Karl Pearson 
257 
leading, according to our horizontal division, to ?; = "319 and 77' = "337 with 
mean value '328, in excellent agreement vs^ith the two row table value •321. 
Lastly the whole data has been taken as a contingency table : 
Severity of Attack. 
Haemorrhagic 
Confluent 
Abundant 
Sparse 
Very 
Sparse 
Totals 
0—10 
1 
6 
11 
12 
30 
10—25 
5 
37 
114 
165 
136 
457 
25— 
29 
155 
299 
268 
181 
932 
Over 1(5 
11 
35 
48 
33 
28 
155 
Unvaccinated 
4 
61 
41 
7 
2 
115 
Totals 
49 
289 
508 
484 
359 
1689 
Worked out as a 25-fold contingency table by mean square contingency we have 
G, = -335, 
which lies between the two values of found from the three rowed table by the 
new method, and is within the probable error of either. This agreement is very 
satisfactory evidence that it was legitimate to apply the new method to a case of 
this kind, and that in this particular instance the material was closely normal. 
(6) The illustrations given will, I think, show that a wide range of problems 
can be dealt with by the new method, and that results found by it are closely 
comparable with results obtained by other processes. It does not assume linearity 
of regression, and in fact as in the examples on alcoholism it brings out in a very 
effective way the deviations from linearity. The hypothesis used is : that in the 
case of the variable with alternate categories, we can suppose it continuous and 
sufficiently normal in character to have the mean values found from tables of the 
probability integral. If the arrays are not sufficiently homoscedastic for us to 
replace their standard deviations by their mean value, then we must use if we can 
a three and not a two rowed table. It is always desirable to have if possible such 
a table, because it enables us to test the nature of the variability in the arrays. 
Our last illustration, however, shows that very considerable variations in the 
standard deviations of the arrays do not sensibly modify the result obtained by 
using for those standard deviations their mean value. On the whole this new 
method, which replaces the fourfold table method for cases in which the latter 
does not give a unique answer, is, I think, likely to prove of some service. 
