178 SAMPLING NORMAL POPULATIONS Ch. 6 



It turns out from mathematical analysis that the sampling ratios 

 k = (di — M)/ S di follow the same sampling distribution as the t previ- 

 ously discussed, if n = the true average df, hence Table IV can be 

 used here provided we employ 2 (ft — 1) degrees of freedom for t in- 

 stead of the (n — 1) used before. 



The way is now open to solve the type of problem proposed at the 

 beginning of this section. To illustrate, suppose that 20 steers of the 

 same breed, weight, and previous history are divided into two equal 

 lots by some impartial means such as drawing numbers from a hat. 

 Thereafter, one group is fed a ration 50 per cent of which is peanut 

 meal and 50 per cent is a standard ration. The other group of steers 

 is fed only 20 per cent peanut meal, the remainder being the same 

 standard ration. After an adequate period of time, the average daily 

 gains of the steers were obtained as follows, with A standing for the 

 group of steers whose diet contained 50 per cent of peanut meal: 



Group 



V - f. ft 15.39 17.37 



2(zi 2 ) = 0.0531, S(x 2 2 ) = 0.0608. 



/■"? 



</ 



/ 0.0531 + 0.0608 

 Sd \ 10(10 - 1) 



= V0.001266 



= 0.036, approximately. 



t = (0.20 - m)/0.036 = 5.56 if /x = 0. 

 t has 18 degrees of freedom. 



We learn from Table IV that less than one-half of one per cent 

 of the sample £'s with 18 degrees of freedom are numerically as large 

 as 5.56; therefore, the hypothesis that ju. = is rejected and the two 

 samples are regarded as having been drawn from different normal 



