Module 11: Chi-Square Tests - Statistics Visualizer

Goodness of Fit Test

Interactive Chi-Square Calculator

	Outcome A	Outcome B
Group 1
Group 2

χ² statistic-

Degrees of freedom-

p-value-

Cramér's V-

Enter frequencies and click Calculate

Test for independence between two categorical variables! Enter observed frequencies in the 2×2 contingency table and click Calculate. The calculator computes χ², degrees of freedom, p-value, and Cramér's V effect size. Try different patterns to see when variables are associated vs. independent.

More uneven distributions → Larger χ² → Stronger association

Testing Expected Proportions

Are dice fair? Roll 60 times:

⚀

Obs: 8
Exp: 10

⚁

Obs: 12
Exp: 10

⚂

Obs: 9
Exp: 10

⚃

Obs: 11
Exp: 10

⚄

Obs: 7
Exp: 10

⚅

Obs: 13
Exp: 10

Do observed frequencies match expected (1/6 each)?

Goodness of fit tests whether observed frequencies match expected proportions. If dice are fair, each face should appear equally often. Does the data fit this expectation?

One variable, multiple categories

The χ² Statistic

χ² = Σ (O - E)² / E

O
Observed frequency

E
Expected frequency

(O-E)²/E
Contribution to χ²

For each category, calculate (Observed - Expected)² / Expected. Sum these values. Larger χ² = bigger discrepancy between observed and expected frequencies.

χ² sums squared deviations scaled by expected

Step-by-Step Calculation

Face	O	E	O-E	(O-E)²	(O-E)²/E
1	8	10	-2	4	0.4
2	12	10	2	4	0.4
...	...	...	...	...	...
Sum					χ² = 2.8

df = k - 1 = 6 - 1 = 5 | Critical value at α=.05: 11.07

χ² = 2.8 < 11.07 → Fail to reject H₀ (dice are fair)

With χ² = 2.8 and df = 5, p > 0.05. We fail to reject H₀—the dice appear fair. The observed variation from expected is within normal random fluctuation.

df = number of categories - 1

Test of Independence

Are Two Variables Related?

Is smoking related to lung disease?

	Disease	No Disease	Total
Smoker	90	60	150
Non-Smoker	30	120	150
Total	120	180	300

The test of independence asks: Are two categorical variables related, or are they independent? If independent, knowing one variable doesn't help predict the other.

Two categorical variables in a contingency table

Calculating Expected Frequencies

E = (Row Total × Column Total) / Grand Total

Smoker + Disease

E = (150 × 120) / 300 = 60

Observed was 90

Non-Smoker + Disease

E = (150 × 120) / 300 = 60

Observed was 30

Expected = what we'd see if variables were independent

If smoking and disease were independent, we'd expect 60 smokers with disease (not 90). The big gap between observed (90) and expected (60) suggests a relationship.

E = (Row × Column) / Total

Independence Hypotheses

H₀

Variables are independent

Smoking and disease are unrelated

Hₐ

Variables are associated

There IS a relationship

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

H₀: The variables are independent (no relationship). Hₐ: The variables are associated. This is a two-tailed test—we don't specify the direction of association.

df = (r-1) × (c-1)

Interpreting the Result

Result: χ²(1) = 60.0, p < .001

Reject H₀: Smoking and lung disease are significantly associated

What this means:
• The relationship is statistically significant
• Smokers have higher disease rates than expected
• Does NOT prove causation
• Need effect size for practical significance

χ² = 60.0 is highly significant—there's a clear association between smoking and disease. But chi-square only tells us there IS an association, not how strong it is.

Significant χ² = association exists

Assumptions & Rules

Chi-Square Assumptions

✓ Required

• Random sampling
• Independent observations
• Categorical variables
• Mutually exclusive categories

⚠️ Expected Frequency Rule

• All E ≥ 1
• At least 80% of E ≥ 5
• If violated: use Fisher's exact test or combine categories

Chi-square requires categorical data, independence, and adequate expected frequencies. The 'rule of 5' says most expected frequencies should be ≥ 5 for the approximation to work well.

