I am trying to solve this problem: Assume that each child is male with probability $p$ independently of all other children.
We observed 19711 male births out of a total of 38562 births in American families with two children each. Use the likelihood ratio test statistic to test the hypothesis $H_0: p=1/2$ against a suitable alternative, which you should specify.
We also found 17703 males out of 35042 similar births in Finland. Use the generalised likelihood ratio test to test the hypothesis that $p$ has the same value in each country versus a suitable alternative.
For the first part we have $X_1,..X_{38562}$ I.I.D. r.v. with distribution $\mathrm{Ber}(p)$.
The alternative hypothesis should be $H_1 : p \in [0,1]$.
Now as we know that the sample mean is the MLE for Bernoulli distributions, we have likelihood ratio, with $n= 38562$: $$ \displaystyle \lambda(X)= \frac{(1/2)^n}{\overline x^{\sum x_i} \cdot (1-\overline x)^{(n-\sum x_i)} } $$ So the likelihood ratio statistic is: $$ \Lambda(X) = -2 \log(\lambda(X)) $$ Hence our test is that we reject $H_0$ iff $\Lambda(x) \geq k$ for a constant $k$ such that $$ P(\chi^2_1 \geq k) = \alpha. $$ Where $\alpha$ is the required size. We can do that as $n$ is large and we know that the likelihood ratio statistic converges in distribution to the chi squared with 1 degree of freedom
(QUESTION: the degrees of freedom should be $\mathrm{Dim}([0,1])- \mathrm{Dim}({1/2})= \mathrm{Dim}([0,1])$ why do we assign dimension 1 to this? I think we do, because otherwise 0 degrees of freedom wouldn't make much sense...)
I thought a sensible choice of $\alpha$ would be 0.05 so I worked out $k= 3.841$.
How do I proceed now? Putting the values in the likelihood ratio statistic leads to impossible calculations...
Any advice on this last bit or on the following part would be really appreciated!!
