This is a particularly ill-formed question.
If by "alpha" you mean Type I error, you need to go back to Square One and get definitions straight. Type I error is not something inherent in the data, or even in the hypothesis; it's a subjectively and externally applied measure of risk. And without the Type I error, you have no reference point from which to calculate Type II error, the complement of power.
Worse yet, it's not clear--after Adam's question--whether you have JUST TWO OBSERVATIONS (10 and 25), or two distributions with means of 10 and 25, and you're looking for a suitable sample size for a balanced test comparing the means. In the first case, all you can do is a likelihood ratio test that gives an approximate p-value; there's no more information to be had in two observations. In the second case, simulation can give some useful results, but you still need a value for the Type I error to get started.
This is a binomial test. The hypothesis is that if you choose a skull you have a 50% chance it is female. Kind of like a coin flip, except instead of testing heads and tails you're testing male and female.
$X = 110$
$N = 110 + 162 = 272$
$p = .5$
$\alpha = .05$
Formally stated the hypotheses are:
$H_0: p = .5$
$H_1: p \ne .5$
To find the test statistic assuming the above alpha value and given that we have a two tailed test we consult this z table: http://lilt.ilstu.edu/dasacke/eco148/ztable.htm
Because it's a two-tailed test we need to find the z-value of alpha/2 which is 1.96. So that's our test statistic.
Then we set up a binomial test and evaluate it like so (refer to http://www.elderlab.yorku.ca/~aaron/Stats2022/BinomialTest.htm):
$$z = \frac{\frac{X}{N} - p}{\sqrt{\frac{pq}{N}}}=\frac{\frac{110}{272} - .5}{\sqrt{\frac{.5*.5}{272}}} = -\frac{.0956}{.0303} = -3.1551$$
Since the absolute value of the statistic we calculated is greater than the test statistic of 1.96 we reject the null hypothesis that the proportion of female skulls is .5.
As far as likelihood goes, we can find the probability of heads given the number of female heads found in the population of 272 total heads found.
$$L(p | n, y) = {n \choose y}p^y{(1-p)}^{n-y}$$
Where n is the total number of heads found, y is the number of female heads found, p is the anticipated ratio that $y \over n$ may represent.
$$L(.5 | 272, 110) = {272 \choose 110}{.5}^{110}{(1-.5)}^{272-110}= 0.00033$$
I computed this in Wolfram Alpha here:
http://www.wolframalpha.com/input/?i=choose%28272%2C110%29*%28.5%5E110%29%281-.5%29%5E%28272-110%29
So both results agree! We can conclude it is very unlikely the ratio of male and female skulls is in fact 1:1.
Best Answer
Now that any homework deadlines are long past, assuming a continuous random variable $X$ whose densities $f_0(x)$ and $f_1(x)$ enjoy the property that $f_1(x) > 0 \implies f_0(x) > 0$, (that is, the likelihood ratio does not "blow up" at any $x$), we have that $$E[\Lambda(X)\mid H_0] = \int_{\mathbb R}\frac{f_1(x)}{f_0(x)}\cdot f_0(x)\, \mathrm dx = \int_{\mathbb R}f_1(x)\, \mathrm dx = 1.$$
$$E[\Lambda^n(X)\mid H_1] = \int_{\mathbb R}\left[\frac{f_1(x)}{f_0(x)}\right]^n\cdot f_1(x)\, \mathrm dx = \int_{\mathbb R}\left[\frac{f_1(x)}{f_0(x)}\right]^{n+1}\cdot f_0(x)\, \mathrm dx = E[\Lambda^{n+1}(X)\mid H_0].$$