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Independence on a contingency table or goodness of fit, with every expected count exact.
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The Coddy math course teaches the method itself - you work each step on an interactive board and get told exactly where a move went wrong.
Every chi-square test compares a table of observed counts with the counts some hypothesis predicts. For a test of independence the hypothesis is that the row variable and the column variable have nothing to do with each other, so each expected count is the row total times the column total over the grand total. For a goodness-of-fit test the hypothesis is a set of shares, equal shares for a fair die or 9:3:3:1 for Mendel's peas, scaled to your total. Either way the statistic is the same sum: for every cell, observed minus expected, squared, divided by expected.
That sum is exact. A 2 by 2 table with counts 20, 30, 30, 20 has every expected count equal to 25 and a chi-square of exactly 4; a messier table has expected counts like 350/3, and this page keeps them as 350/3 rather than 116.67, so the statistic you read is the true one and not a sum of five rounded terms. What cannot be exact is the p-value, which is the area under the chi-square curve beyond the statistic; the page draws that tail and labels the number as the area it is.
The table under the answer is the part a student is marked on and the part most calculators skip: each cell's observed count, its expected count and its contribution to the statistic. It also shows where the association lives. In a large table one or two cells often carry most of the chi-square, and the contribution column points at them. Cramér's V, given for contingency tables, turns the statistic into an effect size between 0 and 1 that can be compared across tables of different sizes.
Rows for one variable's categories, columns for the other's. Total each row, each column, and the whole table.
Row total times column total, divided by the grand total. Keep the fractions; they are about to be squared.
Observed minus expected, squared, over expected. Add them all up; the total is the chi-square statistic.
(rows − 1)(columns − 1). The p-value is the right-tail area of the chi-square curve with that many degrees of freedom; compare with alpha, or the statistic with the critical value.
The statistic must exceed these for the result to be significant at the given level.
| Degrees of freedom | α = 0.10 | α = 0.05 | α = 0.01 |
|---|---|---|---|
| 1 | 2.706 | 3.841 | 6.635 |
| 2 | 4.605 | 5.991 | 9.210 |
| 3 | 6.251 | 7.815 | 11.345 |
| 4 | 7.779 | 9.488 | 13.277 |
| 5 | 9.236 | 11.070 | 15.086 |
| 9 | 14.684 | 16.919 | 21.666 |
20, 30 / 30, 20Row totals 50 and 50, column totals 50 and 50, grand total 100, so every expected count is 50 × 50 / 100 = 25. Each cell contributes 25/25 = 1, the chi-square is exactly 4 on one degree of freedom, and the p-value is about 0.046. Significant at 5%, just; Cramér's V is 0.2, a weak association.
10, 12, 8, 11, 9, 10 over six facesSixty rolls, so ten expected per face. The contributions are 0, 0.4, 0.4, 0.1, 0.1 and 0, giving a chi-square of exactly 1 on five degrees of freedom and a p-value of about 0.96. Nothing here suggests the die is loaded.
315, 108, 101, 32 against 9 : 3 : 3 : 1The total is 556, so the expected counts are 556 × 9/16 = 312.75, 104.25, 104.25 and 34.75. The chi-square comes to about 0.47 on three degrees of freedom, p about 0.93: the counts fit the 9:3:3:1 prediction almost suspiciously well, which is a famous argument in its own right.
