134 SAMPLING FROM BINOMIAL POPULATIONS Ch. 5 



a 50:50 sex ratio purely as a consequence of the chance element in 

 all sampling. On the other hand, if the sampling x 2 na d been 15, 

 say, we see from Figure 5.31 that sampling variation almost never 

 produces such a large value of x 2 - We would then conclude that the 

 sex ratio was not 50:50. It is clear that such conclusions as these 

 are valid only if a representative sample has been drawn. If just 

 a few guinea pigs in one particular laboratory have been the basis 

 for the sample, the conclusions drawn would not apply, without more 

 sampling evidence, to guinea pigs in general. 



As a matter of fact, Figure 5.31 can be used as above for samples 

 of various sizes just as long as there are only two classes of attributes 

 involved. Under these circumstances, x 2 is said to have one degree 

 of freedom. In this connection it should be noted that when the 

 expected number has been calculated for one of the two classes, the 

 other number follows automatically so as to keep the sum of the 

 expected numbers equal to the sum of the observed numbers. Like- 

 wise, if there are 3 too many males compared to expectation there 

 must be 3 too few females ; that is, there really is but one basic dif- 

 ference between the observed numbers and the expected numbers. 

 Basically, that is the reason there is only one degree of freedom for 

 the x 2 . 



Table V makes it possible to determine more easily and accurately 

 the probability that a sample x 2 will exceed the observed value when 

 the H is correct. Actually this mathematical distribution is not 

 exactly right for the x 2 as defined in this chapter, but the loss of 

 accuracy is negligible for most sample sizes which would cause peo- 

 ple to have faith in the conclusions drawn therefrom. 



Problem 5.31. It was stated in Chapter 3 that the A-B blood groups were 

 considered to be inherited in a simple Mendelian manner so that AO X AO 

 should produce offspring three-fourths of whom test to be type A and one- 

 fourth are type 0. Suppose that among a random sample of 400 children from 

 such parents, 312 are A and 88 are O ; that is, 78 per cent are A and 22 per cent 

 are O. Does this sampling evidence justify rejection of the hypothesis that 

 75 per cent should be A, or, in more symbolic terminology, should the hy- 

 pothesis H (p = 3/4 for A) be rejected? 



The expected numbers are 300 A and 100 O; therefore, x 2 = 

 U2) 2 /300 + (12)7100 = 1.92, with 1 D/F. By Table V it is found 

 by interpolation that P = .17. By any usual standards H (p = 3/4 

 for A) is accepted, especially if the hypothesis seems to be well 

 founded theoretically. In some circumstances we would wish to 



