1922] DENNY—FRUITS 55 
or lot will have the same value (compare acidity of two grapefruit 
trees, table VI). When two trees or lots of fruit are found to have 
the same value for P.E. sing with respect to one constituent, it must 
not be assumed that they agree also with respect to other constitu- 
ents (compare trees no. 1 and no. 2, table VI, with respect to 
brix and ratio). 
TABLE VII 
VARIABILITY IN COMPOSITION OF INDIVIDUAL LEMONS, EUREKA VARIETY 
EvuREKA STRAIN* SHADE TREE STRAIN* 
Acidi Acidity 
P t-) Refrac. * P t-| Refrac. er: 
Lemon no. Spe “age ‘odex of aay he es 7 ae “ae index of — 
of juice | 10H rind — NaOH 
Leas 0.92 40 | 42.9 | 27.2 Pre 0.96 Al |49;1 424.3 
RE 0.96 AO 1 Ad, 27) 28.5 7 SS 0.94 59 | 45 24.4 
Sie el. 0.04 SOM AS A P2876 8. ia 0.96 501-4558 | 2459 
Ass ipccr ts 2.96 AQ) AS-O (20.7 I vat ois 0.98 36 | 44 26.8 
See: 0.95 AOL ASO 199.0 Lo Ree 0.98 47 | 46.8 | 21.8 
Co ease °.94 Ot 44:6 .4°20.7 7 8.550... - 0.95 62 | 47.0 | 24.0 
(ee ee 0.06 ASA ASO 1008 Pe Pere oas 0.95 56 | 45.8 | 24.5 
ae 0.95 BS AAO F267 eee oes 0.96 54 | 45.6 | 22.9 
2 ahs 0.94 401 49.64 ABP ho gue i 0.98 47 | 45.4 | 22.0 
BO es 0.96 50} 40.5 4.25 2 IO. ea, 0.99 54 | 47-6 | 20.8 
2 oe ee eae 0.95 S01 45.0 | 27.8 tt 0.96 | 48 | 45.7 | 26.6 
eR LEY. .95 a8 | 48.0] 90.61 190.0 3: 0.98 | 39] 45-7 | 26.3 
ee ae 0.96 67 1.45.6{ 06.8 Figs... 0.98 51 | 46.2 | 22.6 
fe ee 0.96 39 143.9 |, 25-0 f 14...455.; 0.98 St loss 20.9 
eS ul 0.97 40145-4 1 20.7 1 25........ 0.07 SO 22.0 
Mean....| 0.95 46 | 44.6 | 28.5 | Mean..... - 0.97 50 | 45-7 | 23-7 
P.E. mean |+0.002/+1.0 |+0.2 |+0.2 | P.E.mean.|+0.003/+1.3 |#0.3 |+0.4 
P.E. sing. .|=0.007|+4.0 |+0.6 |#o 9 | P.E.sing. .|0.010/+*5.0 |+0.9 |*1.3 
* Strains described by SHamEL, Scott, and Pomeroy (8). 
Comparison of standard formula with Peter’s formula for 
calculating probable error of single observation 
‘Two general methods for calculating the value of P.E. sing. 
are as follows: 
STANDARD FORMULA PETER’S FORMULA 
2d? 
n—I 
zd 
P.E. sing. = +0.6745 \ P.E. sing. = 0.845317 s(n —2) 
Thus, to use the standard formula, the sum of the squares of 
the deviations must be found, while with Peter’s formula only the 
