Learning Objectives

By the end of this chapter, you will be able to:

When to Use Chi-Square Test for Independence

Use chi-square test when:

  1. Data is categorical (not numerical)
  2. You want to test if two variables are related/associated
  3. Data is organized in a contingency table
  4. Sample size is adequate (expected frequencies ≥ 5)
flowchart TD
    A["Type of data?"]
    B{"Categorical?"}
    C{"Testing association?"}
    D["Chi-square test"]
    E[Use other tests]
    F{"Testing distribution?"}
    G[Goodness of fit]

    A --> B
    B -->|Yes| C
    B -->|No| E
    C -->|Yes| D
    C -->|No| F
    F -->|Yes| G

Key Concepts

Contingency Table

A contingency table (cross-tabulation) shows frequency distribution of two categorical variables.

Example:

  Support Oppose Total
Male 30 20 50
Female 25 25 50
Total 55 45 100

Independence

Two variables are independent if knowing one doesn’t help predict the other.

Expected Frequency Formula

If variables were independent, expected frequency for each cell:

\[E_{ij} = \frac{(\text{Row Total}_i) \times (\text{Column Total}_j)}{\text{Grand Total}}\]

Test Statistic

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

Where:

Degrees of Freedom

\[df = (r - 1)(c - 1)\]

Where:

Step-by-Step Example 1: 2×2 Table

Problem: A survey asked 200 people about their opinion on a policy:

  Support Oppose Total
Urban 60 40 100
Rural 45 55 100
Total 105 95 200

Test at α = 0.05 if opinion is associated with residence.

Solution:

Step 1: State hypotheses

Step 2: Calculate expected frequencies

For Urban-Support: \(E_{11} = \frac{100 \times 105}{200} = 52.5\)

For Urban-Oppose: \(E_{12} = \frac{100 \times 95}{200} = 47.5\)

For Rural-Support: \(E_{21} = \frac{100 \times 105}{200} = 52.5\)

For Rural-Oppose: \(E_{22} = \frac{100 \times 95}{200} = 47.5\)

Expected Table:

  Support Oppose Total
Urban 52.5 47.5 100
Rural 52.5 47.5 100
Total 105 95 200

Step 3: Calculate chi-square

Cell O E (O-E)² (O-E)²/E
Urban-Support 60 52.5 56.25 1.071
Urban-Oppose 40 47.5 56.25 1.184
Rural-Support 45 52.5 56.25 1.071
Rural-Oppose 55 47.5 56.25 1.184
Total       4.510
\[\chi^2 = 4.510\]

Step 4: Find critical value

Step 5: Decision

Step 6: Conclusion At the 0.05 level of significance, there is sufficient evidence to conclude that opinion on the policy is associated with residence (urban/rural).

Chi-Square Critical Values Table

df α = 0.10 α = 0.05 α = 0.025 α = 0.01 α = 0.005
1 2.706 3.841 5.024 6.635 7.879
2 4.605 5.991 7.378 9.210 10.597
3 6.251 7.815 9.348 11.345 12.838
4 7.779 9.488 11.143 13.277 14.860
5 9.236 11.070 12.833 15.086 16.750
6 10.645 12.592 14.449 16.812 18.548

Step-by-Step Example 2: 3×2 Table

Problem: Job satisfaction by department:

  Satisfied Not Satisfied Total
Finance 40 20 60
HR 30 30 60
IT 50 10 60
Total 120 60 180

Test at α = 0.01 if satisfaction differs by department.

Solution:

Step 1: State hypotheses

Step 2: Calculate expected frequencies

\[E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\]
Cell Calculation E
Finance-Satisfied (60×120)/180 40
Finance-Not Satisfied (60×60)/180 20
HR-Satisfied (60×120)/180 40
HR-Not Satisfied (60×60)/180 20
IT-Satisfied (60×120)/180 40
IT-Not Satisfied (60×60)/180 20

Expected Table:

  Satisfied Not Satisfied Total
Finance 40 20 60
HR 40 20 60
IT 40 20 60
Total 120 60 180

Step 3: Calculate chi-square

Cell O E (O-E)² (O-E)²/E
Finance-Sat 40 40 0 0
Finance-Not 20 20 0 0
HR-Sat 30 40 100 2.5
HR-Not 30 20 100 5.0
IT-Sat 50 40 100 2.5
IT-Not 10 20 100 5.0
Total       15.0
\[\chi^2 = 15.0\]

Step 4: Find critical value

Step 5: Decision

Step 6: Conclusion At the 0.01 level of significance, there is strong evidence that job satisfaction differs significantly across departments.