Expected frequencies should be ≥ 5

When Expected < 5

Problem: Expected frequency < 5

Chi-square approximation becomes unreliable

Solutions:

1. Fisher's Exact Test
Calculates exact p-value, no approximation needed

2. Combine Categories
Merge sparse categories to increase expected counts

3. Collect More Data
Larger sample increases expected frequencies

When expected frequencies are small, the chi-square distribution is a poor approximation. Fisher's exact test is the gold standard for small samples or sparse tables.

Fisher's exact test for small samples

Continuity Correction

χ²_Yates = Σ (|O - E| - 0.5)² / E

When to Use

2×2 tables with small samples (some argue always for 2×2)

Effect

Makes test more conservative (reduces χ², increases p)

⚠️ Controversial: Many statisticians prefer uncorrected or Fisher's

Yates' correction subtracts 0.5 from each |O-E| before squaring. It makes the test more conservative. For 2×2 tables, many prefer Fisher's exact test over Yates' correction.

Yates: subtract 0.5 from |O-E|

Effect Size

Measuring Association Strength

V = √(χ² / (n × min(r-1, c-1)))

V = .10

Small

V = .30

Medium

V = .50

Large

Cramér's V ranges from 0 (no association) to 1 (perfect association). Unlike χ², it doesn't depend on sample size. Use it to describe how STRONG the relationship is.

V ranges 0-1, independent of sample size

Phi for 2×2 Tables

φ = √(χ² / n)

Only for 2×2 tables

Interpretation (same as correlation):

• φ = 0: No association
• φ = ±0.1: Small
• φ = ±0.3: Medium
• φ = ±0.5: Large
• φ = ±1.0: Perfect association

φ = Cramér's V when table is 2×2

Phi (φ) is a special case of Cramér's V for 2×2 tables. It equals Cramér's V when both dimensions have 2 categories. Interpreted like a correlation coefficient.

φ = V for 2×2 tables

Odds Ratio for 2×2 Tables

	Disease	No Disease
Smoker	a=90	b=60
Non-Smoker	c=30	d=120

OR = (a × d) / (b × c) = (90 × 120) / (60 × 30) = 6.0

Smokers have 6× the odds of disease compared to non-smokers

The odds ratio shows how much more likely an outcome is in one group vs another. OR = 6 means smokers have 6 times the odds of disease. Very useful for medical/epidemiological interpretation.

OR = (a×d)/(b×c) for 2×2 tables

Complete Reporting

Example write-up:

"A chi-square test of independence was performed to examine the relation between smoking status and lung disease. The relation between these variables was significant, χ²(1, N = 300) = 60.0, p < .001. Smokers were more likely to have lung disease (60%) than non-smokers (20%). The effect size was large (Cramér's V = .45, OR = 6.0, 95% CI [3.5, 10.3])."

Report χ²(df, N) = value, p-value, direction of effect (percentages), and effect size (V or φ). For 2×2 tables, odds ratios with confidence intervals are especially informative.

χ²(df, N) = value, p, effect size

★

Real-World Examples

Apply your knowledge to practical scenarios

Testing Casino Dice

A casino regulator suspects a die is loaded. They roll it 600 times and record: Face 1: 85, Face 2: 95, Face 3: 105, Face 4: 100, Face 5: 110, Face 6: 105. Is the die fair?

Sample Data

{
  "observed": [
    85,
    95,
    105,
    100,
    110,
    105
  ],
  "expected": [
    100,
    100,
    100,
    100,
    100,
    100
  ],
  "n": 600
}

Q What type of chi-square test is appropriate? +

A Goodness of fit test—we're comparing observed frequencies to expected frequencies under the hypothesis that each face has 1/6 probability.

Q Calculate the expected frequency for each face. +

A E = 600 / 6 = 100 for each face. Under H₀ (fair die), each face should appear 100 times.

Q Calculate χ² for this data. +

A χ² = Σ(O-E)²/E = (85-100)²/100 + (95-100)²/100 + (105-100)²/100 + (100-100)²/100 + (110-100)²/100 + (105-100)²/100 = 2.25 + 0.25 + 0.25 + 0 + 1 + 0.25 = 4.0

Q With df = 5 and α = 0.05, critical χ² = 11.07. Is the die loaded? +

A χ² = 4.0 < 11.07. We fail to reject H₀. There's insufficient evidence that the die is unfair—the variation is within normal random fluctuation.

