Chi-Square Tests

📊 Chi-Square Goodness-of-Fit Test

Part 1 of 7 — Testing Categorical Distributions

When to Use Chi-Square Goodness-of-Fit

Use when you want to test whether observed frequencies match expected frequencies for a categorical variable.

Example: A die is rolled 60 times. Do the results suggest it is fair?

The Chi-Square Statistic

$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$

where $O_i$ = observed count and $E_i$ = expected count.

For the die example: $\chi^2 = \frac{(8-10)^2}{10} + \frac{(12-10)^2}{10} + \cdots = \frac{4+4+9+25+1+1}{10} = 4.4$

Hypotheses

• $H_0$: The observed distribution matches the expected distribution
• $H_a$: The observed distribution does NOT match the expected

Conditions

1. Random sample or random assignment
2. Expected counts ≥ 5 for all categories
3. Independence — observations are independent

🔑 Chi-square tests are always right-tailed — larger $\chi^2$ values provide more evidence against $H_0$.

Goodness-of-Fit Check 🎯

Chi-Square Calculation 🧮

A bag should contain equal numbers of 4 colors. From 80 candies: Red=24, Blue=18, Green=22, Yellow=16.

1) Expected count for each color = 80/4 = ?

2) Compute $\chi^2 = \frac{(24-20)^2}{20} + \frac{(18-20)^2}{20} + \frac{(22-20)^2}{20} + \frac{(16-20)^2}{20}$

3) Degrees of freedom = $k - 1$ = ?

📊 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: All three chi-square tests (GoF, Independence, Homogeneity) require the same three conditions: Random, 10%, and Large Counts.

The Three Conditions

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

Degrees of Freedom Summary

Why Degrees of Freedom Matter

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

Higher $df$ shifts the distribution to the right and increases the mean ($\mu = df$).

What If Conditions Fail?

🔑 AP Tip: On the AP exam, if an expected count is below 5, you should note this but may still be asked to proceed with the test. State the concern and continue.

Conditions & df Concepts 🎯

Degrees of Freedom Practice 🧮

1) GoF test with 8 categories: $df =$

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: Goodness-of-Fit

Problem: A company claims its candy mix is 30% red, 20% blue, 20% green, 15% yellow, 15% orange. A random sample of 200 candies yields:

Step 1 — Hypotheses: $H_0$: The distribution of colors matches the company claim. $H_a$: The distribution of colors does not match the company claim.

Step 2 — Conditions:

• Random: Random sample stated ✓
• 10%: $200 < 10\%$ of all candies produced ✓
• Large Counts: All expected counts $\geq 5$ (smallest is 30) ✓

Step 3 — Calculate: $\chi^2 = \frac{(75-60)^2}{60} + \frac{(35-40)^2}{40} + \frac{(32-40)^2}{40} + \frac{(28-30)^2}{30} + \frac{(30-30)^2}{30}$

$= \frac{225}{60} + \frac{25}{40} + \frac{64}{40} + \frac{4}{30} + 0 = 3.75 + 0.625 + 1.60 + 0.133 + 0$

$\chi^2 = 6.108, \quad df = 5-1 = 4$

From the table: $p$ is between $0.10$ and $0.25$ (since $6.108 < 7.779$).

Step 4 — Conclude: Since the p-value is greater than $\alpha = 0.05$, we fail to reject $H_0$. There is not convincing evidence that the distribution of candy colors differs from the company claim.

🔍 Follow-up: The red category had the largest contribution (3.75), suggesting there may be more red candies than claimed.

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. $H_a$: Education level and exercise frequency are associated.

Step 2 — Conditions:

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

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.

⚠️ Common AP Mistakes on Chi-Square FRQs

Workshop Concept Check 🎯

Quick Calculations 🧮

A GoF test: categories A, B, C with observed = 30, 25, 45 and expected = 33.3, 33.3, 33.3 (total = 100, equal proportions).

1) Contribution from category A: $(O-E)^2/E$ (round to 2 decimal places)

2) Contribution from category C: $(O-E)^2/E$ (round to 2 decimal places)

3) $df$ for this test?

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 all three chi-square tests. Know when to use each, how to check conditions, and how to write full AP-quality solutions.

Three Chi-Square Tests Compared

Universal Formula

$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$

Universal Conditions

Expected Counts

Decision Guide: Which Test?

Key AP Reminders

• $\chi^2$ is always $\geq 0$ and always right-tailed
• Large Counts 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) GoF test, 4 categories, $n = 100$, equal proportions. Expected count per category?

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 ✅