Carrying Out a Chi-Square Test for Homogeneity or Independence

The statistic is $\chi^2 = \sum \frac{(\text{observed} - \text{expected})^2}{\text{expected}}$, one non-negative contribution per cell, scaled so a fixed gap matters more where less was expected. For the three-school table the six contributions are about 3.84, 3.87, 1.07, 1.08, 1.14, and 1.15, totaling 12.15 with $df = (3-1)(2-1) = 2$ and an upper-tail p-value near 0.0023. The tail is always the upper one, because only a large statistic indicates disagreement with the null.

A correct calculation then gets over-read. The test is taken to identify which cell differs, though the alternative says only that the distributions differ somewhere; it is taken as causal, though that depends on random assignment; and a large p-value is taken as proof of independence, which accepts the null. Squaring also erases direction, so the statistic cannot say which group runs high. The follow-up supplies what the test cannot: the largest components point to the cells contributing most, and the conditional distributions, 60% against 45% and 44%, give the direction and the size.