hypothesis testinglikelihood-ratiop-valueself-study
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.
I post this as an answer, because I do not have the privilege to comment.
The LR-test compares two different models, which could differ not only in one predictor but possibly in many. The result of your test tells you, that the model including the variable temperature better fits the data compared to the model without it. The test has two dfs because you dropped two betas.
If you are interested in comparing the different levels of the variable temperature, summary(J_mod.TEMP) will give you the p-values. Note that Ruses dummy coding by default, so the first coefficient contrasts level 2 vs. 1 and the second coefficient level 3 vs. 1. You may use contrasts to adapt this to your personl needs.
Maybe by understanding (and using) that $\hat p = x/n.$ Also, by relying on your intuition that you're going to reject for observed integers $x$ far from 2.5.
Your criterion is to reject if $Q = \hat p^x(1-\hat p)^{n-x}$ is large. Make a plot (I'm using R, but calculator, pencil, and paper will do.)
n = 5; x = 0:n; p = x/n
q = p^x*(1-p)^{n-x}
plot(x, q, pch=19)
The first thing to try would be to reject only for $x = 0$ or $5.$ Using the binomial formula, that gives rejection probability under the null hypothesis $H_0: p = 1/2$ as $2/32 = 1/16 = 0.0625.$
sum(dbinom(c(0,5), 5, .5))
[1] 0.0625
So the smallest size for a non-randomized test is 6.25%. For a test at exactly 5%, you would have to use a randomized test. I'll leave that part to you.
Best Answer
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.