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The chi-square p value is showing BMI of students is independent of their snack eating habit but Cramer's V shows some association in BMI level and snack eating habit. Why are the results of the two tests different?

> t=table(data$BMI,data$habit)
> t

           No Yes
  Abnormal 19  13
  Normal   59  17

> assocstats(t)
                    X^2 df P(> X^2)
Likelihood Ratio 3.5994  1 0.057801
Pearson          3.7412  1 0.053086

Phi-Coefficient   : 0.186 
Contingency Coeff.: 0.183 
Cramer's V        : 0.186 
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    $\begingroup$ Your question is unclear. Please explain what parts of the output you're comparing and why you expect to see something other than what you see, $\endgroup$
    – Glen_b
    Jul 19, 2019 at 10:44
  • $\begingroup$ In my case the chi-square p value is showing BMI of students is independent of their snack eating habit but cramer's V shows some association in BMI level and snack eating habit. why results by both test are varing? $\endgroup$ Jul 23, 2019 at 12:42
  • $\begingroup$ I have edited your comment into your question. $\endgroup$
    – Glen_b
    Jul 23, 2019 at 23:14

1 Answer 1

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Your comment refers to 'both tests' but Cramér's V only has a statistic, there's not a test there to compare with the chi-squared.

Both the chi-squared statistic and Cramér's V show some suggestion of association -- and since the two are monotonically related, they would be equally suggestive of association.

If you tested Cramér's V, you should get the same p-value as you did for the chi-squared test (again, because they're monotonically related).

Note also that failure to reject the null does not mean the null is exactly true (you don't demonstrate independence by failing to reject it -- absence of evidence is not evidence of absence). Rather, you only know that whatever dependence there might have been was too weak (at least in the sample) to distinguish it from independence + random variation.

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  • $\begingroup$ yes thanks , what does it means monotonically related ? $\endgroup$ Jul 24, 2019 at 4:59
  • $\begingroup$ They incrrease or decrease together (but not in proportion). In fact Cramer's V is just a multiple of the square root of the chisquare. If you tell be the number of rows and columns (the shape of the table) and the total number of observations, you can calculate Cramer's V directly from the chi-square -- they contain the same information. $\endgroup$
    – Glen_b
    Jul 24, 2019 at 5:13
  • $\begingroup$ thanks, In such case is there any alternative test for it ? $\endgroup$ Jul 25, 2019 at 12:25
  • $\begingroup$ To what purpose? If they could yield different decisions there might arguably be some point, but any test based on using V as a statistic or any monotonic transformation of it should yield the same decision; differences will boil down to slight differences in choices of approximations (such as continuity corrections, for example, or exact rather than asymptotic tests) $\endgroup$
    – Glen_b
    Jul 25, 2019 at 12:43
  • $\begingroup$ to get the results with strong evidence eg. that would say there is high association or no association with strong evidence. In my case Chi square is not giving perfect results with strong evidence $\endgroup$ Jul 25, 2019 at 12:52

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