Chi-Square Tests

Perform chi-square tests for goodness of fit, homogeneity, and independence.

Written and reviewed by the Study Mondo Education TeamLast updated April 30, 2026

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📊 Chi-Square Tests

When to Use Chi-Square

Chi-square tests work with categorical data (counts in categories).

$\chi^2 = \sum \frac{(O - E)^2}{E}$

Where $O$ = observed count, $E$ = expected count

Three Flavors of Chi-Square Tests

1. Goodness-of-Fit Test

Question: Does sample match hypothesized distribution?

Example: Is a die fair? (each face should appear 1/6 of rolls)

Degrees of freedom: $df = k - 1$ (where $k$ = number of categories)

2. Test of Independence

Question: Are two categorical variables independent?

Example: Are smoking status and lung cancer independent?

Degrees of freedom: $df = (r - 1)(c - 1)$ (rows and columns)

3. Test of Homogeneity

Question: Do multiple populations have the same distribution?

Example: Do males and females have the same opinion distribution?

Degrees of freedom: $df = (r - 1)(c - 1)$ (same formula as independence)

Worked Example: Test of Independence

Data on 400 students (exercise habit vs. GPA):

	Low GPA	High GPA	Total
Exercise	40	160	200
No exercise	100	100	200
Total	140	260	400

Expected (if independent): $E = \frac{\text{row total} \times \text{column total}}{n}$

$E_{\text{Exercise, Low}} = \frac{200 \times 140}{400} = 70$

$\chi^2 = \frac{(40-70)^2}{70} + \frac{(160-130)^2}{130} + \frac{(100-70)^2}{70} + \frac{(100-130)^2}{130}$

$\chi^2 = 12.86$ ; $df = (2-1)(2-1) = 1$ ; p-value $< 0.001$

Conclusion: Exercise and GPA are NOT independent.

Conditions for Chi-Square Tests

Random sample
Independence between observations
Expected counts $\geq 5$ in all cells

If condition 3 fails, don't use chi-square!

Common Mistakes

❌ Using observed counts instead of expected in formula ❌ Not checking expected count condition ❌ Mixing up df formulas ❌ Using chi-square with continuous data

Decision Rule

If $\chi^2 > \chi^2_{df, \alpha}$ , reject $H_0$
If p-value $< \alpha$ , reject $H_0$

AP Exam Tip

Always create a table. Show calculation of at least two expected counts. Name the test: "chi-square test of independence" (or goodness-of-fit/homogeneity). Check all conditions before concluding.

📚 Practice Problems

1Problem 1easy

❓ Question:

A candy company claims their candy bags contain equal proportions of red, blue, green, and yellow candies. A bag contains 30 red, 25 blue, 20 green, and 25 yellow candies. What type of chi-square test should be used, and what are the expected counts?

💡 Show Solution

Test type: Chi-square goodness-of-fit test (Testing if observed distribution matches a claimed distribution)

Total candies = 30 + 25 + 20 + 25 = 100

If proportions are equal, each color should be 25% of total: Expected count for each color = 100 × 0.25 = 25

Expected counts: • Red: 25 • Blue: 25 • Green: 25 • Yellow: 25

All expected counts ≥ 5, so conditions are met.

2Problem 2medium

❓ Question:

A researcher surveys 200 people about their exercise habits and stress levels. The results are shown below. Calculate the chi-square test statistic.

           Low Stress | High Stress

Exercise 60 | 40 No Exercise 30 | 70

💡 Show Solution

Step 1: Calculate expected counts Row totals: Exercise = 100, No Exercise = 100 Column totals: Low Stress = 90, High Stress = 110 Grand total = 200

Expected = (row total × column total) / grand total

Expected counts: • Exercise & Low: (100 × 90)/200 = 45 • Exercise & High: (100 × 110)/200 = 55 • No Exercise & Low: (100 × 90)/200 = 45 • No Exercise & High: (100 × 110)/200 = 55

Step 2: Calculate χ² χ² = Σ[(Observed - Expected)² / Expected]

χ² = (60-45)²/45 + (40-55)²/55 + (30-45)²/45 + (70-55)²/55 χ² = 225/45 + 225/55 + 225/45 + 225/55 χ² = 5 + 4.09 + 5 + 4.09 χ² ≈ 18.18

df = (rows - 1)(columns - 1) = (2-1)(2-1) = 1

3Problem 3medium

❓ Question:

A die is rolled 120 times with the following results: 1(15), 2(18), 3(22), 4(25), 5(20), 6(20). Test at α = 0.05 if the die is fair.

