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Let's say I have a categorical variable. I try to test the null hypothesis that each category has the same count (of something) using a Pearson's Chi-Squared test. I may not be able to reject null hypothesis using just the categorical variables, but if I group the categories together in the right way, I can reject the null hypothesis. (For example $\{a,b,c\}$ have a higher count than $\{d,e,f\}$.) It seems though, that if I choose my groupings based on my sample distribution, then I'm overfitting. In simulations, I've been able to group categories of counts from a uniform distribution in the correct way to reject the null hypothesis too often for my significance level. However, I want to be quantitative about this error/abuse I'm committing. For example, I may be willing to group $\{a,d,e\},\{b,c,f\}$ but no other partition would make sense in my context. In this case I would be more confident in making the choice to group or not group then if I considered all possible partitions.

Is there some way to quantify this type of overfitting? I thought it might be hiding in the degrees of freedom, or maybe it's a type of parameter and something like AIC or BIC might be useful.

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  • $\begingroup$ Generally the only reason to combine classes is because your expected counts are very low (>5). Do you have low data or are you just trying to 'find significance'? Why are you combining groups? $\endgroup$ – cdeterman Dec 4 '14 at 20:38
  • $\begingroup$ @cdeterman Thanks for your response. In the group I work with we might look at something like age. (Say we're trying to predict likelihood they watch television.) If you want to make a good predictive model, you probably shouldn't consider each year or each day as its own category. But say one person in my group wants to consider 5 year bands, but another wants to consider 10 year bands, and another 15 year bands. Maybe the Chi-Squared test says that age is a predictor of television-viewing, but only with the 10 year bands, should we still use it? What if we take any n year bands? $\endgroup$ – T.J. Gaffney Dec 5 '14 at 0:41
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This procedure is basically the idea behind "CHi-squared Automated Interaction Detection", or "CHAID" described by G.V. Kass in 1980. The general setting is very similar to your television watching prediction example: You want to best predict the occurrence of a categorical variable by a combination of other categorical variables. You do this by finding the split with the maximal $\chi^2$ value.

A description of the algorithm and the issues around adjusting for statistical significance are given in (Kass, 1980). In that paper the Bonferroni correction is used to adjust for the selection of the maximal $\chi^2$ value.

Some actual theory is available for the case of reduction to a $2\times2$ table (Kass, 1975).

There is an R package called CHAID which implements the algorithm and is available on R-Forge.

Although it is a little different from your question, there is a similar situation that arises when dichotomizing a continuous variable to predict another dichotomous variable. Namely, where should you put the cut-point? This is discussed in (Miller and Siegmund, 1980) and (Halpern, 1982), among others.

Yet another setting where this type of question comes up is in change-point estimation or segmentation, though it has been too long since I looked at those papers to recall authors.

References:

Halpern, J. (1982). Maximally selected chi square statistics for small samples. Biometrics, 1017-1023.

Kass, G. V. (1975). Significance testing in automatic interaction detection (AID). Applied Statistics, 178-189.

Kass, G.V. (1980). An Exploratory Technique for Investigating Large Quantities of Categorical Data. Applied Statistics, 29(2), 119-127.

Miller, R. and Siegmund, D. (1980). Maximally Selected Chi-Squares. Technical Report 64. Stanford, Calif, Division of Biostatistics, Stanford University.

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    $\begingroup$ Kass' paper looks like what I'm looking for. (Thank you!) However, I'm concerned about his use of Bonferroni correction. Bonferroni's Inequality is equality when the sets are disjoint. Here, because there is so much overlap, wouldn't the inequality be very inequal? $\endgroup$ – T.J. Gaffney Dec 5 '14 at 17:32
  • $\begingroup$ @TJGaffney That is a good question. It looks like the usage is tuned to the algorithm at least to some degree in this case, so is not just a blind application to all possible subsets. In this type of problem extreme value distributions also tend to pop up; speculating, it might be a wash to go through the effort. As Kass says, it's also possible to enumerate the distribution for smaller cases --- it would be interesting to compare. $\endgroup$ – jvbraun Dec 5 '14 at 21:57

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