For categorical variable testing, we have chi-square test. For example, we want to test if a dice is fair (assuming sample has [6,5,7,7,6,5] and [6,6,6,6,6,6]).

But from the previous post, the testing results are sensitive to the magnitudes of counts, say test statistic of [600,500,700,700,600,500] and [600,600,600,600,600,600] is different from that of [6,5,7,7,6,5] and [6,6,6,6,6,6]. Furthermore, if two independent samples have different sizes, chi-square test will be subtle too.

My question is whether there exists multiple categorical proportional testing, which is an extension of single proportion testing?

For example, we test [1/6, 5/36, 7/36, 7/36, 1/6, 5/36] and [1/6,1/6,1/6,1/6,1/6, 1/6]. In this case we don't worry about data magnitude and unequal sample size.

Can we just naively have individual proportion test 1/6 = 1/6, 5/36 = 1/6, 7/36=1/6, ...., and once all null hypotheses are right, the sample is from the population? Is it Chi-Square Test of Homogeneity?

  • 1
    $\begingroup$ The extension is called the chi-squared test. $\endgroup$
    – whuber
    Jan 11, 2019 at 20:14
  • $\begingroup$ It is typically approached by the chi-squared goodness-of-fit test. $\endgroup$
    – Michael M
    Jan 11, 2019 at 21:25

1 Answer 1


Firstly, for anything like rolling dice, ignoring how many dice rolls your proportion estimate is based on is absurd. With an outcome of (1,0,0,0,0,0) you should not yet be worried about the fairness of a dice, at all, with (10000,0,0,0,0,0) you should be.

Secondly, many would consider the analysis of such data as a (simple?) extension of the single proportion case. E.g. for an exact test you similarly "simply" enumerate all possible outcomes and "just" need to define what counts as more extreme. On the Bayesian side there's a conjugate analysis with a beta-prior for a single proportion and with a Dirichlet prior for more outcome categories.

If you are looking for models/distributions for a single proportion or on a simplex there's the beta and Dirichlet distributions. Whenever possible it is better to consider the underlying binomial/multinomial (if out exists and these data are available).

  • $\begingroup$ I found the answer to be more like my case is the chi-squared test for homogeneity, so in some textbooks it is called chi-squared contingency Tables, chi-squared tests for two samples. But thanks. $\endgroup$
    – TripleH
    Jan 23, 2019 at 6:03

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