222 LINEAR REGRESSION AND CORRELATION Ch. 7 



was reduced, on the average, to only a one-half-pound advantage at 

 28 weeks of age. Thus /? was 1/2. It might be that the size of /? 

 could be increased by superior breeding and handling. 



If two samples — from two methods of breeding and raising turkeys, 

 for example — have resulted in the computation of bi, b 2 , r lf and r 2 , 

 the testing of the two hypotheses i/ (/8i = /? 2 ) an d H {pi = po) can 

 be carried out as follows: 



For H Q (ft =#>): 



(a) Pool the 2(F,- - f ,-) 2 and the 2(F,- - Yj) 2 from the two sam- 

 ples, pool the degrees of freedom, and calculate 



2(F, - Y { ) 2 + 2(F y - Yj) 2 



pooled 8 y . x = x ■ 



n\ + n 2 — 4 . 



1 1 



(b) Compute x j — + 



2(x, 2 ) S(x/) 



(c) Multiply the standard deviation from a by the result obtained 

 in b. This is the estimated standard deviation of (6i — b 2 ), which 

 will be called s^-hf 



(d) Compute t — (&i — b 2 )/s bl _ bi , assign it n x + n 2 — 4 degrees of 

 freedom, and interpret as before with respect to the acceptance or the 

 rejection of # o (0i = fo) • 



For i/ (pi = P2): 



(a) Transform the r x and the r 2 to Z\ and 2 2 , respectively, in the 

 manner described earlier in this chapter. 



1 

 (6) Compute <r Zl - Zi = x f h 



n\ — 3 n 2 — 3 



(c) Calculate y = | Z\ — z 2 \/<r Zl -z 2 and consider this ratio as a 

 normally distributed quantity in deciding whether or not it is so large 

 that the hypothesis #o(pi = P2) should be rejected. 



If it seems appropriate after a hypothesis has been rejected, con- 

 fidence intervals can be determined for the difference /?i — (3>, but 

 not for pi — p2- 



It is useful at times to have a convenient tabular procedure for 

 computing b and r when the data are sufficiently numerous to justify 

 the use of frequency distribution tables. Such data rarely would 



