IIL.—ASSOCIATION. 1 
The advantage of the forms (2) over the form (1) is that they 
give expressions for the second-order frequency in terms of the 
frequencies of the first order and the whole number of observa- 
tions alone ; the form (1) does not. 
Example i.—I1f there are 144 4’s and 384 B’s in 1024 observa- 
tions, how many 4B’s will there be, 4 and B being independent ? 
144 x 384 
ras = OL, 
1024 
There will therefore be 54 AB's. 
Example ii.—1If the A’s are 60 per cent., the B’s 35 per cent., of 
the whole number of observations, what must be the percentage 
of ABs in order that we may conclude that 4 and B are 
independent ? 
60 x 35 
rp 2 3] 
100 
and therefore there must be 21 per cent. (more or less closely, cf. 
§§ 7, 8 below) of 4B’s in the universe to justify the conclusion 
that 4 and B are independent. 
3. It follows from § 1 that if the relation (2) holds for any one 
of the four second-order frequencies, e.g. (4B), similar relations 
must hold for the remaining three. Thus we have directly 
from (1)— 
(4B) _(AB)+(4B) _ (4) 
> B) @B+B ¥ 
giving 
4)(B) 
Prml 
(4B) ="=4 
and so on. This is seen at once to be true on consideration 
of the fourfold table on p. 26. For if (4B) takes the value 
(4)(B)/N, (AB) must take the value (4)(B3)/N to keep the total 
of the row equal to (4), and so on for the other rows and columns, 
The fourfold table in the case of independence must in fact have 
the form— 
Attribute. 
Attribute. Total. 
- (4XBIN ~~ (4)B)N (4) 
a (a)(B)/N (a)(B)/N (a) 
Total {rs a i 
2k