Introduction to Probability and Statistics for Engineers and Scientists

12.3The Signed Rank Test 521

Var(T)=Var





∑n

j= 1

jIj





=

∑n

j= 1

j^2 Var(Ij)

=

∑n

j= 1

j^2

4

=

n(n+1)(2n+1)

24

(12.3.2)

where the fact that the variance of the Bernoulli random variableIjis^12 (1−^12 ) =^14
is used.
It can be shown that for moderately large values ofn(n>25 is often quoted as being
sufficient)Twill, whenH 0 is true, have approximately a normal distribution with mean
and variance as given by Equations 12.3.1 and 12.3.2. Although this approximation can be
used to derive an approximate levelαtest ofH 0 (which has been the usual approach until
the recent advent of fast and cheap computational power), we shall not pursue this approach
but rather will determine thep-value for given test data by an explicit computation of the
relevant probabilities. This is accomplished as follows.
Suppose we desire a significance levelαtest ofH 0. Since the alternative hypothesis is
that the median is not equal tom 0 , a two-sided test is called for. That is, if the observed
value ofTis equal tot, thenH 0 should be rejected if either

PH 0 {T≤t}≤

α

2

or PH 0 {T≥t}≤

α

2

(12.3.3)

Thep-value of the test data whenT=tis given by

p-value=2 min(PH 0 {T≤t},PH 0 {T≥t}) (12.3.4)

That is, ifT =t, the signed rank test calls for rejection of the null hypothesis if the
significance levelαis at least as large as thisp-value. The amount of computation necessary
to compute thep-value can be reduced by utilizing the following equality (whose proof
will be given at the end of the section).

PH 0 {T≥t}=PH 0

{

T≤

n(n+1)

2

−t

}

Using Equation 12.3.4, thep-value is given by

p-value=2 min

(

PH 0 {T≤t},PH 0

{

T≤

n(n+1)

2

−t

})

= 2 PH 0 {T≤t∗}