Page 106 - Becoming Metric Wise

96 Becoming Metric-Wise

2
2
2
ðm 1 nÞðm 1 n 1 1Þ mðm 1 1Þ m 1 mn 1 m 1 nm 1 n 1 n 2 m 2 m
2 5
2 2 2
nðn 1 1Þ
5 mn 1
2
(4.25)
If the ranks of these two groups are mixed, T2 will be smaller than
this maximum. This is the basic idea for considering the statistic U2 cal-
culated as:

nðn 1 1Þ
U2 5 mn 1 2 T2 (4.26)
2
U2 is small when groups differ greatly and large when groups differ
little. A symmetry argument shows that U2 can be large when the ranks
of the elements in the second group are the lowest, but in this case the
roles of the first and the second group are interchanged. The null hypoth-
esis of this test is that both groups do not differ. Consider

mðm 1 1Þ
U1 5 mn 1 2 T1 (4.27)
2
where T1 is the sum of the ranks of the elements in the first group. In
practice one uses U 5 min (U1, U2). However, one does not have to cal-
culate U1 and U2 as it is easy to show that their sum is mn. Indeed:

mðm 1 1Þ nðn 1 1Þ
U1 1 U2 5 2mn 1 1 2 ðT1 1 T2Þ
2 2
(4.28)
mðm 1 1Þ nðn 1 1Þ ðm 1 nÞðm 1 n 1 1Þ
5 2mn 1 1 2 5 mn
2 2 2
Applying this procedure to Table 4.5 with m 5 8 and n 5 10 yields
T1 5 80 and T2 5 91, and thus U1 5 36 and U2 5 44 (and we note that

36 1 44 5 8 10). So we use the value 36.
If m and n are large (say at least 10), then U is approximately normal
with mean mn/2 and a variance of nm(n 1 m 1 1)/12. Standardizing leads
to:

U 2 mn
2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi BNð0; 1Þ (4.29)
q
mnðm 1 n 1 1Þ
12