STAT 203 Solutions HWA09

 

 

16.6: 

 

Goodness of Fit table

 

 

O

E

Yellow

6022

(3/4)(8023)=6017.25

Green

2001

(1/4)(8023)=2005.75

    TOTAL

8023

8023

 

 

ChiSqStatistic = Sum(O-E)^2/E = .0150, df = c-1 = 1

 

.90 < P-Value < .95

 

H0:  Yellow/Green follow a ratio of 3 to 1

HA:  They don’t

Alpha = .01

ChiSqStatistic = .015

.90 < P-Value < .95 (from Table 6)

Conclusion:  The data follows Mendel’s model

 

16.10:

 

H0:  Gender of second child is independent of gender of first child

HA:  Gender of second child is related to gender of first child

Alpha = not stated (let’s assume 5%)

 

Goodness of Fit table

 

 

O

E

O-E

FF

15109

15162

-53

FM

14906

15162

-256

MF

14886

15162

-276

MM

15747

15162

585

    TOTAL

60648

60648

0

 

 

 

ChiSqStatistic = (-53)^2/15162 + (-256)^2/15162 + (-276)^2/15162 + (585)^2/15162 = 486746/15162 = 32.103,  df = c – 1 = 3

 

 P-Value < .001 (From Table 6)

Conclusion:  There is overwhelming evidence that the sex of a second child is not independent of the sex of the first child