranks.
Thus an appropriate decision rule is

RejectH 0 in favor ofH 1 ifT≥c, (10.3.3)

wherecis determined by the levelαof the test.

Givenα, we need the null distribution ofTto determine the critical pointc.The
set of integers{−n(n+1)/ 2 ,−[n(n+1)/2] + 2,...,n(n+1)/ 2 }form the support
ofT. Also, from Section 10.2, we know that the signs are iid with support{− 1 , 1 }
and pmf

p(−1) =p(1) =

1

2

. (10.3.4)

A key result is the following lemma:

Lemma 10.3.1.UnderH 0 and symmetry about 0 for the pdf, the random variables
|X 1 |,...,|Xn|are independent of the random variables sgn(X 1 ),...,sgn(Xn).

Proof: BecauseX 1 ,...,Xnis a random sample from the cdfF(x), it suffices to
show thatP[|Xi|≤x,sgn(Xi)=1]=P[|Xi|≤x]P[sgn(Xi)=1]. DuetoH 0 and
the symmetry off(x), this follows from the following string of equalities

P[|Xi|≤x,sgn(Xi)=1] = P[0<Xi≤x]=F(x)−

1

2

=[2F(x)−1]

1

2

=P[|Xi|≤x]P[sgn(Xi)=1].

Based on this lemma, the ranks of the|Xi|s are independent of the signs of
theXis. Note that the ranks are a permutation of the integers 1, 2 ,...,n.By
the lemma this independence is true for any permutation. In particular, suppose
we use the permutation that orders the absolute values. For example, suppose the
observations are− 6. 1 , 4. 3 , 7. 2 , 8. 0 ,− 2 .1. Then the permutation 5, 2 , 1 , 3 ,4orders
the absolute values; that is, the fifth observation is the smallest in absolute value,
the second observation is the next smallest, etc. This permutation is called the
anti-ranks, which we denote generally by byi 1 ,i 2 ,...,in. Using the anti-ranks,
we can writeTas

T=

∑n

j=1

jsgn(Xij), (10.3.5)

where, by Lemma 10.3.1, sgn(Xij) are iid with support{− 1 , 1 }and pmf (10.3.4).