K. Pearson 
297 
These are the general equations for fitting the sine curve (xvii) by the method 
of moments to any series of observations. We see how simple is this method as 
compared with that of least squares ; we must first find a root of (xviii) and this 
value of z substituted in (xix) will give us A and B. 
For the special case of the barometer observations I suppose our ordinates 
placed at the middle of the elements so that the range 2^=11. I then find 
the moments by an application of (X) on p. 276, using the values of P and Q 
given for |) = 11 on p. 277. This gives 
if„ = 10-979,4240, = 4-420,0564, M, = 86-783,0235, 
and thence /3 = - -369,353. 
To solve (xviii), the hyperbola 
y = -- -369,3532, 
^ z 
and the transcendental curve y = cot z 
were roughly plotted and observed to intersect about z = 1-2. Using Newton's 
method of approximation, I found with Miss M. A. Lewenz's aid : 
2 = 1-1867, 2= 1-1844, 2 = 1-184,4132, 
which last value is practically exact. This gave 
?i = -2 15,348 = 12° 20' 19", 
.4 = -213,545, 5 = 1-276,288. 
Whence the required curve is 
y = -213,545 sin ( 215,348a') + 1-276,288 cos (-215,348^), 
or y = 1-29403 sin {(12° 20' 19") x + 80° 30' 5"), 
the latter form allowing of easy calculation of the ordinates. 
We have 
X 
Observed y 
Calculated y 
— 5 
•382 
-417 
-4 
•674 
-669 
-3 
-923 
-891 
-2 
1104 
1-071 
- 1 
1-214 
1-201 
0 
1-273 
1-27.6 
+ 1 
1-270 
1-292 
+ 2 
1-215 
1-250 
+ 3 
1-137 
1-148 
+ 4 
•989 
-993 
+ 5 
•819 
-793 
The root mean square error of the ordinates is -0233. Thus the fit is by no 
means so good as that of the parabola of the third order y = cto + chx + a^so^ + cijX^ 
Biometrika i 30 
