In a chi-square test of independence on a 3x4 contingency table, how do you get expected counts and degrees of freedom?

## The question a test of independence asks Given a cross-tabulation of two categorical variables, is the distribution of one variable the same regardless of the level of the other? Take a survey of N respondents cross-tabbed as **job role** (engineer / analyst / manager — 3 rows) by **preferred work mode** (fully remote / hybrid-two-days / hybrid-four-days / fully on-site — 4 columns). That is a 3x4 contingency table with 12 cells. The null hypothesis is independence: knowing someone's role tells you nothing about their work-mode preference. ## Expected counts under independence Probability theory does the work. If two events are independent, P(A and B) = P(A) * P(B). Estimating the marginals from the data, the probability of row i is row_i_total / N and the probability of column j is col_j_total / N. So under the null the probability of cell (i, j) is their product, and the expected number of people there is E_ij = N * (row_i_total / N) * (col_j_total / N) = (row_i_total * col_j_total) / N Hence the familiar rule: **row total times column total over grand total**. Worked concretely, if 240 of 600 respondents are engineers and 150 of 600 prefer fully remote, the engineer/fully-remote cell expects 240 * 150 / 600 = 60 people. Expected counts are usually fractional and should be left that way — never round them before computing the statistic. By construction the expected counts reproduce both sets of margins exactly. That is the whole point: the test holds the marginal distributions fixed at their observed values and asks only whether the *interior* of the table is arranged as independence would predict. ## The statistic chi-square = sum over all cells of (O_ij - E_ij)^2 / E_ij summed over every one of the 12 cells. Each term is a squared deviation scaled by the noise you expect in a cell of that size. Larger totals mean larger absolute deviations are normal, and the division by E accounts for it. ## Where (r-1)(c-1) comes from Degrees of freedom count how many cell deviations are genuinely free once the constraints are imposed. There are r * c cells. The margins are not free — they were used to build the expected counts — so: - fixing the row totals costs r constraints, - fixing the column totals costs c constraints, - but one of those is redundant, since both sets must sum to the same N. That leaves r*c - (r + c - 1) = (r-1)(c-1). For the 3x4 table: 12 - (3 + 4 - 1) = 12 - 6 = **6**, matching (3-1)(4-1) = 6. A hands-on way to see it: in a 3x4 table with all margins known, you can fill any 2x3 block of cells freely, and every remaining cell is then forced by subtraction. Six free cells, six degrees of freedom. ## Assumptions and health checks - **Counts, one cell per subject.** Each respondent contributes exactly one tally. If people can pick two work modes, the totals exceed N and the test is invalid. - **Independent observations.** Surveying five people per team, or the same person twice, breaks the sampling assumption; the test has no idea and will happily return a small p-value. - **Adequate expected counts.** The chi-square reference distribution is asymptotic. Sparse expectations degrade it, and the usual remedy is to merge substantively similar categories — for example collapsing the two hybrid columns into one — or to use an exact method. ## Reading the outcome A significant result says only "the variables are associated in this sample". It does not name a direction, a magnitude, or a cause, and it is symmetric — the test cannot tell you whether role shapes preference or the reverse. To describe the association, go back to the table: compare row percentages across columns, and inspect standardised residuals `(O - E)/sqrt(E)` to see which cells over- and under-shoot. Interpret those descriptively, because you are now scanning twelve cells. Also note what independence testing and goodness of fit share and where they differ. Both use the same statistic and the same reference family; goodness of fit compares one variable's counts against *externally specified* probabilities with df = k - 1, while independence compares a two-way table against margin-derived expectations with df = (r-1)(c-1). Candidates who memorise a single df formula for "the chi-square test" get caught here. Finally, sample size cuts both ways. With N in the millions, an association far too weak to act on will still reject; with a few dozen respondents, a strong association can go undetected. Always report the table alongside the p-value.