Key Insights

Goodness of fit tests compare observed to theoretical proportions
Expected values are calculated from H₀ assumptions
Even 'fair' processes produce variation—that's randomness

Gender and Career Choice

A university studies whether gender is associated with STEM vs. Humanities major choice. Data from 400 students:

Sample Data

{
  "table": {
    "headers": [
      "",
      "STEM",
      "Humanities",
      "Total"
    ],
    "rows": [
      [
        "Male",
        120,
        80,
        200
      ],
      [
        "Female",
        90,
        110,
        200
      ],
      [
        "Total",
        210,
        190,
        400
      ]
    ]
  }
}

Q What are the expected frequencies for each cell under independence? +

A E = (Row × Column) / Total. Male+STEM: (200×210)/400 = 105. Male+Hum: (200×190)/400 = 95. Female+STEM: 105. Female+Hum: 95.

Q Calculate χ² for this table. +

A χ² = (120-105)²/105 + (80-95)²/95 + (90-105)²/105 + (110-95)²/95 = 2.14 + 2.37 + 2.14 + 2.37 = 9.02

Q With df = (2-1)(2-1) = 1, critical χ² at α = 0.05 is 3.84. What's the conclusion? +

A χ² = 9.02 > 3.84. We reject H₀. There IS a significant association between gender and major choice. Males are overrepresented in STEM, females in Humanities.

Q Calculate Cramér's V for effect size. +

A V = √(χ²/n) = √(9.02/400) = √0.0226 = 0.15. This is a small effect. While statistically significant, the association is weak.

Key Insights

Independence test examines association between two categorical variables
Significant χ² means variables are associated, not independent
Effect size (Cramér's V) may be small even when χ² is significant

Treatment Effectiveness Study

A clinical trial compares a new treatment to placebo for recovering from illness within 7 days:

Sample Data

{
  "table": {
    "headers": [
      "",
      "Recovered",
      "Not Recovered",
      "Total"
    ],
    "rows": [
      [
        "Treatment",
        80,
        20,
        100
      ],
      [
        "Placebo",
        50,
        50,
        100
      ],
      [
        "Total",
        130,
        70,
        200
      ]
    ]
  }
}

Q What is the recovery rate in each group? +

A Treatment: 80/100 = 80%. Placebo: 50/100 = 50%. The treatment appears to increase recovery rate by 30 percentage points.

Q Chi-square test gives χ²(1) = 19.78, p < 0.001. What does this mean? +

A There's a highly significant association between receiving treatment and recovery. The treatment group has significantly higher recovery rates than placebo.

Q Calculate the odds ratio. +

A OR = (80×50)/(20×50) = 4000/1000 = 4.0. Patients receiving treatment have 4× the odds of recovery compared to placebo.

Q A patient asks: 'Does this mean I have 4× the probability of recovering?' How do you respond? +

A Not exactly. OR = 4 means odds ratio, not probability ratio. The risk ratio (RR) = 80%/50% = 1.6. So probability of recovery is 1.6× higher, not 4×. Odds ≠ probability!

Key Insights

Chi-square tests association but doesn't quantify effect direction
Odds ratio (OR) and risk ratio (RR) give different perspectives on effect
Odds ratios can be misleading—always consider absolute differences too

← ANOVA Next: Non-parametric Methods →

Chi-Square Tests

Chi-Square Applications

Goodness of Fit

Independence

Expected Frequencies

Interpretation

Goodness of Fit Test

Interactive Chi-Square Calculator

Testing Expected Proportions

The χ² Statistic

Step-by-Step Calculation

Test of Independence

Are Two Variables Related?

Calculating Expected Frequencies

Independence Hypotheses

Interpreting the Result

Assumptions & Rules

Chi-Square Assumptions

When Expected < 5

Continuity Correction

Effect Size

Measuring Association Strength

Phi for 2×2 Tables

Odds Ratio for 2×2 Tables

Complete Reporting

Real-World Examples

Testing Casino Dice

Sample Data

Key Insights

Gender and Career Choice

Sample Data

Key Insights

Treatment Effectiveness Study

Sample Data

Key Insights

Chi-Square Tests Quiz

Quiz Results