💡 Show Solution

H₀: The die is fair (all outcomes equally likely) Hₐ: The die is not fair

Expected count for fair die: 120/6 = 20 for each outcome

χ² = Σ[(O - E)² / E] χ² = (15-20)²/20 + (18-20)²/20 + (22-20)²/20 + (25-20)²/20 + (20-20)²/20 + (20-20)²/20 χ² = 25/20 + 4/20 + 4/20 + 25/20 + 0/20 + 0/20 χ² = 1.25 + 0.2 + 0.2 + 1.25 + 0 + 0 χ² = 2.9

df = 6 - 1 = 5

P-value: P(χ² > 2.9) ≈ 0.715

Decision: Since p-value (0.715) > α (0.05), fail to reject H₀

Conclusion: There is insufficient evidence to conclude the die is unfair.

4Problem 4hard

❓ Question:

A study examines the relationship between smoking status and lung disease in 500 people:

           Disease | No Disease

Smoker 80 | 120 Non-smoker 20 | 280

Perform a chi-square test at α = 0.01 to determine if smoking and lung disease are independent.

💡 Show Solution

H₀: Smoking status and lung disease are independent Hₐ: Smoking status and lung disease are associated

Step 1: Calculate expected counts Row totals: Smoker = 200, Non-smoker = 300 Column totals: Disease = 100, No Disease = 400 Total = 500

Expected counts: • Smoker & Disease: (200×100)/500 = 40 • Smoker & No Disease: (200×400)/500 = 160 • Non-smoker & Disease: (300×100)/500 = 60 • Non-smoker & No Disease: (300×400)/500 = 240

Step 2: Calculate χ² χ² = (80-40)²/40 + (120-160)²/160 + (20-60)²/60 + (280-240)²/240 χ² = 1600/40 + 1600/160 + 1600/60 + 1600/240 χ² = 40 + 10 + 26.67 + 6.67 χ² ≈ 83.34

df = (2-1)(2-1) = 1

P-value: P(χ² > 83.34) < 0.0001

Decision: Reject H₀

Conclusion: There is very strong evidence (p < 0.01) that smoking status and lung disease are associated.

5Problem 5hard

❓ Question:

A school surveys students from three grades about their favorite subject. Results:

Grade 9: Math(40), Science(30), English(30) Grade 10: Math(35), Science(35), English(30) Grade 11: Math(25), Science(45), English(30)

Test if the distribution of favorite subject is the same across grades at α = 0.05.

💡 Show Solution

H₀: Distribution of favorite subject is the same across grades Hₐ: Distribution differs by grade

Step 1: Set up table Math | Science | English | Total Grade 9 40 | 30 | 30 | 100 Grade 10 35 | 35 | 30 | 100 Grade 11 25 | 45 | 30 | 100 Total 100 | 110 | 90 | 300

Step 2: Calculate expected counts E = (row total × column total) / grand total

For each cell: Grade 9 & Math: (100×100)/300 = 33.33 Grade 9 & Science: (100×110)/300 = 36.67 Grade 9 & English: (100×90)/300 = 30 [Continue for all cells...]

Step 3: Calculate χ² χ² = (40-33.33)²/33.33 + (30-36.67)²/36.67 + ... (all 9 cells) χ² ≈ 1.33 + 1.21 + 0 + 0.09 + 0.08 + 0 + 2.08 + 1.89 + 0 χ² ≈ 6.68

df = (3-1)(3-1) = 4

P-value: P(χ² > 6.68) ≈ 0.154

Decision: Fail to reject H₀

Conclusion: There is insufficient evidence that the distribution of favorite subject differs across grades.

Explain using:

⚠️ Common Mistakes: Chi-Square Tests

Avoid these 3 frequent errors

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Chi-Square Tests for Independence and Homogeneity

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📌 Related Topics in Unit 8: Inference for Categorical Data — Chi-Square

Chi-Square Tests for Independence and Homogeneity

❓ Frequently Asked Questions

What is Chi-Square Tests?▾

Perform chi-square tests for goodness of fit, homogeneity, and independence.

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Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Chi-Square Tests

Try the Interactive Version!

📊 Chi-Square Tests

When to Use Chi-Square

Three Flavors of Chi-Square Tests

1. Goodness-of-Fit Test

2. Test of Independence

3. Test of Homogeneity

Worked Example: Test of Independence

Conditions for Chi-Square Tests

Common Mistakes

Decision Rule

AP Exam Tip

📚 Practice Problems

1Problem 1easy

❓ Question:

2Problem 2medium

❓ Question:

3Problem 3medium

❓ Question:

4Problem 4hard

❓ Question:

5Problem 5hard

❓ Question:

⚠️ Common Mistakes: Chi-Square Tests

Practice Lab

Practice with Flashcards

Browse All Topics

📌 Related Topics in Unit 8: Inference for Categorical Data — Chi-Square

❓ Frequently Asked Questions