AGE OF COW AND BUTTER-FAT PERCENTAGE 89 



The last constant needed is summation (XY). This is deter- 

 mined by multiplying the summation of each row, S(X) for each 

 row, by the value of Y, thus: 



(918 X 1) + (11063 X 2) + (6374 X 3) + (5314 X 4) .... (5 X 31) + (4 X 33) = 



418940 



From these data it is now possible to obtain the correlation of 

 age with milk yield. This coefficient is obtained as follows. Each 

 of the constants are divided through by the total number of indi- 

 viduals in the table, 10644. 



61258 

 10644 



= 5.755167 



S(X) 

 n 



406680 

 10644 " 



= 38.207441 



S(X 2 ) 

 n 



65190 

 10644 



= 6.124577 



_S(Y) 

 n 



607450 

 10644 " 



= 57.069710 



8(Y*) 



n 



418940 



= 39.359263 



S(XY) 



10644 



To transfer our data to the true mean of the table, the following 

 calculations are made. 



38.207441 - (5.755167) 2 = 5.085494 = a~* 

 57.069710 - (6.124577) 2 = 19.559267 = <r y 2 

 39.359263 - (5.755167 X 6.124577) = 4.111300 



In symbols this relation is 



S(X>) /S(X) 



n \ n 



S(Y*) : . (S(Y)\* _ 

 n \ n J 



2 



= ax 2 



ay' 



S(XY) S(X) S(Y) fQ /™ I+1 # 



X = value ot S(XYJ around the true mean. 



n n n 



The value of c x is equal to the square root of cr' 2 x or yjb. 085494 = 2.255104 



