1
$\begingroup$

I have a categorical variable with binary outcomes 0,1 (say Y), I want to study the association with other variables X1,X2,X3,X4,X5..... most of Xi's are categorical and few of them are continuous. I used Chisquare test to study the same, which tell wheather I can consider two variables independent or not.

Further to rank the Xi's in order of association with Y . Now my question is can I use the pvalue from the ChiSquare test or standardized residual of the chi Square Test to measure the degree of association between the variables ?

As far as I could understand chi sqaure test can be used just to tell whether the variable are independent or not but not the degree of association. Is there any other test to do so or chi square can do that?

$\endgroup$
3
  • 1
    $\begingroup$ Are you familiar with some of the many measures of association, such as phi coefficient and Spearman rank-order correlation coefficient? If not, look into them. $\endgroup$
    – Joel W.
    Jun 10, 2014 at 0:22
  • $\begingroup$ @JoelW.Thanks for the help. Thanks for the help, just have question that Pearson's chi-squared test for independence i.e the statistics : $$ \sum\limits_{i=1}^n \frac{(Oi-Ei)^2}{Ei} $$ won't measure the degree of association.Right ? $\endgroup$
    – Vijay
    Jun 10, 2014 at 5:49
  • 1
    $\begingroup$ Yes, the value of Chi-sq will reflect the degree of association, but it is not on a scale that is usually used to express the degree of association. Typically, correlation coefficients are used for that purpose. They have a range of -1 to +1 (nominally) and zero indicates no correlation. $\endgroup$
    – Joel W.
    Jun 10, 2014 at 10:27

1 Answer 1

2
$\begingroup$

There are several statistics related to the chi-square that measure association in contingency tables.

For example, there's Cramer's $\phi$ (also $\phi_C$ or Cramer's $V$), which in 2x2 tables is also called the phi coefficient.

For a $r\times c$ table,

$V = \sqrt{ \frac{\chi^2/n}{\min(c - 1,r-1)}}$

In your case, if one of your variables is Y, which you state to be 0-1, then that will reduce to:

$V = \sqrt{ \chi^2/n}$

where $n$ is the total number of observations.

Wikipedia:

Cramer's Phi/Cramer's V

Phi coefficient

There are a number of other ways of measuring association in contingency tables.

$\endgroup$
2
  • $\begingroup$ Thanks for the help, just have question that Pearson's chi-squared test for independence i.e the statistics : $$ \sum\limits_{i=1}^n \frac{(Oi-Ei)^2}{Ei} $$ won't measure the degree of association.Right ? $\endgroup$
    – Vijay
    Jun 10, 2014 at 5:46
  • 1
    $\begingroup$ Well, it sort of depends on what you need it to do -- they're directly a measure of squared deviation from independence; for a given set of margins they get bigger the more the variables become associated -- the difficulty is they're hard to compare across different variables. In the 2x2 table, by comparison, $\phi$ is directly a correlation. $\endgroup$
    – Glen_b
    Jun 10, 2014 at 6:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.