Introduction to Probability and Statistics for Engineers and Scientists

530 Chapter 12:Nonparametric Hypothesis Tests

From this it follows that choosing thenrankings of the first sample is probabilis-

tically equivalent to randomly choosingnof the (possible rank) values 1, 2,...,

n+m. Using this, it can be shown thatThas a mean and variance given by

EH 0 [T]=

n(n+m+1)

2

VarH 0 (T)=

nm(n+m+1)

12

In addition, it can be shown that when bothnandmare of moderate size (both

being greater than 7 should suffice)Thas, underH 0 , approximately a normal

distribution. Hence, whenH 0 is true

T−

n(n+m+1)

√^2

nm(n+m+1)

12

∼ ̇N(0, 1) (12.4.6)

If we letddenote the absolute value of the difference between the observed

value ofTand its mean value given above, then based on Equation 12.4.6 the

approximatep-value is

p-value=PH 0 {|T−EH 0 [T]|>d}

≈P

{

|Z|>d/

√

nm(n+m+1)

12

}

whereZ∼N(0, 1)

= 2 P

{

Z>d/

√

nm(n+m+1)

12

}

EXAMPLE 12.4e In Example 12.4a,n=5,m=6, and the test statistic’s value is 21. Since

n(n+m+1)

2

= 30

nm(n+m+1)

12

= 30

we have thatd=9 and so

p-value≈ 2 P

{

Z>

9

√

30

}

= 2 P{Z>1.643108}