The Chi-Squared Test

Every test so far has concerned a mean. The chi-squared test concerns counts: it asks whether the frequencies you observed differ from what a hypothesis predicts by more than chance would explain.

The idea of a chi-squared test

Compare observed frequencies $O$ with the expected frequencies $E$ that $H_0$ predicts. If they are close, $H_0$ survives; if they diverge, it does not.

Two standard uses:

• Goodness of fit — do the data follow a stated distribution? $H_0$: they do.
• Independence (a contingency table) — are two categorical variables associated? $H_0$: they are independent.

Expected frequencies

For goodness of fit, $E$ comes from the hypothesised distribution. If a fair die is rolled 60 times, each face is expected $\frac{60}{6} = 10$ times.

For a contingency table:

$$E = \frac{\text{row total} \times \text{column total}}{\text{grand total}}$$

$$E(\text{A, Pass}) = \frac{60 \times 70}{120} = 35 \qquad E(\text{A, Fail}) = \frac{60 \times 50}{120} = 25$$

and by the same calculation $E(\text{B, Pass}) = 35$, $E(\text{B, Fail}) = 25$.

The test statistic

$$\chi^2 = \sum \frac{(O-E)^2}{E}$$

For the table above:

$$\frac{(40-35)^2}{35} + \frac{(20-25)^2}{25} + \frac{(30-35)^2}{35} + \frac{(30-25)^2}{25}$$ $$= 0.714 + 1 + 0.714 + 1 = 3.43$$

Each term measures a squared discrepancy relative to what was expected: being 5 out of 25 matters more than being 5 out of 35.

Degrees of freedom

Goodness of fit:

$$\nu = (\text{number of categories}) - 1 - (\text{parameters estimated from the data})$$

Six die faces with nothing estimated gives $\nu = 5$.

Contingency table:

$$\nu = (r-1)(c-1)$$

A $2 \times 2$ table gives $\nu = 1$; a $3 \times 4$ table gives $\nu = 2 \times 3 = 6$.

Some critical values at the 5% level:

Contingency tables

Test at the 5% level whether pass rate is associated with teaching method, for the table above.

1. Hypotheses. $H_0$: method and outcome are independent. $H_1$: they are associated.

2. Expected frequencies. $35, 25, 35, 25$ as computed.

3. Statistic. $\chi^2 = 3.43$.

4. Degrees of freedom and critical value. $\nu = (2-1)(2-1) = 1$, so $\chi^2_{0.05} = 3.841$.

5. Compare and conclude. $3.43 < 3.841$, so do not reject $H_0$: at the 5% level there is insufficient evidence of an association between teaching method and pass rate.

Conditions and interpretation

All expected frequencies must be at least 5. Small expected values make the $\frac{(O-E)^2}{E}$ term unstable, and the chi-squared approximation breaks down.

If some are too small, combine adjacent categories until every expected value reaches 5 — and recalculate the degrees of freedom for the reduced number of categories.

Other requirements: the data must be frequencies from a random sample, and the observations independent — each item counted in exactly one cell.