To find association between peer's support (independent variable) and work satisfaction (dependent variable) I wish to apply chi-square test. Peer's support is categories in four groups according to the extent of support: 1=very less extent, 2=to some extent, 3=to great extent and 4=to very great extent. Work satisfaction is categories into two: 0=not satisfied and 1=satisfied.

The SPSS output says than 37.5 percent cell frequencies are less than 5. My sample size is 101 and I don't want to reduce categories in independent variable into lesser number. In this situation is there any other test that can be applied to test this association?

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    $\begingroup$ I'm not entirely sure how it's handled in higher dimensional tables like yours, but in the 2x2 case, the small sample analog to the chi-square is the Fisher Exact Test. I'd heard it's possible to use the FET in arbitrary r x c contingency tables, but that it was computationally intensive. Another option would be to do a permutation test. $\endgroup$ Sep 4, 2012 at 9:23
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    $\begingroup$ Given that both categories are ordinal, you could use a test that exploits that. See Agresti, Analysis of Ordinal Categorical Data for various possibilities. $\endgroup$
    – Peter Flom
    Sep 4, 2012 at 10:44
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    $\begingroup$ @Michael Because it is not an answer: it is merely a hint followed by a (vague) pointer to an answer elsewhere. Please see the SE FAQ about answers. $\endgroup$
    – whuber
    Sep 4, 2012 at 12:56
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    $\begingroup$ You're welcome to discuss this on meta, @Michael, but not here. If you do open a discussion, I will maintain that "a form of" and "other alternatives" are too vague to be considered answers, as MånsT was gently trying to suggest. Sure, there is a gray area between answer status and comment status. As a moderator and reviewer I constantly am called to determine when would-be answers are really functioning as comments: this test of vagueness is one I attempt to consistently apply. $\endgroup$
    – whuber
    Sep 4, 2012 at 14:31
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    $\begingroup$ @Braj-Stat, one thing to note is that the "requirement" (such as it is) for the chi-squared test is that expected values are >5 in all cells, not raw counts, although you may still violate that rule of thumb, &/or want to run a different test anyway. $\endgroup$ Sep 4, 2012 at 15:32

2 Answers 2


Conover (1999:202) suggested that the expected values can be "as small as 0.5, as long as most are greater than 1.0, without endangering the validity of the test."

He also provides a "rule of thumb" from Cochran (1952) which suggested that if expected values are less than 1 or if more than 20% are less than 5, the test may perform poorly. However, Conover (1999) provides some evidence that Cochran's "rule of thumb" is overly conservative.


Cochran, W. G. 1952. The $\chi^2$ test of goodness of fit. Annals of Mathematical Statistics 23:315-345.

Conover, W. J. 1999. Practical nonparametric statistics. Third Edition. John Wiley & Sons, Inc., New York, New York, USA.


The $\chi^2$-test was originally devised by Pearson as an approximation to the log-likelihood ratio, due to the fact that log-likelihoods were too computationally intensive for the time.

Pearson's G is defined as $G = 2\sum_{ij}O_{ij}\ln(O_{ij}/E_{ij})$. It follows the same distribution as the corresponding $\chi^2$-test.

(Forgot to mention originally: G is much less sensitive to expected cell counts < 5).


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