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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!!

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What's "impossible"? If you mean you're trying to calculate a numerical value for the likelihood & then take its logarithm, then the numbers involved will be stupidly big. Work out an expression for the log-likelihood first - using $\log(x^y)=y\log(x)$ &c. - & then put the numbers in.

The degrees of freedom are 1 because under the alternative you have one free parameter ($p$), & under the null no free parameters - $p$ is constrained to be $\frac{1}{2}$.

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  • $\begingroup$ yeah I've been stupid thank you! i calculated the value and i got 8.3, hence I should reject the null hypothesis... it is strange as the data were pretty close isn't it? also my concern was about [0,1] not being a vector space! $\endgroup$ – gianluca Dec 11 '12 at 13:13
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    $\begingroup$ Well, in a large sample you can detect very small differences from the null hypothesis. That doesn't mean they're necessarily of any practical significance; or even that they're real, given that there will also be unknown systematic errors. $\endgroup$ – Scortchi Dec 11 '12 at 13:22
  • $\begingroup$ ah ok, i understand now! for the second part, I Was thinking of considering p_1 - p_2 as the parameter and to test the null of it being zero against the alternative of it being free, do you think this is the right track? $\endgroup$ – gianluca Dec 11 '12 at 13:36

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