# Is this the correct likelihood equation?

Consider a 2x2 contingency table with X having two categories {men,women} and Y having categories {yes,no}. For each category of X the observations was fixed (so the rows had fixed totals).The observations for each row are independent. Suppose that there where a total of $n_1$ observations for category {women} and that $f_1$ observations was 'yes'.A total of $n_2$ observations for category {men} with $f_2$ observations on 'yes'.

So therefore I have two independent random variables which has a binomial distribution say $Z_1 \sim Bin(n_1,\pi_1)$ , $Z_2 \sim Bin(n_2,\pi_2)$

I want to make a test using the Pearson statistic $X^2$. I want to test $H_0= \pi_1 = \pi_2 = \pi$ , so therefore I need to estimate the parameter $\pi$ for example by maximizing the likelihood equation. But my question is how should i set up this equation?

• Should the ML equation be the product $L_1(\pi)= Pr(Z_1 = f_1 ) Pr(Z_2 = f_2 )$

or

• Should i treat all observations as 'one' , that is , don't care about gender. So we will actually have a random variable $Z_3 \sim bin (n_1 + n_2, \pi)$ and the likelihood equation will be $L_2(\pi) = Pr(Z_3 = f_1 + f_2)$
• Sorry guys the two alternatives are equivalent in this case . Should I close this question ? – Danny Jan 30 '16 at 20:04
• You can post an answer yourself explaining how you answered your question :) – JohnK Jan 30 '16 at 20:53
• Yes, better to answer the question yourself than close it or delete it - a good answer will help others in a similar situation. – Glen_b -Reinstate Monica Jan 31 '16 at 0:45

$$L_1(\theta)= Pr(Z_1=f_1)\cdot Pr(Z_2 =f_2) = \binom{n_1}{f_1}\pi^{f_1}(1-\pi)^{n_1-f_1}\cdot \binom{n_2}{f_f}\pi^{f_2}(1-\pi)^{n_2-f_2} \propto \pi^{f_1+f_2}(1-\pi)^{n_1+n_2-(f_1+f_2)}$$
Since under the null hypothesis men and women have the same "opinion" ($\pi_1 = \pi_2 = \pi$), we can treat everything as 'one sample' as opposed to two samples (one sample for men, and one for women). Setting up this likelihood equation we get:
$$L_2(\theta)=Pr(Z_3=f_1+f_2) = \binom{n_1+n_2}{f_1+f_2}\pi^{f_1+f_2}(1-\pi)^{n_1 +n_2 -(f_1 +f_2)}\propto \pi^{f_1+f2}(1-\pi)^{n_1 +n_2 -(f_1 +f_2)}$$
Inference about $\pi$ will be the same in both cases.(The ML equations are the equivalent.)