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.
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.
Comments
Leave a comment