308 Probable Error of a Correlation Coefficient 
In this case the grouping has had less influence and the largest contributions 
to (in the second, sixth, eighth, and twelfth compartments) are due to 
differences of opposite sign on opposite sides, and may therefore be supposed to be 
entirely due to random sampling. 
My equation then fits the two series of empirical results about as well as could 
be expected. I will now show that it is in accordance with the two theoretical 
cases n "large" and n = 2, for a — —2=^ which approximates sufficiently closely to 
V?l— 1 
1 — 
Pearson and Filon's - — when r = 0 and n is large. Also when n is large /Sj 
vn 
becomes 3 and the distribution is normal. 
And if n = 2, the equation becomes y = yo{l — x^)~^* where 
N 
yo^—Fi ■ 
2 (1 - x'Y^dx 
Jo 
Put X = sin 6. Then dx = cos 6d6, 
N . N 
2/o = f//Jsec^rf^=f/« = 0, 
I.e. there is no frequency except where {1—x'^Y^ is infinite, all the frequency is 
equally divided between x= \ and x = — l which we know to be actually the case. 
n—i 
Consequently I believe that the equation _y = 2/o(l — x-y^ probably represents the 
theoretical distribution of r when samples of n are drawn from a normally distri- 
buted population with no correlation. Even if it does not do so, I am sure that it 
will give a close approximation to it. 
Let us consider Mr Hooker's limit of "50 in the light of this equation. For 
cc -— sill 0 ") 
21 cases the equation becomes ^ _^ cos"^ I proportion of the area lying 
beyond x = ± "50 will be 
J e = siii-i oO 
- cos'8 ddO 
0 
I find this to be '02099, or we may expect to find one case in 50 occurring 
outside the limits + "50 when there is no correlation and the sample numbers 21. 
* If a Pearson curve be fitted to the distribution whose moment coefficients are |U2 = 1=M4 S'Hd 
/i3=0 we have /32=1, ^i — O, hence the curve must be of Type II. and the equation is given by 
y = yo[l-;.,) where a3 = .--^^=l and '" = 2^3^^) or 2/ = y„(l-.r-)-', 
agreeing with the general formula. 
