Inference for Categorical Data: Chi-Square

When data are counts spread across several categories, we test whether the observed counts differ from what a claim predicts. The tool is the chi-square 卡方 ($\chi^2$) statistic, which adds up the standardized gaps between observed and expected counts:

A large $\chi^2$ means the observed counts are far from expected – evidence against the claim. The chi-square distribution is right-skewed and depends on its degrees of freedom 自由度.

A goodness-of-fit (GOF) 拟合优度 test checks whether one categorical variable follows a claimed distribution (e.g. "the die is fair"). Hypotheses:

Expected count for each category $=n\times(\text{claimed proportion})$. Conditions: random sample, all expected counts $\ge 5$, and the 10% condition.

The chi-square distribution is right-skewed. A large statistic lands in the shaded right tail past the critical value – that is where you reject the model.

Compute $\chi^2=\sum\dfrac{(O-E)^2}{E}$ with $df=(\text{number of categories})-1$. Find the $p$-value from the chi-square distribution (upper tail), compare to $\alpha$, and conclude in context. A large component of the sum points to the category that deviates most.

Chi-square compares observed counts with those expected under the null hypothesis

Worked example. A die rolled $60$ times gives counts $8,10,12,9,11,10$. If it is fair, each expected count is $60/6=10$, so

with $df=5$. This is small (a large $p$-value), so we fail to reject $H_0$ – no evidence the die is unfair.

For a two-way table, the expected count in a cell (under "no association") is

This is the count you would see if the row and column variables were unrelated.

Worked example. In a two-way table a cell's row total is $40$, its column total is $50$, and the grand total is $200$. Its expected count is $E=\dfrac{40\times50}{200}=10$. Repeating for every cell gives the expected table to compare against the observed one.

Two tests use the same $\chi^2$ math but answer different questions:

• Test for homogeneity 同质性: are the distributions of one categorical variable the same across several populations or groups (separate samples/treatments)?
• Test for independence 独立性: are two categorical variables associated within a single population (one sample, two variables measured)?

The design (several samples vs one sample) decides which name and hypotheses to use.

Compute expected counts, then $\chi^2=\sum\dfrac{(O-E)^2}{E}$ over all cells, with

Conditions: random data, all expected counts $\ge 5$, 10% condition. Find the $p$-value, compare to $\alpha$, and conclude in context – evidence of a difference between groups (homogeneity) or of an association (independence).

Decide by the setup: one categorical variable against a claimed distribution $\Rightarrow$ goodness-of-fit; one sample cross-classified by two variables $\Rightarrow$ independence; several samples/groups compared $\Rightarrow$ homogeneity. Comparing just two proportions can use either a two-proportion $z$-test or a chi-square test – they agree.

• Use $\chi^2=\sum\tfrac{(O-E)^2}{E}$ for categorical data; always divide by the expected count.
• Pick the right test: goodness-of-fit (one variable), independence, or homogeneity (two-way table).
• Compute expected counts as $\tfrac{\text{row total}\times\text{column total}}{\text{grand total}}$ and check each is $\ge5$.
• A large $\chi^2$ (small p-value) means observed counts differ from expected by more than chance.
• State degrees of freedom correctly (categories $-1$, or $(r-1)(c-1)$).