In many classification applications, for example, exposure-disease association, we can summarize data in the following $2 \times 2$ table:
| Event: $D$ | No Event: $\bar D$ | ||
|---|---|---|---|
| Exposure: $E$ | $a$ | $b$ | $n_B$ |
| No Exposure: $\bar E$ | $c$ | $d$ | $n_A$ |
| $a+c$ | $b+d$ | $n$ |
The hypothesis to verify: the exposure $E$ and the event $D$ are independent? Actually, we would like to disprove this hypothesis, then we can claim the exposure is associated with the event.
Let $p_B = Pr(D | E)$ be the conditional probability of event, $D$, given the exposure group, $E$, and $p_A = Pr(D | \bar E)$ the conditional probability of event, $D$, given the non-exposure group, A.
The interested hypothesis can be represented as
$$ H_0: p_B = p_A, vs. H_1: p_B \neq p_A $$Chi-squared test statistics under $H_0$:
$$ X_p = \frac{(\widehat {p}_B - \widehat {p}_A)^2}{\widehat{p} (1 - \widehat{p}) \left ( \frac{1}{n_B} + \frac{1}{n_A} \right )} = \frac{n (ad - bc)^2}{(a + b) (a + c) (b + d) (c + d)} = \sum_{i=1}^2 \sum_{j=1}^2 \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \rightarrow^d \chi^2_{(1)} \quad \text{(1)} $$where
By applying (1) we can obtain the corresponding testing procedure or confidence interval.
Let $p_B = \Pr(D | E)$, and $p_A = \Pr(D | \bar E)$,
$$ \widehat p_B = \frac{a}{a+b} = \frac{a}{n_B}, \widehat p_A = \frac{c}{c+d} = \frac{c}{n_A} $$Under $H_0: p_B = p_A = p$,
$$ E(\widehat p_B - \widehat p_A) = 0, Var(\widehat p_B - \widehat p_A) = \frac{1}{n_B} p_B (1 - p_B) + \frac{1}{n_A} p_A (1 - p_A) = p (1 - p) \left [ \frac{1}{n_B} + \frac{1}{n_A} \right ] := \widehat V $$From the Central Limit Theorem,
$$ \frac{\widehat p_B - \widehat p_A}{\sqrt{\widehat V}} = \frac{\widehat p_B - \widehat p_A)}{\sqrt{\widehat p (1 - \widehat p) \left [ \frac{1}{n_B} + \frac{1}{n_A} \right ]}} \rightarrow^d N(0, 1) $$Thus,
$$ \frac{(\widehat p_B - \widehat p_A)^2}{\widehat p (1 - \widehat p) \left [ \frac{1}{n_B} + \frac{1}{n_A} \right ]} \rightarrow^d \chi^2_{(1)} $$Numerator:
$$ (\widehat p_B - \widehat p_A)^2 = \left ( \frac{a}{n_B} - \frac{c}{n_A} \right )^2 = \frac{(ad - bc)^2}{(n_B n_A)^2} $$Denominator:
$$ \widehat p (1 - \widehat p) \left [ \frac{1}{n_B} + \frac{1}{n_A} \right ] = \frac{a+c}{n} \frac{b+d}{n} \frac{n_B + n_A}{n_B n_A} = \frac{(a+c) (b+d)}{n n_B n_A} $$Hence the first identity in equation (1) follows. Alternatively, first focus on cell $O_{11} = a$. Under $H_0:$ exposure and event are independent,
$$ E_{11} = E(O_{11}) = E(a) = n \Pr(E \cdot D) = n \Pr(E) \Pr(D) = n \frac{a+c}{n} \frac{a+b}{n} $$Similar results can be derived for cells $b$, $c$ and $d$. Hence the second identity in equation (1) follows.