Take the 2-minute tour ×
Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It's 100% free, no registration required.

I applied a survey consisting of 12 questions to 120 people and each questions include 4 nominal categories; I want to make comparison of people's answers according to their socio-demographic characteristics such as their educational level or socioeconomic status. All of my comparison criteria consist of nominal categories and number of categories change between 3 to 6. I cannot drop or combine comparisons; categories in other words number of categories are fixed.

My question is when I compare the frequency distribution of people's answers based on their education levels (for example) through chi-squared test, I got a warning that says (for instance) 7 cells have less than count five; minimum expected count is; 43. I got this warning for almost my all questions - demographic characteristics comparisons.

Shall I underestimate this warning and use my test results or is there a different test I should use? If I should use a different test, which one?

share|improve this question
1  
Chi square seems poorly disposed to handling varying numbers of categories within a single test. Could you be more clear about the test you are performing? Chisquare goodness of fit? Independence? How are you handling the varying number of categories? –  rpierce Apr 9 at 18:28
1  
Have you quoted the warning message absolutely literally? There is an important difference between a low count and a low expected count. –  whuber Apr 9 at 19:44
add comment

3 Answers 3

First let me try to clarify the question:

You have a bunch of chi-square tests to run, specifically 12 answers to questions against several different demographic variables. All the demographic variables are nominal, with different numbers of levels (e.g. could be 2 (Male vs. female), 5 for race (or fewer depending on how you classify race) etc.

In most of your tests you get a warning about small expected cell sizes.

If that's the case, the natural thing to do is use an exact test. In SAS there is an EXACT statement in PROC FREQ. In R there is fisher.test in the stats package.

If your questions have ordinal answers there may be better methods.

share|improve this answer
add comment

A lot of the time, you may not need to do anything. The "5" rule is overly conservative, and there are a number of less restrictive (but somewhat more complex) guidelines to be found in the more recent literature (where 'more recent' means 'over the last half century or more').

For example, if all your cells have expected higher than 1 and about 80% are above 5, you're probably safe just treating it as chi-square (in that the p-values will still be roughly correct in instances you'll care to have good accuracy in). If expecteds are close to equal you can go lower.

If you have access to something that can generate random tables with fixed margins (such as can be done in R), you can use simulation to estimate p-values without changing anything else. That's often the easiest to do and is built into chi-square testing in R, as an option.

There are a number of other options (some mentioned in other answers), but my usual preference is to simulate if the null distribution of the test statistic won't be adequately described by the chi-square.

share|improve this answer
add comment

You have two options:

  • ignore the warning
  • combine/merge the bins
share|improve this answer
6  
Those are not the only two options –  Peter Flom Apr 9 at 18:56
1  
... they are also poor options until demonstrated otherwise. In particular, combining bins after the fact can change the distribution of the test statistic (causing it no longer to be chi-squared). –  whuber Apr 9 at 19:42
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.