Chi-Square Tests for Independence and Homogeneity

📊 The Chi-Square Statistic and Its Distribution

Part 1 of 7 — Measuring How Far Counts Are From Expected

Where Chi-Square Tests Fit

Chi-square tests answer questions about two-way tables of categorical data. On the AP exam there are two of them:

Both tests use the same statistic, the same expected counts and the same degrees of freedom. This part builds those tools; Parts 2 and 3 cover each test.

ℹ️ The goodness-of-fit test (one categorical variable against a claimed distribution) is not part of the current AP Statistics course.

Expected Counts

If the null hypothesis is true, each cell of a two-way table has an expected count:

$\boxed{\text{expected count} = \frac{\text{row total} \times \text{column total}}{\text{table total}}}$

Example: 200 students (100 in Grade 10, 100 in Grade 12) report how they get to school.

Expected count for Grade 10, Bus: $\frac{100 \times 50}{200} = 25$. The full set of expected counts is 25, 45, 30 in each row.

The Chi-Square Statistic

$\chi^2 = \sum \frac{(\text{observed} - \text{expected})^2}{\text{expected}}$

The sum runs over every cell of the table. Each term measures the distance between an observed and an expected count relative to the expected count, so a gap of 7 matters more in a cell that expects 25 than in a cell that expects 250.

For the example:

$\chi^2 = \frac{(32-25)^2}{25} + \frac{(40-45)^2}{45} + \frac{(28-30)^2}{30} + \frac{(18-25)^2}{25} + \frac{(50-45)^2}{45} + \frac{(32-30)^2}{30}$

$= 1.960 + 0.556 + 0.133 + 1.960 + 0.556 + 0.133 \approx 5.30$

🔑 $\chi^2 = 0$ only when every observed count equals its expected count. The farther the counts are from expected, the larger $\chi^2$ gets.

The Statistic Check 🎯

📈 Chi-Square Distributions

When the null hypothesis is true, the chi-square statistic follows a chi-square distribution with

$df = (\text{number of rows} - 1)(\text{number of columns} - 1)$

For the transport example, $df = (2-1)(3-1) = 2$.

Shape of the family

Why the p-value is a right-tail area

Only large values of $\chi^2$ are evidence against $H_0$: they mean the observed counts sit far from what $H_0$ predicts. So the p-value is always the area to the right of the observed $\chi^2$ under the chi-square curve with the correct $df$.

For the example, $\chi^2 \approx 5.30$ with $df = 2$ gives a right-tail area of about 0.07 (from a table or technology). Parts 4 and 5 show how to read the table.

Distribution Check 📈

Chi-Square Calculation 🧮

A two-way table has row totals 60 and 90, column totals 50, 70 and 30, and a table total of 150.

1) Expected count for row 1, column 2 = ?

2) The observed count in that cell is 35. What is its contribution $(O-E)^2/E$ to $\chi^2$?

3) Degrees of freedom for this table = ?

📊 Chi-Square Test for Independence

Part 2 of 7 — Are Two Categorical Variables Related?

Topics in This Part

🔑 Key Concept: The chi-square test for independence uses data from one sample to determine whether two categorical variables are associated (related) or independent.

When to Use the Test for Independence

Example: Survey 300 students and record both grade level (freshman, sophomore, junior, senior) and preferred lunch (pizza, salad, sandwich). Is there an association between grade and lunch preference?

Expected Counts

For each cell in a two-way table:

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

This gives the count you would expect if the two variables were truly independent.

Worked Example

Data: A random sample of 200 adults:

$H_0$: Gender and opinion are independent.
$H_a$: Gender and opinion are associated.

