test for independence / homogeneity (df = 1)
chi-square = 10.5263, p = 0.0012
Smallest expected count: 23.75 (all at least 5, condition met).
Steps
- 1.row totals: 70, 50; column totals: 63, 57; grand total: 120
- 2.expected count = (row total x column total) / grand total for every cell
- 3.expected counts: 36.75, 33.25 | 26.25, 23.75
- 4.chi-square = sum of (observed - expected)² / expected = 10.5263
- 5.df = (rows - 1)(columns - 1) = 1; right-tail p-value = 0.0012