What is the explanation that the Chi-Squared Goodness of Fit Test can be used to determine if a observed distribution equals an other distribution unnecessary of the kind of this distribution.

I know that there is a link to the central limit theorem - respectivelly the central limit theorem is used to explain why this is valid -.

I know the essence of the central limit theorem and also of the Chi-Squared test but I dont't get the link.

A thinking barrier for me is that the definition of the Chi-Squared distribution sais that it is the sum of squared normally distributed deviates. So WHY can I use the Chi-Squared test to test NON NORMAL distributed distributions?


1 Answer 1


The chi-squared goodness of fit test is for counts within different intervals.

When set up in this fashion, conditional on the total number of observed values, you have a multinomial distribution (a multi-category version of the binomial).

It is this multinomial distribution which is well approximated by a (multivariate) normal in large enough samples.

More specifically, it has approximately a degenerate multivariate normal distribution (since the sum of counts in each bin will equal the overall number of observations); with $k$ bins the distribution is restricted to a $k-1$ dimensional subset of $k$-space, from which the degrees of freedom arise.

This is not related to the distribution you're testing, except in so far as it impacts the expected proportions in each bin (and hence the expected counts), which in turn affects the quality of the normal approximation to the multinomial.

  • $\begingroup$ Thank you. I do still have some issues: 1. Am I right that the "multinomial Distribution" is smth. like a Distribution which "consists" of af x values where each x consists of y values. Thus x is approximately normal distributed (central Limit theorem)? 2. But HOW is the link to the Chi Squared test because actually if I draw only ONE sample from a e.g. a binomial Distribution X(1) this is not normally distributed. Can you please explain this to me in simple words or at an proctical example $\endgroup$
    – 2Obe
    Sep 4, 2017 at 6:12
  • $\begingroup$ 1. A multinomial variable consists of $k$ counts (as my answer already indicates). Each "+1" in the count in each bin represents an additional observation falling in that bin. 2. You seem to be confusing a single observed value with the distribution from which it is drawn. The distribution of a single binomial random variable can often be well approximated by a normal distribution. Of course when you observe a single value, the sample distribution doesn't look like a normal distribution but neither does a single value drawn exactly from a normal distribution; this is of no consequence $\endgroup$
    – Glen_b
    Sep 4, 2017 at 8:35
  • $\begingroup$ Thank you. For my understanding I will sumarrize: The Chi-Squared test is valid to test the proportions of non-normally distributed samples because the Chi-Squared Distribution consists of "bins" --> the single X in the sum(X(1)**2....X(n)***2) equation where each bin can be approximated to be normally distributed. The Explanation therefore is the central Limit Theorem. Did I get this right? $\endgroup$
    – 2Obe
    Sep 4, 2017 at 9:16
  • $\begingroup$ The chi-squared distribution does not consist of "bins". $\endgroup$
    – Glen_b
    Sep 4, 2017 at 9:18
  • $\begingroup$ With "bins" I meant the variates X in sum(X(1)**2....X(n)**2) equation where each X itself consists of values and here each X can be assumed to be normally distributed. Is my comment therewith correct? $\endgroup$
    – 2Obe
    Sep 4, 2017 at 12:56

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