Chi-Square Test of Independence

Background

The chi-square test of independence asks whether two categorical variables (e.g. arm vs converted) are statistically independent. It compares the observed contingency-table counts to those expected under independence; large discrepancies imply dependence.

Problem statement

Implement chi_square_independence(observed) returning (chi2, df) where:

$$E_{ij} \;=\; \frac{(\text{row total } i) \cdot (\text{col total } j)}{\text{grand total}},\qquad \chi^2 \;=\; \sum_{i, j}\frac{(O_{ij} - E_{ij})^2}{E_{ij}},\qquad \mathrm{df} = (R - 1)(C - 1)$$

Input

• observed — np.ndarray shape $(R, C)$ of non-negative integer counts.

Output

Returns a tuple (chi2: float, df: int). chi2 >= 0; df = (R-1)(C-1).

Examples

Example 1 — independent counts

Input:  [[10, 20], [10, 20]]
Output: (~0.0, 1)

Explanation: rows are scaled copies of each other → exactly independent → $\chi^2 \approx 0$.

Example 2 — strong dependence

Input:  [[100, 0], [0, 100]]
Output: large chi2, df = 1

Example 3 — three-by-three table

Input:  [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
Output: small chi2, df = 4

Constraints

• Cells with $E_{ij} = 0$ are skipped (otherwise dividing by zero).
• Returns plain Python float and int.
• For small expected counts ($E < 5$), the chi-square approximation breaks; this problem does not enforce that — caller's responsibility.

Notes

• Yates correction. For 2×2 tables some statisticians subtract 0.5 from $|O - E|$ before squaring (Yates' continuity correction). Out of scope here.
• Reads with scipy. Combine with scipy.stats.chi2.sf(stat, df) to get a $p$-value.