Question

Records taken from the number of male and 

female births in 800 families having four 

children are given below : 

No. of Births 

Frequency 

Male Female 

0 4 32 

1 3 178 

2 2 290 

3 1 236 

4 0 64 

Test whether the data is consistent with the 

hypothesis that the binomial law holds for 

the chance of a male birth is equal to that of 

a female birth at the 5% level of significance.

Answer 1

Number of boys  0  1 2 3 4 

Number of girls 4 3 2 1 0

Number of families 32 178 290 236 64

P(all boys) $= (\dfrac{1}{2})^4 = \dfrac{1}{16}$

P(3 boys and 1 girl) $= ^4C_3\times (\dfrac{1}{2})^3\dfrac{1}{2} = \dfrac{1}{4}$

P(2 boys and 2 girls) $= ^4C_2\times \dfrac{1}{2}^2\times \dfrac{1}{2}^2 = \dfrac{3}{8}$

P(1 boy and 3 girls) $= ^4C_1 \times \dfrac{1}{2} \times \dfrac{1}{2}^3 = \dfrac{1}{4}$

P(all girls)$= \dfrac{1}{2}^4 = \dfrac{1}{16}$

For 800 families,

Number of Families(all boys) $= \dfrac{1}{16} \times 800 = 50$

Number of Families(3 boys and 1 girl) $= \dfrac{1}{4}\times 800 =200$

Number of Families(2 boys and 2 girls) $= \dfrac{3}{8}\times 800 = 300$

Number of Families(1 boy and 3 girls) $= \dfrac{1}{4}\times 800 = 200$

Number of Families(all girls)$= \dfrac{1}{16}\times 800 = 50$

$H_0:$ The male & female births are equally probable.

Based on the information provided, the significance level is $\alpha = 0.05$ the number of degrees of

freedom is $= 5-1 = 4$ so then the rejection region for this test is 

$R=$ { $\chi^2: \chi^2>9.488$ }

The Chi-Squared statistic is computed as follows:

$x^2 = \sum \dfrac{(f_0-f_c)^2}{f_c}$

$= \dfrac{(32-50)^2}{50}+\dfrac{(178-200)^2}{200}+\dfrac{(290-300)^2}{300}+\dfrac{(236-200)^2}{200}+\dfrac{(64-50)^2}{50} = \dfrac{589}{30} = 19.633$

Since it is observed that $\chi^2 = 19.633>9.488 = \chi_c^2$ it is then concluded that the null

hypothesis is rejected.

Therefore, there is enough evidence to claim that the male & female births are not equally probable, at the $\alpha = 0.05$ significance level.