Introduction to Probability and Statistics for Engineers and Scientists

12.3The Signed Rank Test 523

we see that

Pk(i)=^12 Pk− 1 (i−k)+^12 Pk− 1 (i) (12.3.6)

Starting with Equation 12.3.5, the recursion given by Equation 12.3.6 can be successfully
employed to computeP 2 (·), thenP 3 (·), and so on, stopping when the desired valuePn(t∗)
has been obtained.

EXAMPLE 12.3b For the data of Example 12.3a,

t∗=min

(

3,

4 · 5

2

− 3

)

= 3

Hence thep-value is 2P 4 (3), which is computed as follows:

P 2 (0)=^12 [P 1 (−2)+P 1 (0)]=^14

P 2 (1)=^12 [P 1 (−1)+P 1 (1)]=^12

P 2 (2)=^12 [P 1 (0)+P 1 (2)]=^34

P 2 (3)=^12 [P 1 (1)+P 1 (3)]= 1

P 3 (0)=^12 [P 2 (−3)+P 2 (0)]=^18 sinceP 2 (−3)= 0

P 3 (1)=^12 [P 2 (−2)+P 2 (1)]=^14

P 3 (2)=^12 [P 2 (−1)+P 2 (2)]=^38

P 3 (3)=^12 [P 2 (0)+P 2 (3)]=^58

P 4 (0)=^12 [P 3 (−4)+P 3 (0)]= 161

P 4 (1)=^12 [P 3 (−3)+P 3 (1)]=^18

P 4 (2)=^12 [P 3 (−2)+P 3 (2)]= 163

P 4 (3)=^12 [P 3 (−1)+P 3 (3)]= 165 ■

Program 12.3 will use the recursion in Equations 12.3.5 and 12.3.6 to compute the
p-value of the signed rank test data. The input needed is the sample sizenand the value
of test statisticT.

EXAMPLE 12.3c Suppose we are interested in determining whether a certain population
has an underlying probability distribution that is symmetric about 0. If a sample of size 20
from this population results in a signed rank test statistic of value 142, what conclusion
can we draw at the 10 percent level of significance?