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Can chi-square test be performed on data that does not have normal or gaussian distribution?

One of the chi-square tests I am performing results in percentage points less than the 12.59 (α=0.05, v=6) using cross-tabulation. Therefore, I cannot reject the null hypothesis.

Is there a test that I can perform to confirm the alternative hypothesis?

Added clarification 30nov2011:

I will attempt to clarify some of the issues raised in these comments. Please be gentle on me as I am not trained in statistics. In fact, thanks to the unimaginative curriculum and teaching, I abhorred statistics taught in engineering school. Just trying to help my wife with a project.

So, we are trying to establish a weak (if at all) relationship/dependence between 2 different attributes of a dataset. To accomplish this, we have generated a cross-tabulation.

One of the attributes matches perfectly with a lognormal distribution (acc. to Minitab) as the data points and the lines are almost coincident. We have used the k-means clustering algorithm to group the lognormal attribute into 3 broad categories (large, medium and small).

Next, we are trying to perform a chi-square test on the cross-tabulation. df = 6. But, not having sufficient insight into this, I am not sure what the result of 6.8 percentage points supposed to mean for the null hypothesis. Per our understanding, the null hypothesis is "two attributes are not related".

I am wondering what the next step should be. Further progress hinges upon confirming or denying the existence of a relationship. (Note, we do not have to determine the relationship. However, I wouldn't mind getting some insight into the relationship in this process.)

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Could you be more specific about which chi-square test you are performing and how? Is this is a goodness of fit test? If it's a cross-tabulation, how does distribution enter into consideration? – whuber♦ Nov 30 '11 at 4:41
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Why are you trying to make your analysis reject the null hypothesis at a level below 0.05? Maybe you cannot reject the null because you shouldn't? Maybe you have a rounding error. If it's really 12.592 or higher than p < 0.05 at 6 digits out. Have you considered that that's kind of an absurd situation to be in and perhaps you should think about it differently? – John Nov 30 '11 at 5:08
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Additionally, suppose someone on here proposes a test and the analysis does come out at p < 0.05. What you've done is just take two passes at the same data looking for the same thing with an alpha of 0.05 so now you've actually inflated it and have to correct for the multiple tests. Stop thinking of your statistics and research that way. It's the path to Type I erros. – John Nov 30 '11 at 5:11
Re the edit: For insight, draw pictures of the data. A scatterplot of the two attributes is a good place to start. If the second one is just a small number of categories, then side-by-side boxplots of the first one (one for each category of the second) can be illuminating. Due to the lognormality, use a log scale for the first attribute (or use its logarithms directly). – whuber♦ Dec 2 '11 at 4:21

3 Answers

up vote 5 down vote

The assumption of normality is not generally one of the assumptions of a Pearson chi-square test. Typically the assumptions are that you must have a large enough n in each cell of the test, that the sample is selected randomly, and that the samples are independent. That's it.

As to the implied way you're thinking about statistics, you might want to read this. It may or may not be in your field but the principles apply broadly.

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That paper seems to be getting a lot of press recently. – cardinal Nov 30 '11 at 9:39
up vote 1 down vote

Suggestion: take the lognormally-distributed variable and show how it is distributed (obtain a histogram) separately for each level of the other, categorical variable. That will show you more about the nature of the relationship than truncating into 3 categories and doing a "binary," rather sterile hypothesis test.

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up vote -1 down vote

Does the Pearson chi square test not implicitly assume a normal distribution? Having fallen foul of the random simple sampling assumption I tend to make sure I have normal data (or a lot of non-normal data) before setting too much stall by the simple chi squared test? As I understand it that's why it only works with a sufficient sample size. Maybe I'm being pedantic or just plain wrong although it seems to follow from the definition of the distribution.

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Incorrect or too much like a comment? I can remove the question mark so it's a more definite answer. – Bowler Dec 2 '11 at 7:55
Welcome to the site @Bowler. I agree that whomever left the downvote should be more explicit in their critique via a comment. Although to be fair, I have no idea what you are trying to say! – Andy W Dec 2 '11 at 21:02
I was simply commenting on the fact that the chi squared distribution with n degrees of freedom is formed from the sum of squares of n normal measurements. This is why tests of the chi squared kind, Pearson or otherwise, usually have a statement regarding the number of samples. – Bowler Dec 4 '11 at 23:07
normality is not an assumption, implicit or otherwise, of chi-square... perhaps that's why the down vote. – John Dec 7 '11 at 13:18
Ah ok fair enough. My understanding of the test was that data sampled from a Poisson distribution where the mean is the same as the variance will, given sufficient samples, approximate to a normal distribution then SUM(O-E)^2/E approaches a chi squared distribution. Mentally that makes sense to me because with non-categorial data that equation gives amusing anomalous results. If you have a good ref it would be appreciated, its rarely relevant to me outside of histograms, but ignorance is never a good thing. – Bowler Dec 7 '11 at 17:44

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