where:

• Observed = actual count in each cell of the contingency table
• Expected = count predicted under the null hypothesis

The test statistic follows a chi-square distribution with degrees of freedom depending on the context.

Test for Independence

Purpose: Test whether two categorical variables in a single population are independent.

Hypotheses:

• $H_0$: The two variables are independent (no association)
• $H_a$: The two variables are associated (not independent)

Expected Cell Counts (under $H_0$ of independence): $E_{ij} = \frac{(\text{Row total}_i) \times (\text{Column total}_j)}{\text{Grand total}}$

Degrees of Freedom: $df = (r - 1)(c - 1)$ where $r$ = number of rows, $c$ = number of columns in the contingency table.

Test for Homogeneity

Purpose: Test whether the distribution of a categorical variable is the same across multiple independent groups (samples).

Hypotheses:

• $H_0$: The distribution of the variable is the same in all groups (homogeneous)
• $H_a$: The distribution of the variable differs across groups (not homogeneous)

Expected Cell Counts (under $H_0$ of homogeneity): $E_{ij} = \frac{(\text{Row total}_i) \times (\text{Column total}_j)}{\text{Grand total}}$

Note: Formula is identical to the independence test, but interpretation differs.

Degrees of Freedom: $df = (r - 1)(c - 1)$ where $r$ = number of categories in the variable, $c$ = number of groups/samples.

Conditions for Chi-Square Tests

Worked Example 1: Test for Independence

A study of 200 college students asks about Gender (M/F) and Exercise Habit (Regular/Irregular). The data:

Test $H_0$: Gender and Exercise are independent, at $\alpha = 0.05$.

Calculate expected counts:

• $E_{M,R} = \frac{70 \times 100}{200} = 35$
• $E_{M,I} = \frac{70 \times 100}{200} = 35$
• $E_{F,R} = \frac{130 \times 100}{200} = 65$
• $E_{F,I} = \frac{130 \times 100}{200} = 65$

All expected counts $\geq 5$ ✓

Calculate $\chi^2$: $\chi^2 = \frac{(40-35)^2}{35} + \frac{(30-35)^2}{35} + \frac{(60-65)^2}{65} + \frac{(70-65)^2}{65}$ $= \frac{25}{35} + \frac{25}{35} + \frac{25}{65} + \frac{25}{65} = 0.714 + 0.714 + 0.385 + 0.385 = 2.198$

Degrees of freedom: $df = (2-1)(2-1) = 1$

p-value: For $\chi^2 = 2.198$ with $df = 1$, $P(\chi^2 > 2.198) \approx 0.138$.

Since $p\text{-value} \approx 0.138 > 0.05$, fail to reject $H_0$. No significant evidence that gender and exercise are associated.

Worked Example 2: Test for Homogeneity

Compare the distribution of Political Party (Democrat/Republican/Independent) across three Age Groups (18–30, 31–50, 50+). Sample of 600 people:

Test $H_0$: Political party distribution is the same across age groups, at $\alpha = 0.05$.

Calculate expected counts (sample):

• $E_{D,1} = \frac{240 \times 150}{600} = 60$
• $E_{D,2} = \frac{240 \times 220}{600} = 88$
• ... (all $\geq 5$) ✓

Calculate $\chi^2$ (simplified; full calculation omitted): $\chi^2 \approx 15.38$

Degrees of freedom: $df = (3-1)(3-1) = 4$

p-value: For $\chi^2 \approx 15.38$ with $df = 4$, $P(\chi^2 > 15.38) \approx 0.0036$.

Since $p\text{-value} \approx 0.0036 < 0.05$, reject $H_0$. Significant evidence that political party distribution differs across age groups.

Common Pitfalls

⚠️ Using Observed Instead of Expected Counts: The chi-square statistic compares observed counts to expected counts under $H_0$. Always compute expected counts carefully; a mistake here invalidates the entire test.

⚠️ Ignoring the Large Counts Condition: If expected counts are too small (typically $< 5$), the chi-square distribution is not a good approximation. Combine categories or use Fisher's exact test if appropriate.

⚠️ Confusing Independence and Homogeneity: Both use the same $\chi^2$ statistic and formula, but test different hypotheses. Independence: one sample, two variables. Homogeneity: multiple samples, one variable.

Calculator Tip

💡 TI-84 / TI-Nspire: Use $\chi^2$ Goodness-of-Fit or $\chi^2$ Test. Enter the observed counts in a matrix, then calculate. The calculator computes expected counts, $\chi^2$, df, and p-value automatically. (Some calculators use "Chi2 Test" for homogeneity/independence and "Chi2 GOF" for goodness-of-fit.)

Expected count: $E_{1,2} = \frac{(\text{Row 1 Total}) \times (\text{Column 2 Total})}{\text{Grand Total}} = \frac{100 \times 110}{250} = \frac{11000}{250} = 44$

Answer: Expected count = 44.

All $\geq 5$ ✓

Chi-square: $\chi^2 = \frac{(30-25)^2}{25} + \frac{(20-25)^2}{25} + \frac{(25-32.5)^2}{32.5} + \frac{(40-32.5)^2}{32.5}$ $= \frac{25}{25} + \frac{25}{25} + \frac{56.25}{32.5} + \frac{56.25}{32.5} = 1 + 1 + 1.73 + 1.73 = 5.46$

Degrees of freedom: $df = (2-1)(2-1) = 1$

p-value: $P(\chi^2 > 5.46 \mid df=1) \approx 0.019$

Since $0.019 < 0.05$, reject $H_0$. Significant evidence that Size and Color are associated.

Decision: Since the observed $\chi^2 = 8.45 > 7.815$, we reject $H_0$.

Alternatively, using p-value: $P(\chi^2 > 8.45 \mid df=3) \approx 0.038$

Since $p\text{-value} \approx 0.038 < 0.05$, we reject $H_0$.

Conclusion: There is significant evidence that the distribution of preferences (Yes/No) differs across the four groups. At least one group has a different preference distribution than the others.