# Likelihood ratio test for sex ratio

I am tryig to solve this problem: Assume that each child is male with probability $p$ indipendently 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 likeligood 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 $Ber(p)$

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 l. 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 degree of freedom shuold be $Dim([0,1])- Dim({1/2})= Dim([0,1])$ why do we assign dimension 1 to this?, i think we do otherwise 0 degrees of freedom whouldn'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!!

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