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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    $\begingroup$ 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? $\endgroup$ – whuber Nov 30 '11 at 4:41
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    $\begingroup$ 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? $\endgroup$ – John Nov 30 '11 at 5:08
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    $\begingroup$ 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. $\endgroup$ – John Nov 30 '11 at 5:11
  • $\begingroup$ 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). $\endgroup$ – whuber Dec 2 '11 at 4:21

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.

  • $\begingroup$ That paper seems to be getting a lot of press recently. $\endgroup$ – cardinal Nov 30 '11 at 9:39

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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