Chi-Squared Test of Independence

Tests if two Random Variables $(X,Y)$ are independent.

• Partition both random variables into sets (already done for discrete random variables)
• Calculate Joint Class Probabilities
• Calculate Margin Class Probability

If $X,Y$ are independent, then the Joint Class Probability is the same as the product of the Margin Class Probabilities. Thus the Null Hypothesis is: $H_0:p_{ij}=p_{i\bullet }\cdot p_{\bullet j}.$

Given $H_0$ the Test Statistic $T_n={\sum }_{i=1}^r{\sum }_{j=1}^s\frac{{\left(N_{ij}-n\frac{N_{i\bullet }}{n}\frac{N_{\bullet j}}{n}\right)}^2}{n\frac{N_{i\bullet }}{n}\frac{N_{\bullet j}}{n}}$ should be small.

We can simulate the test statistic by drawing with replacement from box with $r\times s$ types of balls with Relative Frequencies $p_{i\bullet }\cdot p_{\bullet j}$ . For large $n$ the distribution is actually independent on the probabilities.

The distribution converges to the Chi-Squared Distribution with $(r-1)\cdot (s-1)$ Degrees of Freedom.

Hence the Acceptance Domain is asymptoically given by $I=\left[0,{\chi }_{(r-1)(s-1);1-\alpha }^2\right]$ Without knowing the probabilities the Class Condition can be checked approximately with $n_{ij}\ge 5$

Another test of independence is the so-called Exact Test of Fisher

Application

One can best perform this test using a Contingency Table.