7.10 Chi-Square Goodness-of-Fit and Tests of Independence

The common statistic

All chi-square count tests use the same statistic:

$\chi^2 = \sum_{\text{all cells}} \frac{(O_i - E_i)^2}{E_i}$

where $O$ is the observed count and $E$ the expected count under the null hypothesis. Large values mean observed counts are far from expectation, so the test is always right-tailed regardless of the direction of any difference.

Three applications appear in the Body of Knowledge and on the exam.

Independence and homogeneity use identical arithmetic and differ only in how the data was sampled: one sample cross-classified two ways gives independence; several fixed-size samples from different populations gives homogeneity.

Goodness of fit

Tests whether observed frequencies match those expected from a hypothesized distribution.

Worked example. A team believes defects are equally likely across four shifts. Observed counts over a month: 52, 38, 61, 49, total 200. Under equal likelihood each expected count is $200/4 = 50$.

Degrees of freedom: $k - 1 = 3$ (no parameters were estimated, because the equal-proportions hypothesis was specified in advance). $\chi^2_{0.05,3} = 7.815$. Since $5.40 < 7.815$, do not reject: the data are consistent with defects being uniformly distributed across shifts.

Note the practical reading. Shift 3 has 61 defects against 49 on shift 4, a difference that looks substantial, but with these counts it is within what random variation produces. Acting on the apparent shift-3 problem would be chasing noise.

Degrees of freedom when parameters are estimated

Each parameter estimated from the same data costs one degree of freedom:

Forgetting this correction inflates the degrees of freedom, understates the critical value, and makes rejection too easy.

Chi-square versus other normality tests

For testing normality specifically, the chi-square goodness-of-fit test is available but weak: it requires binning continuous data, and the result depends on the arbitrary bin choice. The Anderson-Darling, Shapiro-Wilk, and Kolmogorov-Smirnov tests work on the raw data and have substantially more power. Chi-square goodness of fit is the right tool for genuinely categorical data and for count distributions such as the Poisson.

Test of independence

Tests whether two categorical variables are associated.

Worked example. Defect type by supplier:

Expected count for each cell:

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

$E_{11} = (60)(60)/200 = 18$, $E_{12} = (60)(80)/200 = 24$, $E_{13} = 18$, and so on.

$\chi^2 = \frac{(30-18)^2}{18} + \frac{(20-24)^2}{24} + \frac{(10-18)^2}{18} + \frac{(20-24)^2}{24} + \frac{(40-32)^2}{32} + \frac{(20-24)^2}{24} + \frac{(10-18)^2}{18} + \frac{(20-24)^2}{24} + \frac{(30-18)^2}{18}$

$= 8.00 + 0.67 + 3.56 + 0.67 + 2.00 + 0.67 + 3.56 + 0.67 + 8.00 = 27.78$

Degrees of freedom $= (3-1)(3-1) = 4$, and $\chi^2_{0.05,4} = 9.488$. Since $27.78 > 9.488$, reject independence: defect type is associated with supplier.

Reading the cell contributions is where the practical value lies. The largest contributions come from Supplier A / dimensional (8.00) and Supplier C / assembly (8.00), so those are the specific supplier-defect combinations driving the association and the ones to investigate.

Assumptions and limits

When expected counts are too small, either collapse adjacent categories into meaningful groups or use Fisher's exact test, which computes the exact probability without a large-sample approximation and is the standard choice for 2x2 tables with small counts.

Interpretation cautions

• Association is not causation. A significant result says defect type and supplier are related; it does not say the supplier causes the defect. Both could depend on the material grade each supplier happens to be assigned.
• A significant result does not identify which cells matter. Inspect the individual $(O-E)^2/E$ contributions, or standardized residuals $(O-E)/\sqrt{E}$, to find them.
• Large samples make trivial associations significant. Chi-square grows in proportion to sample size, so pair the test with an effect size measure such as Cramer's V:

$V = \sqrt{\frac{\chi^2}{n \cdot \min(r-1,\, c-1)}}$

For the example, $V = \sqrt{27.78 / (200 \times 2)} = \sqrt{0.0695} = 0.264$, a moderate association.