Discussion
In an \(r \times s\) contingency table for testing H0 : Two attributes A and B are independent against the alternative H1 : A and B are not independent, the expected frequency \(e_{ij}\) of (i − j)th cell is given by :
Answer: C
The expected frequency is ni. n.j / n.
Under independence, \(P(A_{i} \cap B_{j}) = P(A_{i})P(B_{j})\), estimated by \(\frac{n_{i.}}{n} \cdot \frac{n_{.j}}{n}\). Multiplying by the total n gives the expected count
\[ \begin{aligned} e_{ij} &= n \cdot \frac{n_{i.}}{n} \cdot \frac{n_{.j}}{n} \\ &= \frac{n_{i.}\, n_{.j}}{n} \end{aligned} \]
The product \(n_{i.}n_{.j}\) without dividing is far too large, dividing by \(n^{2}\) gives the estimated probability rather than the expected count, and \(n_{ij}/n\) is the observed proportion, not an expected frequency.
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