Two Categories Walk into a Contingency Table

Eighty fictional commuters report whether they drive or use a train, and whether they frequently arrive late.

Mode	Late	Not late	Total
Car	18	22	40
Train	10	30	40
Total	28	52	80

The observed late rates are 45% for drivers and 25% for train riders. Are these variables associated?

What independence predicts

Under a model in which the row and column classifications are independent, the expected count in row $i$, column $j$ is

$$E_{ij}=\frac{(\text{row total}_i)(\text{column total}_j)}{n}.$$

For the car-and-late cell, the expected count is $40(28)/80=14$. The other expected counts are 26, 14, and 26.

Pearson’s chi-square statistic sums squared observed-minus-expected differences divided by the corresponding expected counts. Here:

$$X^2=16/14+16/26+16/14+16/26\approx3.516.$$

For a $2\times2$ table, the usual reference degrees of freedom are $(2-1)(2-1)=1$ under appropriate conditions.

Interpretation before ritual

If the test indicates an association, that does not establish that switching a commuter from car to train would reduce lateness. Route distance, schedules, weather, job flexibility, and other factors might differ between the groups.

If the same commuter appears repeatedly, the assumed independent observational units may also be wrong.

When expected counts are sparse, a different procedure, such as Fisher’s exact test for a suitable $2\times2$ reference model, can be preferable.

Ask R the same question

tab <- matrix(c(18,10,22,30), nrow=2,
  dimnames=list(Mode=c("Car","Train"),
                Late=c("Yes","No")))
chisq.test(tab)
prop.table(tab,1)

The row proportions are usually much easier to interpret than a bare p-value.

Check your understanding

1. What is the expected car-and-late count under independence?
2. What is the usual degrees-of-freedom formula for an $r\times c$ table?
3. Why doesn’t an association imply that one classification caused the other?

Worked responses

2. $(r-1)(c-1)$.

3. Confounding and selection, among other causes, may create or distort the association.

Derive the expected table and its constraints

Under independence, $P(R=i,C=j)=P(R=i)P(C=j)$. Estimating the marginal probabilities by row and column proportions gives $E_{ij}=R_iC_j/n$. These expected counts preserve the observed row and column totals. For an $r$ by $c$ table with positive relevant margins, fitting the margins leaves $(r-1)(c-1)$ independent directions of discrepancy, the usual asymptotic degrees of freedom.

For the synthetic table $[[18,22],[10,30]]$, row totals are 40 each, column totals 28 and 52, and expected counts are $[[14,26],[14,26]]$. Pearson’s statistic is

$$X^2=2\left(\frac{16}{14}+\frac{16}{26}\right)\approx3.51648.$$

The uncorrected one-degree-of-freedom p-value is approximately $0.06076$. R’s default for a two-by-two table may apply Yates’s continuity correction, so specify correct=FALSE to reproduce this exact uncorrected calculation. Neither answer licenses hiding the correction choice.

A measure of association is Cramér’s $V=\sqrt{X^2/[n\min(r-1,c-1)]}$. Here $V\approx0.20966$. The first-row event proportion is $18/40=0.45$ and the second is $10/40=0.25$, a risk difference of $0.20$. Those directional, interpretable rates convey information that a nonnegative global statistic does not.

A contingency table of independent people is not interchangeable with paired measurements on the same people or clustered respondents. For sparse two-by-two tables, Fisher’s exact test conditions on margins and tests a corresponding odds-ratio null under its model. “Exact” refers to the finite reference calculation; it does not guarantee the data-collection design is appropriate.

Additional practice

1. If a three-by-four table meets the test assumptions, how many degrees of freedom does the independence statistic use?
2. Show that expected counts for each row sum back to its observed row total.
3. Why would repeated visits by the same person undermine the ordinary independent-count reference?

Worked additional solutions

1. $(3-1)(4-1)=6$.
2. $\sum_jR_iC_j/n=R_i(\sum_jC_j)/n=R_i$.
3. Visits share person-level characteristics and are not independent units. The reference variance and effective information may differ from the naive count.