= Solution
Treat the four fixed row totals as four independent samples from <binomial distributions>. Under independence, each has the same unknown probability $p$ of a third-child boy. The pooled <maximum-likelihood estimator> is $\widehat p=100/200=1/2$, so each expected cell count is $25$. The <Pearson chi-squared test of homogeneity> statistic is
$$
X^2=\sum_{i=1}^4\left[\frac{(B_i-25)^2}{25}+\frac{(G_i-25)^2}{25}\right]
=\frac{2}{25}(9^2+3^2+0^2+6^2)
=\boxed{10.08}.
$$
There are four independently varying row probabilities under the alternative, and one estimated common probability under the null; the <degrees of freedom> are therefore $4-1=3$. Equivalently this is the $(4-1)(2-1)$ contingency-table count. The fact that the pooled estimate happens to equal one half does not make it a specified null probability and does not change the <degrees of freedom> to four.
Since $10.08>7.8147$, \b[reject independence at the 5% level] using the stated chi-squared calibration. All expected counts are sufficiently large for the usual asymptotic test.
Back to article page
