Learning Objectives

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

  • Understand when to use Kruskal-Wallis test
  • Rank data and calculate test statistic
  • Perform Kruskal-Wallis test step by step
  • Compare results with parametric alternatives
  • Interpret results correctly

When to Use Kruskal-Wallis Test

Use Kruskal-Wallis test when:

  1. Comparing more than two groups
  2. Data is ordinal or not normally distributed
  3. Variances are unequal across groups
  4. Sample sizes are small and normality is questionable
flowchart TD
    A[Comparing 3+ groups]
    B{"Data normal?<br/>Variances equal?"}
    C["One-way ANOVA"]
    D["Kruskal-Wallis Test"]
    E{"Two groups?"}
    F["Mann-Whitney U Test"]

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

In municipal and ministry work you often meet ordinal Likert ratings (1–5 service quality) or performance scores that are skewed and fail normality checks. Kruskal–Wallis converts those scores to ranks and compares groups without demanding a normal distribution or equal variances—useful when comparing satisfaction across wards, branches, or policy pilots.

Kruskal-Wallis vs. One-Way ANOVA

Aspect Kruskal-Wallis One-Way ANOVA
Data type Ordinal or non-normal Interval/ratio, normal
Compares Median ranks Means
Assumptions No normality needed Normality required
Power Lower Higher (if assumptions met)
Sample size Works with small n Needs larger n

The Kruskal-Wallis Test Procedure

Step 1: Combine and Rank All Data

Rank all observations from all groups together (1 = smallest)

Step 2: Calculate Rank Sums

Sum the ranks for each group: $R_1, R_2, R_3, …, R_k$

Step 3: Calculate Test Statistic

Intuition before the formula: If all groups come from the same distribution, their ranks should be “mixed” across the ordered list, so average ranks stay similar. If one ward or department systematically sits higher (or lower) on the ranked scale, its rank sum pulls away and H grows large.

\[H = \frac{12}{N(N+1)} \sum_{i=1}^{k} \frac{R_i^2}{n_i} - 3(N+1)\]

Where:

  • N = total sample size
  • k = number of groups
  • $n_i$ = size of group i
  • $R_i$ = sum of ranks in group i

Step 4: Compare with Chi-Square Distribution

For large samples, H follows χ² distribution with df = k - 1

Step-by-Step Example 1: Three Groups

Problem: Compare satisfaction scores (1-10) across three departments:

Finance HR IT
7 5 9
6 4 8
8 6 7
5 3 10
6 5 8

Test at α = 0.05 if satisfaction differs across departments.

Solution:

Step 1: State hypotheses

  • H₀: All groups have the same distribution
  • H₁: At least one group differs

Step 2: Rank all data combined

Value Group Rank
3 HR 1
4 HR 2
5 Finance 3.5
5 HR 3.5
5 HR 3.5
6 Finance 6
6 HR 6
6 Finance 6
7 Finance 8.5
7 IT 8.5
8 Finance 10.5
8 IT 10.5
8 IT 10.5
9 IT 13
10 IT 15

Wait, let me recount. We have 5 values per group = 15 total values.

Sorted data with ranks:

Rank Value Group
1 3 HR
2 4 HR
3.5 5 Finance
3.5 5 HR
5 5 HR
6.5 6 Finance
6.5 6 HR
8 6 Finance
9.5 7 Finance
9.5 7 IT
11.5 8 Finance
11.5 8 IT
13 8 IT
14 9 IT
15 10 IT

Note: For tied values, use average rank.