Expected counts:

$\chi^2$ calculation:

$\chi^2 = \frac{(60-52.5)^2}{52.5} + \frac{(40-47.5)^2}{47.5} + \frac{(45-52.5)^2}{52.5} + \frac{(55-47.5)^2}{47.5}$

$= \frac{56.25}{52.5} + \frac{56.25}{47.5} + \frac{56.25}{52.5} + \frac{56.25}{47.5}$

$= 1.071 + 1.184 + 1.071 + 1.184 = 4.510$

$df = (r-1)(c-1) = (2-1)(2-1) = 1$

Using a $\chi^2$ table: $p \approx 0.034$

Conclusion: Since $p = 0.034 < 0.05$, we reject $H_0$. There is convincing evidence of an association between gender and opinion on this issue.

Independence Test Concepts 🎯

Expected Count Practice 🧮

A two-way table has row totals of 80 and 120, column totals of 90 and 110, and a grand total of 200.

1) Expected count for the top-left cell (row 1, column 1)?

2) Expected count for the bottom-right cell (row 2, column 2)?

3) Degrees of freedom for this $2 \times 2$ table?

Independence Test Decisions 🔍

Exit Quiz — Test for Independence ✅

📊 Chi-Square Test for Homogeneity

Part 3 of 7 — Comparing Distributions Across Populations

Topics in This Part

🔑 Key Concept: The test for homogeneity uses data from two or more independent samples (or treatment groups) to determine whether the distribution of a single categorical variable is the same across all populations.

Independence vs. Homogeneity

⚠️ AP Exam: The math is identical. The difference is in the hypotheses and context. Read the problem carefully to determine which test is appropriate.

Hypotheses for Homogeneity

$H_0: \text{The distribution of [variable] is the same for all [groups].}$ $H_a: \text{The distribution of [variable] is NOT the same for all [groups].}$

Worked Example

Problem: Two schools were surveyed about favorite subject. Is the distribution of preferences the same?

$H_0$: The distribution of favorite subject is the same for School A and School B.

Expected counts: (each row total = 100, grand total = 200)

$\chi^2$:

$\chi^2 = \frac{(45-40)^2}{40} + \frac{(30-35)^2}{35} + \frac{(25-25)^2}{25} + \frac{(35-40)^2}{40} + \frac{(40-35)^2}{35} + \frac{(25-25)^2}{25}$

$= 0.625 + 0.714 + 0 + 0.625 + 0.714 + 0 = 2.678$

$df = (2-1)(3-1) = 2$

Using a $\chi^2$ table with $df = 2$: $p \approx 0.262$

Conclusion: Since $p = 0.262 > 0.05$, we fail to reject $H_0$. There is not convincing evidence that the distribution of favorite subject differs between the two schools.

Homogeneity Concepts 🎯

Independence or Homogeneity? 🔍

Identify the correct test for each scenario.

Homogeneity Calculation 🧮

Three brands of cereal are compared on sugar level (Low, Medium, High). Samples: Brand A: 50, Brand B: 60, Brand C: 40. Grand total: 150. Column totals: Low = 60, Medium = 50, High = 40.

1) Expected count for Brand A, Low sugar?

2) Expected count for Brand C, High sugar?

3) $df$ for this $3 \times 3$ table?

Exit Quiz — Test for Homogeneity ✅

📊 Conditions and Degrees of Freedom

Part 4 of 7 — When Can You Use the $\chi^2$ Test?

Topics in This Part

🔑 Key Concept: Both chi-square tests (homogeneity and independence) require the same three conditions: Random, 10%, and Expected Counts.

The Three Conditions

⚠️ Critical AP Detail: For chi-square, the expected counts condition uses expected counts, NOT observed counts. This is different from the Large Counts condition for proportions ($np \geq 10$ and $n(1-p) \geq 10$).

Degrees of Freedom

Both tests use the same formula:

$df = (\text{number of rows} - 1)(\text{number of columns} - 1)$

Count only the rows and columns of categories; the total row and total column are not part of the table's dimensions.

Why Degrees of Freedom Matter

The $\chi^2$ distribution changes shape with $df$:

As $df$ increases, the skew becomes less pronounced and the curve spreads to the right, so the same $\chi^2$ value gives a larger p-value when $df$ is larger.

