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I'm dealing with a population where there are $5$ categories $A, B, C, D, E$ and I would like to make the claim that $A$ is the most common category in the population (i.e., the most frequent).

I thought about the following, use a proportion difference test (e.g., z-test) $4$ times to assure that the proportion of $A$ is larger than all others which would hence imply that it has the highest proportion.

However, I read that the the z-test applies on different populations. When here all proportions are part of the same population and I'm at loss as what to do since then.

Thank you for your time.

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If I were asked to "prove" (i.e., statistically confirm using some null hypothesis statistical testing (NHST) protocol) that one category for a categorical variable was maximal for a sample (from a single population), my first exploration approach would be the following:

First, run a chi-square goodness-of-fit test with all 5 categories where we would start by assuming equiprobability. If you have a statistically significant p-value (which you would want in this case if the goal/desire is to see that the first category is the mode), then you can reject $$H_o : \pi_A = \pi_B = \pi_C = \pi_D = \pi_E$$ If this comes out to not be statistically significant, then you really can't claim that any one category is more frequent than the other with the current sample size.

Let's assume you do find a statistically significant result at the first step and let's assume the first category $A$ is both the one you wish to show is the mode and indeed has the largest observed frequency in the sample. My next step would be to run a follow-up chi-square test with all the values for the categorical variable EXCEPT for $A$. If this NHST comes up to be not statistically significant, then we can assume the probabilities for the remaining categories are (nearly) identical.

Next, estimate this common probability $$\pi_\text{pooled} = \frac{k_B+k_C+k_D+k_E}{N} \div 4$$ and run a single sample proportion NHST to confirm that $\pi_A$ is bigger than this pooled proportion estimate $\pi_\text{pooled}$.

The key here would be to decide what to do if at the second step the probabilities are not the same. In that case, I would probably try an iterative/recursive process to obtain subsets of nearly equiprobable categories.

Again, this is how I would explore the given data to see if there was any evidence to suggest one of the categories was indeed the modal category. If you wish to prove this more formally, a different strategy may be required.

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  • $\begingroup$ Thank you, Greg. This is clear and intuitive to me till we get to the common probability. What exactly is $k_{x}$, is it the count $N\pi_{x}$? I would highly appreciate if you can help me understand and justify this step. $\endgroup$
    – Essam
    May 12, 2023 at 17:29
  • $\begingroup$ The $N \pi_X$ would be the population mean for the count for that category. In this case, $k_X$ is the sample count for that category, and $N$ is just the total sample size. $\endgroup$
    – Gregg H
    May 12, 2023 at 18:37
  • $\begingroup$ So $\pi_{pooled}$ is the average sample proportion of the rest of the categories. Does this carry an inherent assumption that this quantity is a good estimate for the population's "average proportion of the rest of the categories" and that hence, if $\pi_{A}$ is larger than it then its likely larger than each proportion $\pi_{x}$ where $x\in {{B,C,D,E}}$? $\endgroup$
    – Essam
    May 12, 2023 at 18:58
  • $\begingroup$ The assumption is less about the quality and more about the fact that if the goodness-of-fit is not significant, then we can assume them to be equal (and thus use all of them as a pooled estimate). $\endgroup$
    – Gregg H
    May 12, 2023 at 20:36
  • $\begingroup$ Aha! Now that makes sense. Thanks a lot. $\endgroup$
    – Essam
    May 12, 2023 at 20:57

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