# How to find the exact and likelihood ratio p-value for a hypothesis test that a population sex ratio is 1:1?

Skulls were excavated from a dig. The sex of each skull was determined by anatomical appreciation. For Sites A, the data was Male=162, Female=110.

a) Find the exact p-value for a hypothesis test about whether the sex ratio of the population buried at this sites was 1:1.

b) Instead use a likelihood ratio test. Find the likelihood ratio test statistic, and resulting p-value. Compare with (a).

Any help would be appreciated.

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To me, the interesting part of this question concerns what assumptions would be needed to justify the binomial distribution in the first place! –  whuber Oct 14 '12 at 15:05

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You only know the number of the females and males in site A? Do you want to test all the sites have the same ratio or site A has ratio 1:1?

I guess this is a problem wrt to Contingency table, for example, you may write it down in the form: \begin{align*} & A\ B\ C\ sum\\ F\\ M\\ sum\\ \end{align*} Then you may use some formulas to calculate p values,etc.

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Welcome to our site, Julie! Could you please be more explicit about what the rows and columns of this contingency table would be and what "formulas" you would propose? The only attributes evident in the question are gender--male or female--so I'm having trouble seeing how you would make a two-way table out of this. –  whuber Oct 14 '12 at 15:07
@whuber, I think we can formulate in this way if following the hint. Assume that the total number is fixed, which is $n=272$, then we take the hypothesis test: $H_0: p=0.5$ vs $H_1: p\ne 0.5$, where $p$ is the probability that one observation is female. Then we calculate the test statistics $T=C_n^{110} (0.5)^{110}(1-0.5)^{162}$. So that you can compare this with the quantiles of the binomial distribution with parameter n and 0.5. –  Julie Oct 15 '12 at 15:43

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.

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The test statistic is $z$. The value of $1.96$ is usually called a "critical value." In what sense does a likelihood at $p=1/2$ of $0.00033$ "agree" with the hypothesis test? (The answer is that there is no meaningful way to compare these two calculations. This exposition includes some useful computation but otherwise looks rather mixed up.) Do you think you could straighten out your explanations? –  whuber Sep 10 at 1:37
Thanks for the feedback @whuber. Hmm. I think my point was that both methods provide evidence against the skulls having a true proportion of 50/50. Let me think some more and I'll try to update my response. –  Justin Bozonier Oct 19 at 20:07