TITLE: How to find which treatment is most effective in gene data given one standard method and 3 variations
QUESTION [0 upvotes]: Sorry I am a biologist and it appears am not quite confident enough for statistical analysis. I have datasets that represent different treatments on a biological system. It records how many genes have increased expression.
my results are
40/409 genes, 
30/412 genes, 
15/407 genes, 
24/430 genes
How would I go about finding if there is a statistical difference between the tests? If I wanted to know which has the lowest, or highest affect how would I go about it?

REPLY [0 votes]: If you're concerned with whether the chance of increased expression differs at all between the four treatment groups, you might start off with a likelihood ratio test for checking whether $p_1 = p_2 = p_3 = p_4$, where $p_i$ is the probability of increased expression within treatment group $i$.
The likelihood ratio will compare the maximum likelihood of the data when all probabilities are constrained to be equal (the null hypothesis) to the maximum likelihood when all four are allowed to vary (the alternative).  Notice that within group $i$ the number with increased expression can be assumed binomially distributed with parameters $n_i$ and $p_i$, and so our likelihood function will be a product of binomial probabilities.  Specifically, if $x_i$ is the number with increased expression within group $i$, $\hat{p}_i \equiv x_i / n_i$, and $\hat{p} \equiv (x_1 + x_2 + x_3 + x_4) / (n_1 + n_2 + n_3 + n_4)$ (these are the estimates which maximize the likelihood under the two hypotheses) the likelihood ratio becomes,
$$
\begin{align}
\Lambda(x) = \frac{\hat{p}^{x_1 + x_2 + x_3 + x_4} (1 - \hat{p})^{n_1 + n_2 + n_3 + n_4 - x_1 - x_2 - x_3 - x_4}}{\hat{p}_1^{x_1} (1 - \hat{p}_1)^{n_1 - x_1} \hat{p}_2^{x_2} (1 - \hat{p}_2)^{n_2 - x_2} \hat{p}_3^{x_3} (1 - \hat{p}_3)^{n_3 - x_3} \hat{p}_4^{x_4} (1 - \hat{p}_4)^{n_4 - x_4}} .
\end{align}
$$
Once you've computed this value, there's theory that says $- 2 \log [\Lambda(x)] \sim \chi^2_3$ under the null hypothesis, where $\chi^2_3$ refers to a chi-squared distribution with three degrees of freedom.  This will allow you to compute an approximate $p$-value and determine if there's evidence that the probabilities differ.
For trying to determine pairwise differences, you could probably just use standard $z$ tests where your test statistic will take the form,
$$
z = \frac{\hat{p}_i - \hat{p}_j}{\sqrt{\frac{\hat{p}_i(1 - \hat{p}_i)}{n_i} + \frac{\hat{p}_j(1 - \hat{p}_j)}{n_j}}} ,
$$
which approximately follows a standard normal distribution when $p_i = p_j$.
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