Question

8.A survey of 800 families with 4 children each revealed the following distribution:No. of boys:01234No. of girls:43210No. of families:3217829023664Is this result consistent with the hypothesis that the male & female births are equallyprobable?

Answer 1

$\begin{matrix} No.of\ boys & 0 & 1 & 2 & 3 & 4 \\ No.of\ girls & 4 & 3 & 2 & 1 & 0 \\ No.of\ families & 32 & 178 & 290 & 236 & 64 \end{matrix}$

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

$P(3\ boys\ \& \ 1\ girl)=\dbinom{4}{3}(\dfrac{1}{2})^3(\dfrac{1}{2})=\dfrac{1}{4}$

$P(2\ boys\ \& \ 2\ girl)=\dbinom{4}{2}(\dfrac{1}{2})^2(\dfrac{1}{2})^2=\dfrac{3}{8}$

$P(1\ boy\ \& \ 3\ girls)=\dbinom{4}{1}(\dfrac{1}{2})(\dfrac{1}{2})^3=\dfrac{1}{4}$

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

For 800 families

$No.of\ families(all\ boys)=\dfrac{1}{16}\times800=50$

$No.of\ families(3\ boys\ \& \ 1\ girl)=\dfrac{1}{4}\times800=200$

$No.of\ families(2\ boys\ \& \ 2\ girl)=\dfrac{3}{8}\times800=300$

$No.of\ families(1\ boy\ \& \ 3\ girls)=\dfrac{1}{4}\times800=200$

$No.of\ families(all\ girls)=\dfrac{1}{16}\times800=50$

$H_0:$ 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 $df=5-1=4,$ so then the rejection region for this test is $R=\{\chi^2:\chi^2>9.488\}.$

The Chi-Squared statistic is computed as follows:

$$x^2=\sum\dfrac{(f_o-f_e)^2}{f_e}=$$

$$=\dfrac{(32-50)^2}{50}+\dfrac{(178-200)^2}{200}+\dfrac{(290-300)^2}{300}+$$

$$+\dfrac{(236-200)^2}{200}+\dfrac{(64-50)^2}{50}=\dfrac{589}{30}\approx19.633$$

Since it is observed that $\chi^2=19.6333>9.488=\chi_c^2,$ 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.