Step-by-Step Example 3: 3×3 Table

Problem: Education level vs. voting preference:

  Party A Party B Party C Total
High School 30 45 25 100
Bachelor’s 40 35 45 120
Graduate 30 20 30 80
Total 100 100 100 300

Test at α = 0.05 if education is associated with voting preference.

Solution:

Step 1: State hypotheses

Step 2: Calculate expected frequencies

For each cell: $E = \frac{\text{Row Total} \times \text{Column Total}}{300}$

Expected Table:

  Party A Party B Party C Total
High School 33.33 33.33 33.33 100
Bachelor’s 40.00 40.00 40.00 120
Graduate 26.67 26.67 26.67 80
Total 100 100 100 300

Step 3: Calculate chi-square

O E (O-E)²/E
30 33.33 0.333
45 33.33 4.083
25 33.33 2.083
40 40.00 0.000
35 40.00 0.625
45 40.00 0.625
30 26.67 0.417
20 26.67 1.667
30 26.67 0.417
Total   10.25
\[\chi^2 = 10.25\]

Step 4: Find critical value

Step 5: Decision

Step 6: Conclusion At the 0.05 level of significance, there is sufficient evidence that education level is associated with voting preference.

Checking Assumptions

Minimum Expected Frequency Rule

flowchart TD
    A["Calculate Expected<br/>Frequencies"]
    B{"All E ≥ 5?"}
    C["Proceed with<br/>Chi-square test"]
    D{"Can combine<br/>categories?"}
    E["Combine categories<br/>and recalculate"]
    F["Use Fisher's<br/>Exact Test"]

    A --> B
    B -->|Yes| C
    B -->|No| D
    D -->|Yes| E
    E --> A
    D -->|No| F

Shortcut Formula for 2×2 Tables

For a 2×2 table:

  Column 1 Column 2 Total
Row 1 a b a+b
Row 2 c d c+d
Total a+c b+d n
\[\chi^2 = \frac{n(ad - bc)^2}{(a+b)(c+d)(a+c)(b+d)}\]

Example 4: Using Shortcut Formula

From Example 1:

\(\chi^2 = \frac{200(60 \times 55 - 40 \times 45)^2}{(100)(100)(105)(95)}\) \(= \frac{200(3300 - 1800)^2}{99,750,000} = \frac{200 \times 2,250,000}{99,750,000}\) \(= \frac{450,000,000}{99,750,000} = 4.51\)

Same result as before!

Interpreting Results

Result Interpretation
Reject H₀ Variables are associated (dependent)
Fail to Reject H₀ Variables are independent (no association found)

Note: Chi-square tells us IF there’s an association, not HOW STRONG or the DIRECTION.

Practice Problems

Problem 1

Test whether gender and preference for online/offline shopping are independent:

  Online Offline Total
Male 70 30 100
Female 50 50 100
Total 120 80 200

Use α = 0.05.

Problem 2

Test if age group is associated with technology adoption:

  Adopted Not Adopted Total
Young (18-30) 80 20 100
Middle (31-50) 60 40 100
Senior (51+) 40 60 100
Total 180 120 300

Use α = 0.01.

Problem 3

Employee performance by training status:

  Excellent Good Average Total
Trained 30 40 10 80
Untrained 15 25 40 80
Total 45 65 50 160

Test at α = 0.05 if training is associated with performance.

Problem 4

Use the shortcut formula to verify Problem 1’s chi-square value.

Summary

Component Formula
Expected frequency $E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}$
Chi-square statistic $\chi^2 = \sum \frac{(O-E)^2}{E}$
Degrees of freedom df = (r-1)(c-1)
2×2 shortcut $\chi^2 = \frac{n(ad-bc)^2}{(a+b)(c+d)(a+c)(b+d)}$
Decision Reject H₀ if χ² > critical value

Next Topic

In the next chapter, we will study the Chi-Square Goodness of Fit Test for testing if observed frequencies match an expected distribution.