Introduction to Probability and Statistics for Engineers and Scientists

12.4The Two-Sample Problem 531

=2(1−.9498)

=.1004

which can be compared with the exact value, as given in Example 12.4c, of .1225.
In Example 12.4d,n=9,m=13, and so

n(n+m+1)

2

=103.5

nm(n+m+1)

12

=224.25

SinceT=72, we have that

d=| 72 −103.5|=31.5

Thus, the approximatep-value is

p-value≈ 2 P

{

Z>

31.5

√

224.25

}

= 2 P{Z>2.103509}

=2(1−.9823)=.0354

which is quite close to the exactp-value (as given in Example 12.4d) of .0364.
Thus, in the two examples considered, the normal approximation worked quite well in
the second example — where the guideline that both sample sizes should exceed 7 held —
and not so well in the first example — where the guideline did not hold. ■

(b)Simulation If the observed value of the test statistic isT=t, then thep-value is

given by

p-value=2 min

{

PH 0 {T≥t},PH 0 {T≤t}

}

We can approximate this value by continually simulating a random selection ofn

of the values 1, 2,...,n+m— noting on each occasion the sum of thenvalues.

The value ofPH 0 {T≥t}can be approximated by the proportion of time that the

sum obtained is greater than or equal tot, andPH 0 {T≤t}by the proportion of

time that it is less than or equal tot.

A Chapter 12 text disk program approximates thep-value by performing the

preceding simulation. The program will run most efficiently when the sample

of smallest size is designated as the first sample.