What If a Condition Fails?

🔑 AP Tip: If an expected count is 5 or less, say that the condition is not met. If the problem asks you to proceed anyway, state the concern and continue with caution.

Conditions & df Concepts 🎯

Degrees of Freedom Practice 🧮

1) A two-way table has 2 rows and 6 columns. $df =$

2) Independence test with a $5 \times 3$ table: $df =$

3) Homogeneity test comparing 4 groups on a categorical variable with 3 levels: Table is $4 \times 3$. $df =$

Condition Checking 🔍

Exit Quiz — Conditions & Degrees of Freedom ✅

📊 Interpreting Results

Part 5 of 7 — Reading $\chi^2$ Output and Drawing Conclusions

Topics in This Part

🔑 Key Concept: A large $\chi^2$ value means the observed data differ substantially from what is expected under $H_0$. The p-value tells you how surprising your $\chi^2$ would be if $H_0$ were true.

What the $\chi^2$ Value Tells You

The p-value makes this precise: it gives the probability of getting a $\chi^2$ as large or larger than yours, assuming $H_0$ is true.

Reading the $\chi^2$ Table

The table gives right-tail areas for the $\chi^2$ distribution:

How to use: If $\chi^2 = 8.5$ with $df = 3$: $7.815 < 8.5 < 9.348$, so $0.025 < p < 0.05$.

AP Conclusion Template

If $p \leq \alpha$: "Since the p-value ($p = \text{value}$) is less than $\alpha = 0.05$, we reject $H_0$. There is convincing evidence that [state $H_a$ in context]."

If $p > \alpha$: "Since the p-value ($p = \text{value}$) is greater than $\alpha = 0.05$, we fail to reject $H_0$. There is not convincing evidence that [state $H_a$ in context]."

⚠️ Never say "accept $H_0$" — say "fail to reject $H_0$."

Follow-Up: Which Cells Drive the Result?

After rejecting $H_0$, examine individual cell contributions $(O_i - E_i)^2/E_i$:

Also note the direction: Is $O > E$ (more than expected) or $O < E$ (fewer than expected)?

Interpretation Concepts 🎯

Using the $\chi^2$ Table 🧮

Use the partial table above.

1) $\chi^2 = 6.0$, $df = 2$. Is $p$ less than or greater than 0.05? Enter "less" or "greater".

2) $\chi^2 = 10.5$, $df = 5$. The p-value is between which two table values? Enter the larger $\alpha$ boundary (e.g., "0.10").

3) For $df = 1$, what $\chi^2$ value gives $p = 0.05$?

Conclusion Writing 🔍

Exit Quiz — Interpreting Results ✅

📊 Problem-Solving Workshop

Part 6 of 7 — Full AP Free-Response Practice

Topics in This Part

🔑 Key Concept: Chi-square FRQs follow the same 4-step framework: Hypotheses, Conditions, Calculate, Conclude. Practice writing each step clearly.

Worked Example 1: Test for Homogeneity

Problem: 150 volunteers were randomly assigned to one of three study methods (50 each) and then took the same exam:

Step 1 — Hypotheses: $H_0$: There is no difference in the distribution of exam results (pass/fail) across the three study methods. $H_a$: There is a difference in the distribution of exam results across the three study methods.

Step 2 — Conditions:

• Random: Volunteers were randomly assigned to methods ✓
• 10%: Not needed, because the data come from a randomized experiment ✓
• Expected Counts: Each row expects $\frac{50 \times 99}{150} = 33$ passes and $\frac{50 \times 51}{150} = 17$ failures. All expected counts are greater than 5 ✓

Step 3 — Calculate: $\chi^2 = \frac{(32-33)^2}{33} + \frac{(18-17)^2}{17} + \frac{(40-33)^2}{33} + \frac{(10-17)^2}{17} + \frac{(27-33)^2}{33} + \frac{(23-17)^2}{17}$

$= 0.030 + 0.059 + 1.485 + 2.882 + 1.091 + 2.118 = 7.665, \quad df = (3-1)(2-1) = 2$

From the table: $7.378 < 7.665 < 9.210$, so $0.01 < p < 0.025$ (technology: $p \approx 0.022$).

