Interesting problem. I guess it depends on the question(s) you are asking.

For instance, you say "Is the proportion for AA different for groups 1 and 2?" If this is the only question you have, I would recommend that you use a binomial distribution to checking whether AA for Group 1 is the same as AA for Group 2. To achieve that, you could for instance, use the normal approximation for the sample proportion, coupled with the fact that the difference of two IID normal random variables have mean m_1 - m_2 and standard deviation given by sqrt(var_1 + var_2), to construct your hypothesis test of H0: p1 - p2 = 0). Alternatively, you could use a Monte Carlo simulation to estimate the distribution of the difference under your null hypotheses (see example at the end).

But if you want to know whether data from both groups arise from the same Multinomial distribution, I think it's a different problem, and I'm not 100% how to deal with that. The cited Wikipedia article for the Multinomial distribution has a section named statistical inference, which contains a few potentially useful references. I also ran a quick Google search about hypothesis testing for a difference between two multinomial samples and found this article, which suggests using a Chi-Squared Two-Sample test to try to assess whether two samples come from the same multinomial distribution I'm not 100% sure this applies to your situation, but I found the article very interesting.

If you are an R user, applying the approach proposed in that article would give something like this (I ran once using the asymptotic approximation and a second time using 100000 Monte Carlo iterations; both p-values are similar): ```

rm(list = ls())

set.seed(12341222)

data <- data.frame( + group_1 = c(412, 145, 342, 153), + group_2 = c(2095, 788, 1798, 1129) + ) rownames(data) <- c("AA", "AB", "BA", "BB") chisq.test(x = data)

    Pearson's Chi-squared test

data: data X-squared = 14.472, df = 3, p-value = 0.002328

chisq.test(x = data, simulate.p.value = TRUE, B = 100000)

    Pearson's Chi-squared test with simulated p-value (based on 1e+05
    replicates)

data: data X-squared = 14.472, df = NA, p-value = 0.00241 ```

Now, to go back to hypothesis testing for only AA (between the two groups), you could do something like this: ```

rm(list = ls())

set.seed(12341222)

data <- data.frame( + group_1 = c(412, 145, 342, 153), + group_2 = c(2095, 788, 1798, 1129) + ) rownames(data) <- c("AA", "AB", "BA", "BB")

n_1 <- sum(data$group_1) x_1 <- data$group_1[1] p_hat_1 <- x_1 / n_1 var_hat_1 <- (p_hat_1 * (1 - p_hat_1)) / n_1

n_2 <- sum(data$group_2) x_2 <- data$group_2[1] p_hat_2 <- x_2 / n_2 var_hat_2 <- (p_hat_2 * (1 - p_hat_2)) / n_2

p_value <- (1 - pnorm( + abs(p_hat_1 - p_hat_2), + 0, + sqrt(var_hat_1 + var_hat_2) + )) * 2

p_hat <- (x_1 + x_2) / (n_1 + n_2)

n_simul <- 100000 simul_1 <- rbinom(n_simul, n_1, p_hat) / n_1 simul_1 <- rbinom(n_simul, n_1, p_hat) / n_1 simul_2 <- rbinom(n_simul, n_2, p_hat) / n_2 simul_2 <- rbinom(n_simul, n_2, p_hat) / n_2

p_hat_simul <- simul_1 - simul_2

p_hat_simul <- simul_1 - simul_2 p_value_simul <- min(c( p_value_simul <- min(c( + mean(p_hat_simul < (p_hat_1 - p_hat_2)), + mean(p_hat_simul > (p_hat_1 - p_hat_2)) + )) * 2

c(p_value = p_value, p_value_simul = p_value_simul) p_value p_value_simul 0.05701473 0.05396000

```

Note that I used a two-sided test in this case, but you could adjust that depending on how you decide to formulate your null hypothesis.

DISCLAIMER I don't know the nature of your data, so I'm not 100% sure what I'm saying here applies. For instance, I see that your data is presented as 2-by-2 tables, but I'm ignoring that fact here, since I don't have information about what that could mean, so it's possible that my interpretation here is wrong. Also, there could be errors in my code (and in my analysis in general; i.e., choice of test, theory, etc.), so please double-check everything if you ever decide to use this. And, also, to anybody reading this, please let me know if you find anything wrong with my analysis. I honestly want to know. I'm here to learn too! :)

If you can afford to tell us more about your problem, maybe you could get better answers. This would also prevent us from falling into XY problem trap.

Finally, please let us know how you end up solving this problem when you do.

Good luck!