We can use T test to check if two proportions are significantly different. Similarly is there a way to test if two multinomial distributions or "2 samples with more than 2 unique values" are significantly different from each other.

For example, I have a sample (say sample 1) where it has 100 red balls, 300 green balls and 400 yellow balls and 200 orange balls and sample 2 has 101 red balls, 302 green balls and 399 yellow and 202 orange balls.

  1. Is there a way to check if the above 2 samples are significantly different ( 2. Is this same as checking if 2 multinomial distributions are significantly different ). If so, can you explain how.

I was told in one of interviews that (if I remember correctly) KL divergence can be used to check this. 3. Can I use KL divergence for this (or to check if the sample multinomial distribution is significantly different from expected) ? 4. If so, how to check for significance with KL divergence or what's cutoff value of KLD to say that the difference is significant (like the p values in statistical tests). 5. Can I use ANOVA, chi square for these (if so, can you please explain)

  • $\begingroup$ Are you saying that you have randomly drawn two astonishingly similar independent samples (of slightly different sizes) with replacement from different populations? If so, in R: x1 = c(100, 300, 400, 200); x2 = c(101, 302, 399, 202); TAB = rbind(x1,x2); chisq.test(TAB) gives P-value almost $1,$ so no significant difference btw populations. $\endgroup$
    – BruceET
    Oct 16, 2020 at 6:51
  • $\begingroup$ This is just an example, I want to know how to find if two samples with multiple variables associated with each (say a sample has gender, height, age, demographic etc). can be tested to see if they are significantly different in one or more variables $\endgroup$
    – tjt
    Oct 16, 2020 at 17:39

1 Answer 1


You can perform the goodness of fit test. Given two vectors of data you test, through the chi-squared test, if they are significantly different or, given a vector of data, you test if their frequencies significantly differ from a given vector of probabilities.

Data comparison:

x1 = c(100, 300, 400, 200)
x2 = c(101, 302, 399, 202)

Frequency comparison:

x1 = c(100, 300, 400, 200)
p = x2/(sum(x2))
x2 = c(100, 300, 400, 200)
p = x1/(sum(x1))
  • 2
    $\begingroup$ Are you sure that's the correct input to the chisq.test function? I am not much familiar with R, but I'm pretty sure the chi square test works on the raw counts, not on the proportions. $\endgroup$
    – PedroSebe
    Oct 16, 2020 at 22:02
  • $\begingroup$ Correct, I edited. $\endgroup$
    – Luke
    Oct 19, 2020 at 6:48

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