clxviii Tables for Statisticians and Biometricians [XXXP h , LI a b 



Unless the function cos" 1 p + p Vl p 2 were tabled, it would not be an easy 

 task to find p from this equation. It does not seem worth while tabling it, as the 

 determination of the sign of r in samples of 4 is tedious, when a large number have 

 to be dealt with. 



Table XXXII supplies the constants and ordinates of the distribution of r for 

 n = 3 to 25. 



Tables XXXF XXXP give the frequency constants for n= 25, 50, 100, 200 

 and 400. A Pearson curve to judge by the coincidence of the modal values and 

 by several trials would give the distribution closely, if we knew the /?i and /3 2 

 of the r-distribution. 



These /3's might be found either: 



(i) by using the formulae and determining r, /j, 2 ', Hz , /J>*', and hence /j, z , /"a, /"4 ; 



or, (ii) more roughly by applying an ordinary interpolation formula to the results 

 for 25, 50, 100, 200 and 400 in Tables XXXI d XXXP. 



The changes of argument are so different and so great that no satisfactory 

 formula of direct interpolation has been discovered. A formula for graduation on 

 four points, if x be measured from the foot of the first brdinate, is 



- \ {(2/4 - 2/s) - 3 (2/3 - 2/2) + 2 (2/ 2 - jfc)] x* 



+ i{(2/4-y3)-2(2/ 3 -2/2) + (.Va-2/i)} 3 (xxix), 



where we suppose equal distances between the points. Now our ordinates cor- 

 responding to n are at the abscissae 25, 50, 100, 400. Suppose we plotted to a 



logarithmic scale 



log 25, log 50, log 100, log 400, 



or log 25, log 25+ log 2, log 25 + 2 log 2, log 25 + 4 log 2. 



Then if we took our units of x as log 2, we should have 5 points if we intro- 

 duced a table for 200, the logarithm of which = log 25 + 3 log 2. This table has 

 been computed* since the publication of the original memoir, and admits for most 

 practical purposes of fairly adequate interpolation between n = 25 and n = 400. 



Illustration (vi). Suppose we need the. constants for a sample of 160; we must 

 express this as 



log 160 = log 25 + log 



= log 25 + log 6-4 = log 25 + ( -pV j log 2 



= log 25 + 'gQiQgQQ log 2 = log 25 + 2-678,072 log 2. 



* By Brenda N. Stoessiger, M.Sc. 



