Association Test

Chi-squared test

\begin{aligned} π_{i j} & = P (j^{t h} outcome and in population i) \\ = P (a sample is from population i and expresses j^{t h} outcome) \\ = π_{i .} π_{. j} \end{aligned}

Where

To test the independence between them (rows x columns):

H_{0} : π_{i j} = π_{i .} π_{. j}, \forall i, j v s H_{a} : \exists i, j s.t. π_{i j} \neq π_{i .} π_{. j}

under $H_{0} :$

\begin{aligned} {\hat{π}}_{i j} & = {\hat{π}}_{i \cdot} {\hat{π}}_{\cdot j} \\ = \frac{n_{i \cdot}}{n} \cdot \frac{n_{\cdot j}}{n} \end{aligned}

We reject $H_{0}$ if

χ_{o b s}^{2} = \sum_{i} \sum_{j} \frac{(f_{i j} - E_{i j})^{2}}{E_{i j}} > χ_{d f, 1 - α}^{2}