Introduction to Probability and Statistics for Engineers and Scientists

524 Chapter 12:Nonparametric Hypothesis Tests

SOLUTION Running Program 12.3 yields that

p-value=.177

Thus the hypothesis that the population distribution is symmetric about 0 is accepted at
theα=.10 level of significance. ■

We end this section with a proof of the equality

PH 0 {T≥t}=PH 0

{

T≤

n(n+1)

2

−t

}

To verify the foregoing, note first that 1−Ijwill equal 1 if thejth smallest value of
|Y 1 |,...,|Yn|comes from a data value larger thanm 0 , and it will equal 0 otherwise.
Hence, if we let

T^1 =

∑n

j= 1

j(1−Ij)

thenT^1 will represent the sum of the ranks of the|Yj|that correspond to data values larger
thanm 0. By symmetry,T^1 will have, underH 0 , the same distribution asT. Now

T^1 =

∑n

j= 1

j−

∑n

j= 1

jIj=

n(n+1)

2

−T

and so

P{T≥t}=P{T^1 ≥t} sinceTandT^1 have the same distribution

=P

{

n(n+1)

2

−T≥t

}

=P

{

T≤

n(n+1)

2

−t

}

REMARK ON TIES

Since we have assumed that the population distribution is continuous, there is no possi-
bility of ties — that is, with probability 1, all observations will have different values.
However, since in practice all measurements are quantized, ties are always a distinct
possibility. If ties do occur, then the weights given to the values less thanm 0 should
be the average of the different weights they could have had if the values had differed
slightly. For instance, ifm 0 =0 and the data values are 2, 4, 7,−5,−7, then the ordered
absolute values are 2, 4, 5, 7, 7. Since 7 has rank both 4 and 5, the value of the test statistic