ERROR TO CASES OP NORMAL DISTRIBUTION AND CORRELATION. 129 
Let the following table represent the proportions, in the original community, ot 
the individuals specified :— 
Values 0 ^ L. 
Values of M. 
M" to M'". 
Remainder. 
L' to L". 
V 
V" 
Remainder .... 
V' 
Y’" 
and let \fj, ijj', xjj", xjj'" be the errors in V, V', V", V'". Thus Uio = n {Y + ^)} 
Pi =: n [Y + V" xjj xjj"), q 2 = n [Y Y' xp xp'). If the distributions are 
independent, V = (V+V') (V+V''); (since V+V'-fV"+V'"= 1), VV'"=V'y". 
Hence (since xp -{• xp' xp" xp'" = 0) 
ni2 ~ -piY-h — n {xp — {Y Y') (xp + xp") — (V + V") (xp + xp')} 
= n {(Y"'xp + Yxp'") - (Y"xp' + Y'r)}- 
By § 15 (v.) it will be found that the mean square of this discrepancy is 
nYY'" = nY'Y" ; and therefore the “ probable discrepancy ” is Q \/ nYY'" 
= Q \/ nY'Y". By calculating this expression for each number in the table, and 
comparing the actual discrepancies, as n^o — piqz/n, with the values so obtained, we 
have data for deciding as to the validity of the hypothesis of independence. 
The following example of a case in which, on a p>7'iori grounds, we should expect 
to find independence, will serve as an illustration. The table is compiled from a 
list of school-teachers who passed a certain examination. 
List. 
First letter of name. 
Total. 
A-D. 
1 
E-J. 
K-R. 
S-Z. 
Men. 
166 
174 
180 
164 
684 
Women, 1st year .... 
427 
379 
411 
366 
1583 
„ 2nd „ . 
549 
493 
577 
492 
2111 
Total .... 
1 
1142 
i 
1046 
1168 
1022 
4378 
By multiplying each total of a row by each total of a column, and dividing each 
product by n = 4378, we get the “ calculated ” table 
VOL. cxcii.—A. 
s 