Step 4 — Conclude: Since the p-value is less than $\alpha = 0.05$, we reject $H_0$. There is convincing evidence that the distribution of exam results differs across the three study methods. Because the methods were randomly assigned, we can conclude that the method affects the pass/fail distribution for volunteers like these.

🔍 Follow-up: The Practice tests / Fail cell had the largest contribution (2.882): far fewer failures than expected.

Worked Example 2: Test for Independence

Problem: A random sample of 400 adults records education level and exercise frequency:

Step 1 — Hypotheses: $H_0$: Education level and exercise frequency are independent among adults. $H_a$: Education level and exercise frequency are associated among adults.

Step 2 — Conditions:

• Random: Random sample stated ✓
• 10%: $400 < 10\%$ of all adults ✓
• Expected Counts: $E_{11} = \frac{200 \times 190}{400} = 95$, $E_{12} = 105$, $E_{21} = 95$, $E_{22} = 105$. All greater than 5 ✓

Step 3 — Calculate: $\chi^2 = \frac{(120-95)^2}{95} + \frac{(80-105)^2}{105} + \frac{(70-95)^2}{95} + \frac{(130-105)^2}{105}$

$= \frac{625}{95} + \frac{625}{105} + \frac{625}{95} + \frac{625}{105} = 6.579 + 5.952 + 6.579 + 5.952 = 25.06$

$df = (2-1)(2-1) = 1$. From the table: $\chi^2 = 25.06 > 6.635$ so $p < 0.01$.

Step 4 — Conclude: Since the p-value is less than $\alpha = 0.05$ (in fact less than 0.01), we reject $H_0$. There is convincing evidence of an association between education level and exercise frequency among adults. This is an observational study, so it does not show that education causes more exercise.

⚠️ Common AP Mistakes on Chi-Square FRQs

Workshop Concept Check 🎯

Quick Calculations 🧮

A random sample of 120 adults is classified by age group (row totals: under 40 = 70, 40 and over = 50) and favorite drink (column totals: coffee = 60, tea = 36, neither = 24). The observed count for under 40 / coffee is 42.

1) Expected count for under 40 / coffee?

2) Contribution of that cell, $(O-E)^2/E$?

3) $df$ for this table?

Which Test? 🔍

Exit Quiz — Problem-Solving Workshop ✅

📊 Review & Applications

Part 7 of 7 — Comprehensive Chi-Square Review

Topics in This Part

🔑 Key Concept: This review covers the two chi-square tests on the AP exam. Know when to use each, how to check conditions, and how to write full AP-quality solutions.

The Two Chi-Square Tests Compared

Universal Formula

$\chi^2 = \sum \frac{(\text{observed} - \text{expected})^2}{\text{expected}}$

Universal Conditions

Expected Counts

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

Decision Guide: Which Test?

Key AP Reminders

• $\chi^2$ is always $\geq 0$ and the p-value is always a right-tail area
• The count condition uses expected counts, not observed
• Never say "accept $H_0$"
• Association ≠ causation (unless randomized experiment)
• Show the formula with substitution on FRQs
• State $df$ explicitly

Comprehensive Concept Check 🎯

Mixed Practice 🧮

1) A two-way table has row total 50, column total 40 and table total 200. Expected count for that cell?

2) Independence test, $3 \times 5$ table. $df =$

3) $\chi^2 = 7.5$, $df = 2$. From the table ($\alpha = 0.05$: 5.991; $\alpha = 0.025$: 7.378). Is $p$ less than 0.025? Enter "yes" or "no".

Quick Decisions 🔍

Final Exam — Chi-Square Unit ✅