Section 9.1 (p.524)

 

1.        

           

 

3.         Q = (10-6)2/6 + (10-12)2/12 + (4-6)2/6 = 3.667. If the principle is correct, Q has a c2 distribution with 2 df. Based on the c2 table, we know the genetic principle should be rejected at a0 = 0.2 but not at a0 = 0.1.

 

4.         (a)

           

 

            (b) When H0 is true (p = p0), we have  and . Therefore,  has a standard normal distribution as n ® ¥ (central limit theorem). This implies Q = Z2 converges to the c2 distribution with 1 df.

 

7.         Let X be the heights, we have

Pr(X < 66) = Pr(Z < -2) = 0.0227   Þ   500 (0.0227) = 11.35

Pr(66 £ X < 67.5) = Pr(-2 £ Z < -0.5) = 0.2858   Þ   500(0.2858) = 142.9

Pr(67.5 £ X < 68.5) = Pr(-0.5 £ Z < 0.5) = 0.3830   Þ   500(0.3830) = 191.5

Pr(68.5 £ X < 70) = Pr(0.5 £ Z < 2) = 0.2858   Þ   500(0.2858) = 142.9

Pr(X ³ 70) = Pr(Z ³ 2) = 0.0.0227   Þ   500 (0.0227) = 11.35

Here we Q = (18-11.35)2/11.35 + (177-142.9)2/142.9 + (198-191.5)2/191.5 + (102-142.9)2/142.9 + (5-11.35)2/11.35 = 27.5. When the null hypotheses (these 500 men form a random sample from all the men who reside in the city) is correct, Q has a c2 distribution with 4 df. Therefore we reject the null hypothesis at a0 = 0.005.