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I'm a beginner in statistics and would like to understand one of the requirements for the chi-square test to work. It is that the random variable being measured has to follow a normal distribution.

My question then is, what other models should I use if the random variables I'm measuring do not follow a normal distribution? This is similar to how the Student-t test should be used instead of the Standard Normal for a small sample, is there an equivalent for chi-square test for a small sample?

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    $\begingroup$ A Chi-square test pops up in all sorts of different situations. For example, testing for a relationship between two categorical variables in a cross tab; or a more general case of comparing any observed distribution with a null hypothesis distribution. Is there a particular use in a small sample case you're interested in? $\endgroup$ – Peter Ellis Aug 12 '12 at 9:47
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    $\begingroup$ What type of chi-square test are you considering using? I ask because the various chi-square tests (there are various types as Peter Ellis and Michael Chernick have already pointed out) are commonly considered to be non-parametric, that is, a class of tests that do not have the assumption of normality. $\endgroup$ – Joel W. Aug 12 '12 at 10:44
  • $\begingroup$ Testing for variance of normal distributions is a special and unique case. I summarized how the chi-squared distribution arises on Insight Things. It's a simplification but should make clear that you cannot obtain a parametric test statistic on all kinds of distributions. $\endgroup$ – Jan Rothkegel Mar 30 '16 at 9:56
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Normality is a requirement for the chi square test that a variance equals a specified value but there are many tests that are called chi-square because their asymptotic null distribution is chi-square such as the chi-square test for independence in contingency tables and the chi square goodness of fit test. Neither of these tests require normality. This agrees with Peter Ellis' comment.

Regarding your question when specific parametric assumptions are not made (normality being just one such assumption) there are nonparametric procedures (rank tests, permutation tests and the bootstrap) that can be applied sith more generality. In regression, robust regression is an alternative to ordinary least squares.

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    $\begingroup$ You say that neither the chi-square test for independence in contingency tables nor the chi square goodness of fit test require normality. That's true, but some may misunderstand that statement, because it is not strong enough. Those tests analyze categorical outcomes, which (by definition) is never Gaussian. $\endgroup$ – Harvey Motulsky Aug 12 '12 at 22:39
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    $\begingroup$ @HarveyMotulsky A good point but not strictly correct. The chi square goodness of fit test applies to data from any probaility distribution including continuous and discrete ones. The test statistic may involve counts for categories but the original data need not be. In fact the test can be used to check normality. $\endgroup$ – Michael Chernick Aug 12 '12 at 23:02
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I learned the chi squared distribution as a special case of a gamma density function. What I have read today online on wikipedia and in texts sometimes says "If Z1, ..., Zk are independent, standard normal random variables, then the sum of their squares is distributed according to the chi-squared distribution with k degrees of freedom." and then in other instances drops the "normal random variables" part-

"By the central limit theorem, because the chi-squared distribution is the sum of k independent random variables with finite mean and variance, it converges to a normal distribution for large k." Perhaps this is an error in wikipedia.

But the normal assumption is really the part I am interested in today. It seems to me that there are NO "requirements for the chi-square test to 'work'" except maybe that the set of random variables not be empty, and be real numbers.

By test I take the asker to mean the squaring of the sum of the squares of the random variables, and then checking that against what would be expected from a standard normal with the given mean and std dev. That is to say the expected versus the observed outcome. Here is a list of so called "chi-squared tests": http://en.wikipedia.org/wiki/Chi-squared_test

The outcomes of a M = Z^2 where Z is a standard normal random variable are different than if M = G^2 where G was a random variable from a gamma distribution.

An example I can think of is in application when there is a small sample size- I suppose this can be defined as less than the amount that a sample size calculator would yield- Here you don't know if your normal assumption is valid, because let's say you have no prior data, and the sample size being small means no central limit theorem application, but not all is lost because a chi-squared test can be done to measure the validity of Gaussian distribution functions being used such as normal probability distribution function, normal cumulative distribution function and their inverses etc.

SO as far as I can tell one of the most useful uses of the chi-squared test in beginning and intermediate practice of statistics is to test the normal assumption on small sample sizes.

But to get to the part of the question that asks about other data types. I think it is good to learn all the different distributions other than normal, and t distribution. There are discrete probability distribution functions- Bernoulli, binomial etc. and there are continuous probability distribution functions- exponential, beta, Poisson, Pareto etc.

Then learn about what a gamma distribution is- how it is an all encompassing distribution function. From there simply looking at the data, graphing the data and measuring observed versus expected in some way i.e. "goodness of fit" etc can help determine what kind of distribution shape your data has. What's great about a gamma or even exponential distribution is that you can make any shape you see.

Due to the central limit theorem and the other versions of that that there are- people often just assume a normal distribution. And this is fine over many many tests and or much prior data.

This has been edited from the original comment.

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    $\begingroup$ I appreciate your effort, but your post is hard to follow because it takes a while to get going and then founders in some misconceptions. Non-central chi-squared distributions are irrelevant here: they appear in power calculations. More importantly, a chi-squared distribution is not simply the distribution of the square of any random variable and indeed the square of a Gamma variate never has a chi-squared distribution. As far as vagueness goes, please heed the comment by @Peter Ellis, who points out the necessity of being clear about what kind of "chi-square test" you are talking about. $\endgroup$ – whuber♦ Feb 17 '14 at 18:54
  • $\begingroup$ Well I'll appeal to the Nirvana Fallacy. and just say that I think it is a good discussion to have. The condition that the independent random variables be normal is the expected that is referenced in the pearson chi^2 test of goodness of fit. Whether there is a name for M = G^2 where G is a set of random variables from Gamma Distribution or not is what I was confused about. This idea would be useful and indeed M = Z^2 has a chi^2 distribution. The fact that people use different chi squared tests to validate assumptions of normal shows how useful the idea is in practice. $\endgroup$ – user40430 Feb 18 '14 at 0:44
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    $\begingroup$ The Poisson is discrete. The exponential is a useful distribution, but its shape is fixed. The answer here has been much revised, but I still struggle to see what point you are making. There seems to be confusion in this thread between (1) the derivation in mathematical statistics of chi-square distributions as the sum of unit Gaussians (2) the particular application of chi-square tests to a hypothesis that some variable is Gaussian, done as a calculation with binned frequencies. In principle similar tests can be applied to other distributions (in practice, there are often much better tests). $\endgroup$ – Nick Cox Feb 18 '14 at 9:19

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