Is there a relationship between p-value and Cramer's V value after conducting Chi-square test for independence... is for Cramer's V value of about 0.13 the associated p-value is of about 0.001 or it depends on the sample size? Or maybe other factors?
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$\begingroup$ An ideal lower confidence limit for the corresponding population value will be in line with the chi-squared test. Otherwise, there is no relevant relation. $\endgroup$– Michael MCommented Mar 22 at 12:39
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Simply look at how Cramér's V and p-value of the chi-square test of independence are computed.
The p-value of the chi-square test of independence is computed from the $\chi^2$ statistic and the degrees of freedom of the contingency table $(r-1)\cdot(c-1)$. Cramér's V is computed from the $\chi^2$ statistic , the sample size, and the smallest dimension of the contingency table.
So if your question is "If I only know the Cramér's V associated to a contingency table, and only that, is it possible to calculate the precise p-value associated to the chi-squared test of independence conducted on this very table?": the answer is no, it's not possible.
For example, if you take two tables of same dimensions, with the same Cramér's V but with a different sample size, the table with the largest sample size will have a smaller p-value. If the same Cramér's V can be associated to multiple p-values, how would it be possible to compute a p-value from Cramér's V alone?
However, given the same sample size and the same contingency table dimensions, a larger Cramér's V will be associated with a smaller p-value, as Cramér's V is partly function of the $\chi^2$ statistic (which itself is associated with smaller p-values as it increases, given the degrees of freedom remain constant).
