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_{i j}=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_{i j}-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}\geq 5

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

Application

One can best perform this test using a Contingency Table.