Chi-Square Test of Independence: 2x2 contingency table and Phi effect size

Expected counts and the Pearson statistic

In a two-way contingency table, the expected count under independence for Row i, Column j is (RowTotal_i * ColTotal_j) / GrandTotal.

For a 2x2 table [[a, b], [cellC, d]] with grand total N = a + b + cellC + d, Pearson's Chi-Square simplifies algebraically to N * (a*d - b*cellC)^2 / ((a+b) * (cellC+d) * (a+cellC) * (b+d)) with 1 degree of freedom (critical value = 3.841 at alpha = 0.05).

2x2 Chi-Square statistic and Phi association coefficient

SQRT(Chi-square/N) gives the magnitude of Phi, from 0 to 1. Signed Phi also depends on category coding; size labels are context-dependent.

=LET(a, B2, b, C2, cellC, B3, d, C3, n, a+b+cellC+d, ROUND(n * (a*d - b*cellC)^2 / ((a+b)*(cellC+d)*(a+cellC)*(b+d)), 2))

Expected count for top-left cell (Row1 * Col1 / GrandTotal)

Verify all four expected cell counts exceed 5 before relying on asymptotic Chi-Square thresholds.

=(SUM(B2:C2) * SUM(B2:B3)) / SUM(B2:C3)

Magnitude of Phi

Use the unrounded statistic in the calculation.

=LET(a, B2, b, C2, cellC, B3, d, C3, chi, (a+b+cellC+d) * (a*d - b*cellC)^2 / ((a+b)*(cellC+d)*(a+cellC)*(b+d)), ROUND(SQRT(chi / (a+b+cellC+d)), 4))

Inputs and interpretation

Use independent observations, nonnegative integer counts and positive row/column totals. This is the uncorrected Pearson statistic, not an exact test or a continuity-corrected test. Expected-count checks concern the asymptotic inference; small sparse tables need a suitable alternative. Association does not establish causation.

Further reading

• Microsoft Support — CHISQ.TEST function overview