QUANTITATIVE INHERITANCE 209 



The standard error of the heritability estimated from a sib correlation is rather 

 laborious to compute. The following method, based on that of Osborne and Pater- 

 son, 971 was devised by Dr. B. Woolf, to whom I am much indebted for explaining it 

 to me and for permission to describe it here. It simplifies the computation considerably 

 and at the same time makes allowance for unequal numbers in the classes. The 

 procedure for computation is set out in table 49 and is illustrated numerically from the 



Table 49 



Procedure for computing the standard error of the heritability estimated from a 

 sib analysis, illustrated numerically from the data of table 48 



II. Sampling variances of the observed mean squares. 



Sires: V A = 2A 2 jm - 1 - 2 x 17.10 2 /70 = 8.35 



Dams: V B = 2B 2 /n - m = 2 x 10.79 2 /118 = 1.97 



Within dams: V c = 2C 2 /N - n = 2 x 2.19 2 /527 = 0.018 



III. Coefficients of components in mean squares, and other quantities required. 



a = 4.16 be = 33.93 



b = 9.75 x = b - a = 5.59 



c = 3.48 y = be - c - x = 24.86 



T = bc(S + D + W) = 33.93 x 5.14 = 174.40 



c 



Half-sib correlation : P = s — = 0.093 



S + D + W 



S + D 



Full-sib correlation: R = -= — = 0.574 



S + D + W 



IV. Sampling variances of the correlation coefficients. 

 Half-sib correlation: 



T 2 a P 2 = (c - cP) 2 V A + (a + xP) 2 V B + (c - a + yP) 2 V c 



9.96 x 8.35 + 21.90 x 1.97 + 2.66 x 0.018 - 126.36 



Full-sib correlation: 



T 2 a R 2 = (c - cR) 2 V A + (* - xR) 2 V B + (x - c + yR) 2 V c 



2.20 x 8.35 + 5.67 x 1.97 + 544.76 x 0.018 = 39.35 



V. Standard errors of correlation coefficients and heritability. 



Half sibs: a P = Vl ^' 36 = 0.064 a h 2 = 4a P = 0.258 



Full sibs: o R = ji; 35 = 0.036 a h 2 = 2a R = 0.072 



