5.9 Chi-Square Test for Independence

Learning Objectives

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

• Understand when to use chi-square test for independence
• Construct and analyze contingency tables
• Calculate expected frequencies
• Perform chi-square test step by step
• Interpret results correctly

---

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.

• Null hypothesis (H₀): Variables are independent
• Alternative hypothesis (H₁): Variables are NOT independent (associated)

---

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:

• O = Observed frequency
• E = Expected frequency

Degrees of Freedom

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

Where:

• r = number of rows
• c = number of columns

---

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

• H₀: Opinion and residence are independent
• H₁: Opinion and residence are associated

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

• df = (2-1)(2-1) = 1
• α = 0.05
• From chi-square table: χ²* = 3.841

Step 5: Decision

• χ² = 4.51 > 3.841
• Reject H₀

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

• H₀: Satisfaction is independent of department
• H₁: Satisfaction is associated with department

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

• df = (3-1)(2-1) = 2
• α = 0.01
• From chi-square table: χ²* = 9.210

Step 5: Decision

• χ² = 15.0 > 9.210
• Reject H₀

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

• H₀: Education and voting preference are independent
• H₁: Education and voting preference are associated

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

• df = (3-1)(3-1) = 4
• α = 0.05
• From chi-square table: χ²* = 9.488

Step 5: Decision

• χ² = 10.25 > 9.488
• Reject H₀

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

• All expected frequencies should be ≥ 5
• If not, combine categories or use Fisher’s Exact Test
• At most 20% of cells should have E < 5

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:

• a = 60, b = 40, c = 45, d = 55, n = 200

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

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.