分类数据的推断：卡方

Do Grade 10 and Grade 11 students study in different places? Ask 200 of them and the counts will never split perfectly evenly, even if grade makes no difference at all. The question this unit answers is how uneven the counts must be before chance stops being a good explanation.

When data are counts 频数 spread across several categories 类别, we test whether the observed counts 观测频数 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 卡方分布 is right-skewed 右偏的 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 分类变量 follows a claimed distribution (e.g. "the die is fair"). Hypotheses:

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

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

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 $df=6-1=5$. Write out every category, including the two that match their expected count exactly and so add $0$ – the sum runs over all six categories, and $df$ counts categories, not just the ones that differ. This $\chi^2$ is small (a large $p$-value), so we fail to reject 不拒绝 $H_0$ – no evidence the die is unfair.

Worked example: unequal claimed proportions. A sweet maker claims its bags are $30\%$ red, $20\%$ green and $50\%$ yellow. A random sample of $200$ sweets has $70$ red, $30$ green and $100$ yellow.

Colour	Red	Green	Yellow
Observed	$70$	$30$	$100$
Expected	$200(0.30)=60$	$200(0.20)=40$	$200(0.50)=100$

The $p$-value is about $0.12>0.05$, so fail to reject $H_0$: there is not convincing evidence that the colour distribution differs from the maker's claim. The expected counts come from the claimed percentages, not from dividing $200$ equally by $3$.

For a two-way table 双向表, 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 (no association 关联).

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 同质性检验: are the distributions of one categorical variable the same across several populations or groups (separate samples/treatments)?
• Test for independence 独立性检验: 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).

Worked example: a test for independence. A random sample of 200 students at one school is classified by grade and by where they usually study.

	Library	Home	Café	Total
Grade 10	30	50	20	100
Grade 11	45	35	20	100
Total	75	85	40	200

• State. $H_0$: grade and study place are independent for students at this school. $H_a$: they are associated.
• Plan. Chi-square test for independence: one sample, two variables. Random sample; 200 is less than 10% of the students if the school has more than 2000; expected counts below are all at least 5.
• Do. Expected counts: $\dfrac{100\times75}{200}=37.5$, then $42.5$ and $20$ in each row. The six components are $1.50,\ 1.32,\ 0,\ 1.50,\ 1.32,\ 0$, so $\chi^2=5.65$ with $df=(2-1)(3-1)=2$, giving $p\approx0.059$.
• Conclude. Because $p\approx0.059>\alpha=0.05$, fail to reject $H_0$. There is not convincing evidence of an association between grade and study place for students at this school.

The library cells contribute most: Grade 11 students chose the library more often than expected. That is worth noting as a pattern, but with this $p$-value it is not convincing evidence.

Common mistakes. Checking the condition with observed counts: it is the expected counts that must all be at least $5$. Running chi-square on percentages or means instead of counts. Using $df=n-1=199$ instead of $(r-1)(c-1)=2$. Writing $H_0$ as “grade and study place are associated”: the null hypothesis always says no association. Calling a one-sample, two-variable test “homogeneity”.

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, but only for a two-tailed 双尾 alternative, where they agree exactly ($\chi^2=z^2$). The chi-square statistic is compared with the upper tail of its reference distribution. For two proportions it tests a non-directional alternative, so if $H_a$ is one-sided 单侧 (say $p_1>p_2$), use the $z$-test.

Answer each before reading the answers below.

1. In a two-way table, a cell's row total is 60, its column total is 50 and the grand total is 200. Find its expected count.
2. A two-way table has 3 rows and 4 columns. Find the degrees of freedom for a chi-square test.
3. A die is rolled 60 times, giving counts $6,\ 10,\ 8,\ 12,\ 14,\ 10$ for faces 1 to 6. Find $\chi^2$ for a goodness-of-fit test of a fair die. Its $p$-value is about $0.55$. Conclude.
4. Name the test: (a) one random sample of 500 adults, classified by phone brand and age group; (b) separate random samples from three cities, comparing favourite sport; (c) the colours in a bag of sweets, compared with the maker's claimed percentages.
5. A student says: “In the worked example $p=0.059$, so grade and study place are independent.” Correct the student.

1. $\dfrac{60\times50}{200}=15$.
2. $(3-1)(4-1)=6$.
3. Each expected count is 10. $\chi^2=\dfrac{16+0+4+4+16+0}{10}=4.0$, $df=5$. With $p\approx0.55$, fail to reject $H_0$: no convincing evidence that the die is unfair.
4. (a) Independence; (b) homogeneity; (c) goodness-of-fit.
5. Failing to reject $H_0$ means the data are not convincing evidence of an association. It does not prove independence: a real but small association could still exist. Write “not convincing evidence of an association”, never “they are independent”.

• 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 $(r-1)(c-1)$).