Step 3: Calculate rank sums

Group Values Ranks Sum (Rᵢ)
Finance 7,6,8,5,6 9.5, 6.5, 11.5, 3.5, 8 39
HR 5,4,6,3,5 3.5, 2, 6.5, 1, 5 18
IT 9,8,7,10,8 14, 11.5, 9.5, 15, 13 63

Check: 39 + 18 + 63 = 120 = N(N+1)/2 = 15(16)/2 = 120 ✓

Step 4: Calculate H statistic

\[H = \frac{12}{15(16)} \left[\frac{39^2}{5} + \frac{18^2}{5} + \frac{63^2}{5}\right] - 3(16)\] \[= \frac{12}{240} \left[\frac{1521}{5} + \frac{324}{5} + \frac{3969}{5}\right] - 48\] \[= 0.05 \times [304.2 + 64.8 + 793.8] - 48\] \[= 0.05 \times 1162.8 - 48 = 58.14 - 48 = 10.14\]

Step 5: Find critical value

  • df = k - 1 = 3 - 1 = 2
  • α = 0.05
  • From chi-square table: χ²* = 5.991

Step 6: Decision

  • H = 10.14 > 5.991
  • Reject H₀

Step 7: Conclusion At the 0.05 level of significance, there is sufficient evidence that job satisfaction differs significantly across departments. Rejecting H₀ means at least one group differs in distribution (or average ranks)—not that every pair of departments differs. Identifying which groups differ requires a post-hoc comparison.

Handling Tied Ranks

When values are tied, assign the average rank:

Original Ranks Tied Values Assigned Rank
3, 4 Both = 5 (3+4)/2 = 3.5
6, 7, 8 All = 6 (6+7+8)/3 = 7

Correction for Ties

For many ties, use correction factor:

\[H_c = \frac{H}{1 - \frac{\sum(t^3 - t)}{N^3 - N}}\]

Where t = number of tied observations in each group of ties.

Step-by-Step Example 2: Policy Effectiveness

Problem: Three different policies were implemented. Effectiveness scores:

Policy A Policy B Policy C
45 52 60
38 48 55
42 55 62
40 50 58

Test at α = 0.05 if policies differ in effectiveness.

Solution:

Step 1: Rank all 12 values

Rank Value Policy
1 38 A
2 40 A
3 42 A
4 45 A
5 48 B
6 50 B
7 52 B
8 55 B
9 55 C
10 58 C
11 60 C
12 62 C

Wait, there’s a tie at 55 (Policy B and C). Average rank = (8+9)/2 = 8.5

Corrected ranks:

Rank Value Policy
1 38 A
2 40 A
3 42 A
4 45 A
5 48 B
6 50 B
7 52 B
8.5 55 B
8.5 55 C
10 58 C
11 60 C
12 62 C

Step 2: Calculate rank sums

  • $R_A$ = 1 + 2 + 3 + 4 = 10
  • $R_B$ = 5 + 6 + 7 + 8.5 = 26.5
  • $R_C$ = 8.5 + 10 + 11 + 12 = 41.5

Check: 10 + 26.5 + 41.5 = 78 = 12(13)/2 = 78 ✓

Step 3: Calculate H

\[H = \frac{12}{12(13)} \left[\frac{10^2}{4} + \frac{26.5^2}{4} + \frac{41.5^2}{4}\right] - 3(13)\] \[= \frac{12}{156} \times \frac{100 + 702.25 + 1722.25}{4} - 39\] \[= 0.0769 \times 631.125 - 39 = 48.55 - 39 = 9.55\]

Step 4: Critical value

  • df = 3 - 1 = 2
  • α = 0.05
  • χ²* = 5.991

Step 5: Decision

  • H = 9.55 > 5.991
  • Reject H₀

Step 6: Conclusion At the 0.05 level of significance, there is sufficient evidence that the three policies differ in effectiveness—i.e., at least one policy’s rank distribution differs from the others.

Step-by-Step Example 3: Four Groups

Problem: Customer wait times (minutes) at four service counters:

Counter 1 Counter 2 Counter 3 Counter 4
5 8 12 15
6 9 11 18
4 7 13 16

Test at α = 0.05 if wait times differ across counters.

