We will be conducting an experiment where a group of $N$ people take a test that has $M$ independent yes-no questions. We compute the number of questions each person $j$ gets correct as a score $m_j$. It seems reasonable to model the score as a binomial random variable $m_j\sim B(n=M, p=p_j)$. However, we have no prior knowledge about $p_j$, and since the test requires skill, it seems that $p_j$ could vary from one person to the next.

Question 1: Is it reasonable to use the average score over the population, $m^* = \frac{1}{N}\sum_1^N m_j$, as the value of $p_j$ for all $j$? Thus, we model each of the $N$ scores as a draw from the same distribution $B(n=M, p=m^*)$.

Then, say we repeat the experiment on a new group of $N$ people with a different test (still $M$ independent yes-no questions). We would like to know if the performance on the second test is better than on the first.

Question 2: What is the best way to determine if the second population had better scores than the first? Does the answer to Q2 change if we can or cannot model each population as coming from $B(n=M, p=m^{\ast})$ with different $m^{\ast}$ as in Q1.

I have considered using a one-sided Welch's t-test on the populations of scores $m_j$ to determine if the average score $m^\ast$ is different between experiment 1 and 2 -- the null hypothesis is that $m^\ast$ is the same for the two experiments, and if we reject the null hypothesis then we can say the second population performed better. The data is unpaired (the $N$ test-takers are completely different in experiment 1 and 2), we do not assume the populations of $m_j$ have equal variance (hence using Welch's test), and we have reason to believe the performance should be better on experiment 2 (hence the one-sided test).

My uncertainty is around whether it is appropriate to use Welch's t-test given that the scores are discrete (they only take integer values from $0$ to $N$), not continuous, and therefore the sample means $m^\ast$ for the two experiments also cannot take on any value ($m^\ast$ can only be a multiple of $1/N$).