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 need a bit help for interpreting this results...

I did a correlation analysis with R with the assocstats function. The result is:

$summary
Call: xtabs(formula = ~MH[, i] + MH[, j], data = MH)
Number of cases in table: 2306 
Number of factors: 2 
Test for independence of all factors:
    Chisq = 2806.6, df = 3318, p-value = 1
    Chi-squared approximation may be incorrect

$object
                    X^2   df P(> X^2)
Likelihood Ratio 1036.1 3318        1
Pearson          2806.6 3318        1

Phi-Coefficient   : 1.103 
Contingency Coeff.: 0.741 
Cramer's V        : 0.637 

$summary
Call: xtabs(formula = ~MH[, i] + MH[, j], data = MH)
Number of cases in table: 2343 
Number of factors: 2 
Test for independence of all factors:
    Chisq = 118.46, df = 73, p-value = 0.000611
    Chi-squared approximation may be incorrect

$object
                    X^2 df   P(> X^2)
Likelihood Ratio 130.83 73 3.8115e-05
Pearson          118.46 73 6.1100e-04

Phi-Coefficient   : 0.225 
Contingency Coeff.: 0.219 
Cramer's V        : 0.225 

But what does the p-value of 1 in this case mean if Cramer's V is 0.637 and therefor show a strong corelation or with 0.000611 with a Cramer's V of 0.225?

And does anyone of you know a good rule of thumb for interpreting Cramer's V?

Thanks for your help.

share|improve this question
add comment

1 Answer 1

Look at the number of df. 3318 df in a crosstabs is problematic. It's especially problematic when n is 2306. But I can't think of a good reason to have a crosstabs table with so many cells. All of the results are, essentially, meaningless. If you describe what you are trying to do and why you have so many cells, and what your research question is, and so on, then perhaps someone (maybe even I) will be able to help

share|improve this answer
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.