Solution:

Step 1: Rank all 12 values

Rank Value Counter
1 4 1
2 5 1
3 6 1
4 7 2
5 8 2
6 9 2
7 11 3
8 12 3
9 13 3
10 15 4
11 16 4
12 18 4

Step 2: Rank sums

  • $R_1$ = 1 + 2 + 3 = 6
  • $R_2$ = 4 + 5 + 6 = 15
  • $R_3$ = 7 + 8 + 9 = 24
  • $R_4$ = 10 + 11 + 12 = 33

Check: 6 + 15 + 24 + 33 = 78 ✓

Step 3: Calculate H

\[H = \frac{12}{12(13)} \left[\frac{36}{3} + \frac{225}{3} + \frac{576}{3} + \frac{1089}{3}\right] - 39\] \[= \frac{12}{156} \times \frac{36 + 225 + 576 + 1089}{3} - 39\] \[= 0.0769 \times 642 - 39 = 49.38 - 39 = 10.38\]

Step 4: Critical value

  • df = 4 - 1 = 3
  • α = 0.05
  • χ²* = 7.815

Step 5: Decision

  • H = 10.38 > 7.815
  • Reject H₀

Step 6: Conclusion Wait times differ significantly across the four counters: rejecting H₀ indicates at least one counter has a different wait-time distribution (in ranks), not that all four pairwise contrasts are significant.

Summary: When to Use Each Test

Test Groups Data Type Assumption
Independent t-test 2 Normal Parametric
Mann-Whitney U 2 Any Non-parametric
One-way ANOVA 3+ Normal Parametric
Kruskal-Wallis 3+ Any Non-parametric

Decision Flow Chart

flowchart TD
    A["How many groups?"]
    B{2 groups}
    C{3+ groups}
    D{"Data normal?"}
    E{"Data normal?"}
    F["t-test"]
    G["Mann-Whitney U"]
    H["One-way ANOVA"]
    I["Kruskal-Wallis"]

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

Practice Problems

Problem 1

Compare productivity scores across three shifts:

Morning Afternoon Night
82 78 70
85 75 72
80 80 68
88 76 74

Test at α = 0.05 if productivity differs across shifts.

Problem 2

Service quality ratings (1-5) at four branches:

Branch A Branch B Branch C Branch D
4 3 5 2
5 2 4 3
4 3 5 2

Test at α = 0.05 if ratings differ across branches.

Problem 3

Response times (seconds) for three software systems:

System X System Y System Z
2.1 3.5 1.8
2.5 3.2 2.0
2.3 3.8 1.5
2.0 3.0 1.9
2.4 3.4 1.7

Test at α = 0.01 if response times differ.

Common Mistakes

  • Using Kruskal–Wallis when one-way ANOVA is fine. If scores are roughly normal with similar variances and adequate n, ANOVA is usually more powerful; KW is the backup when those assumptions fail or data are ordinal.
  • Forgetting ties. Tied values must share the average rank; many ties may also need the tie correction to H. Skipping this understates or misstates the statistic.
  • Over-interpreting a significant H. Reject H₀ ⇒ at least one group differs in distribution/ranks—not that every group differs from every other.

Summary

Component Formula/Details
Purpose Compare 3+ groups (non-parametric)
Test statistic $H = \frac{12}{N(N+1)} \sum \frac{R_i^2}{n_i} - 3(N+1)$
Distribution Chi-square with df = k - 1
Tied ranks Use average rank
Assumptions Independent samples, ordinal data
Decision Reject H₀ if H > χ²*

Key Points to Remember

  1. Rank all data together from smallest to largest
  2. Handle ties by averaging ranks
  3. Check sum of all ranks = N(N+1)/2
  4. Use chi-square table for critical values
  5. Post-hoc tests needed to identify which groups differ

Congratulations!

You have completed Unit 5: Hypothesis Testing. You now understand:

  • ✅ Basic concepts of hypothesis testing
  • ✅ Large sample tests (Z-tests) for means and proportions
  • ✅ Small sample tests (t-tests) for means
  • ✅ Paired t-tests for dependent samples
  • ✅ Chi-square tests for independence and goodness of fit
  • ✅ Kruskal-Wallis test for multiple groups

Course Summary

This completes the MPA 509: Statistics for Public Administration course covering:

Unit Topics
Unit 1 Introduction to Statistics, Central Tendency, Dispersion
Unit 2 Correlation and Simple Linear Regression
Unit 3 Probability Theory and Distributions
Unit 4 Estimation and Sampling
Unit 5 Hypothesis Testing

Good luck with your examinations!