OF ATTRIBUTES IN STATISTICS. 



2 \(a0D) + (BCD)f = (aft) + (0V) - (AD) + (BC) + (CD) - (0D) 

 = (aft) + (BC) + (CD) - (AD) 





.'. (ayS) = (A BCD) 



,.'.. if </,//"/// frequencies are equal in the case <>f (},;,<! order ffrou/)n they are equal 

 in the case of fourth order groujm. 



The method of proof is again quite general in its application, so Theorem III. is proved. 



I have thought it again worth while to illustrate the theorem numerically, and have 

 drawn up Table III. for the purpose. A set of arbitrary fourth order frequencies 

 (set (1) ), with contraries equal, was first written down, and from them the given set 



TABLE III. 



of third order frequencies calculated. Now our theorem tells us that the equality 

 amongst the fourth orders depends solely on equality amongst the thirds ; so that 

 we ought to be able to get any number of sets of fourth order frequencies, all 

 possessing equality of contraries, and all consistent with the given set of third order. 

 That this is so will be at once evident on trial. Take 



(ABCD) = 30 

 for instance. Then we have at once 



(ABCS) = (ABC) - (ABCD) = 58 - 30 = 28 

 (AByS) = (ABS) - (ABCS) = 53 - 28 = 25 

 (Ay6) = (AyS) - (AByS) = 71 - 25 = 46 

 (ay8) = (0yS) - (A/8yS) = 7G - 46 = 30 

 giving (ABCD) = (ayS) = 30. 



VOL. CXCIV. --A. 



2 M 



