576 Nonparametric and Robust StatisticsY 1 Y 2 Yc 1 Yc 1 + 1 Yn–c 1 Ynn – c 1n – 1c 10S( ).........
.........
Figure 10.2.2:The sketch shows the graph of the decreasing step functionS(θ).
The function drops one unit at each order statisticYi.
Based on this lemma, it is easy to show that the power function of the sign test
is monotone for one-sided tests.
Theorem 10.2.1. Suppose Model (10.2.1) is true. Letγ(θ)be the power function
of the sign test of levelα for the one-sided hypotheses (10.2.2). Thenγ(θ)is a
nondecreasing function ofθ.
Proof: Letcαdenote theb(n, 1 /2) upper critical value as defined after expression
(10.2.8). Without loss of generality, assume thatθ 0 = 0. The power function of the
sign test is
γ(θ)=Pθ[S(0)≥cα], for−∞<θ<∞.
Supposeθ 1 <θ 2 .Then−θ 1 >−θ 2 and hence, sinceS(θ) is nonincreasing,S(−θ 1 )≤
S(−θ 2 ). This and Lemma 10.2.1 yield the desired result; i.e.,
γ(θ 1 )=Pθ 1 [S(0)≥cα]
= P 0 [S(−θ 1 )≥cα]
≤ P 0 [S(−θ 2 )≥cα]
= Pθ 2 [S(0)≥cα]
= γ(θ 2 ).This is a very desirable property for any test. Because the monotonicity of the
power function of the sign test holds for allθ,−∞<θ<∞, we can extend the
simple null hypothesis of (10.2.2) to the composite null hypothesis
H 0 :θ≤θ 0 versusH 1 : θ>θ 0. (10.2.11)