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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?

MånsT
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Braj-Stat
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    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. – Christopher Aden Sep 04 '12 at 09:23
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    Given that both categories are ordinal, you could use a test that exploits that. See [Agresti, Analysis of Ordinal Categorical Data](http://www.amazon.com/Analysis-Ordinal-Categorical-Probability-Statistics/dp/0470082895/ref=sr_1_1?ie=UTF8&qid=1346755421&sr=8-1&keywords=agresti+analysis+of+ordinal) for various possibilities. – Peter Flom Sep 04 '12 at 10:44
  • There is a form of Fisher's test for $R\times C$ tables that can be used whether or not the chi square test works asymptotically. There are other alternatives as well. See [Categorical Data Analysis by Agresti](http://www.amazon.com/Categorical-Analysis-Wiley-Probability-Statistics/dp/0471360937/ref=sr_1_1?s=books&ie=UTF8&qid=1346754928&sr=1-1&keywords=categorical+data+analysis). – Michael R. Chernick Sep 04 '12 at 10:36
  • Why was my answer converted to a comment? It answers the question. – Michael R. Chernick Sep 04 '12 at 12:35
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    @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](http://meta.stackexchange.com/questions/118582/what-is-an-acceptable-answer). – whuber Sep 04 '12 at 12:56
  • @whuber I disagree. The answer to the question is to use Fisher's test for RxC tables which I gave. I referred to Agresti for more details. – Michael R. Chernick Sep 04 '12 at 14:27
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    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. – whuber Sep 04 '12 at 14:31
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    @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. – gung - Reinstate Monica Sep 04 '12 at 15:32
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    That's a good point, @gung. I recall someone else indicated in a comment (in another thread a few months ago) that chi-squared tests can remain accurate when a few cell populations are as low as $1$ provided they incorporate a continuity correction. (Comments are hard to search so I'm at a loss to find the exact quotation.) I have not personally verified this advice, but it is plausible. – whuber Sep 04 '12 at 15:35
  • Oops, @whuber, I may have phrased that confusingly. I didn't mean that you *should* violate that rule of thumb, but rather that the *OP's data* may still violate that rule of thumb (ie EV's may be <5 in addition to raw counts <5). Although, I believe it is true that the requirement is much less strict than people believe, & it's possible the test is not badly harmed. – gung - Reinstate Monica Sep 04 '12 at 17:02

2 Answers2

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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.

References

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.

RioRaider
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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).

abaumann
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