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Non-parametric

Kruskal-Wallis H Test — Three Groups Non-parametric ANOVA

Non-parametric alternative to one-way ANOVA. Rank all observations, compute H statistic, compare to chi-square critical value.

📊 H(2) ≥ χ²(2,.05) = 5.991
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🔢 4 · Non-parametric Tests

4a · Mann-Whitney U Test
Non-parametric
Research Question: Do satisfaction scores differ between two departments (n₁=5, n₂=5)? Data is ordinal/skewed.
Dept A68249
Dept B751031
U Statistic
$$U_1 = n_1 n_2 + \frac{n_1(n_1+1)}{2} - R_1, \quad U = \min(U_1, U_2)$$
  1. 1
    Rank all 10 values combined:
    Value12345678910
    Rank12345678910
    GroupBABABABAAB
  2. 2
    $$R_1 = 2+4+6+8+9 = 29 \quad (\text{sum of ranks for Dept A})$$ $$U_1 = 5 \times 5 + \frac{5 \times 6}{2} - 29 = 25 + 15 - 29 = 11$$ $$U_2 = n_1 n_2 - U_1 = 25 - 11 = 14, \quad U = \min(11, 14) = 11$$
  3. 3
    Normal approximation (n>10 each; shown for demonstration):
    $$z = \frac{U - n_1n_2/2}{\sqrt{n_1 n_2(n_1+n_2+1)/12}} = \frac{11 - 12.5}{\sqrt{25 \times 11/12}} = \frac{-1.5}{4.787} = -0.313$$
  4. 4
    Effect size r:
    $$r = \frac{|z|}{\sqrt{N}} = \frac{0.313}{\sqrt{10}} = 0.099 \quad (\text{small})$$
11
U
.754
p (exact)
0.099
r
⚪ Fail to Reject H₀ — No significant difference between departments, U=11, p=.754.
APA-7
A Mann-Whitney U test revealed no significant difference in satisfaction scores between Department A (Mdn=6) and Department B (Mdn=6), U=11, z=−0.31, p=.754, r=.10.
4b · Kruskal-Wallis H Test
Non-parametric
Research Question: Do pain levels differ across three clinics (non-normal data)?
Clinic AClinic BClinic C
3,5,4,2,67,9,8,10,65,4,6,3,5
Kruskal-Wallis H
$$H = \frac{12}{N(N+1)}\sum_{j=1}^{k}\frac{R_j^2}{n_j} - 3(N+1)$$
  1. 1
    Rank all 15 values, sum ranks per group:
    $$R_A = 2+6.5+4+1+9 = 22.5, \quad R_B = 11+14+13+15+9 = 62, \quad R_C = 6.5+4+9+2+6.5 = 28$$
    (ties at value 6 receive average rank (9+10)/2=9.5; value 5: (5+6+7)/3=6)
  2. 2
    $$H = \frac{12}{15 \times 16}\left(\frac{22.5^2}{5} + \frac{62^2}{5} + \frac{28^2}{5}\right) - 3(16)$$ $$= \frac{12}{240}(101.25 + 768.8 + 156.8) - 48 = 0.05(1026.85) - 48 = 51.34 - 48 = 3.34$$
  3. 3
    Compare to χ² distribution with df=k−1=2: χ²(2,0.05)=5.99. Since H=3.34 < 5.99 → fail to reject H₀.
3.34
H
2
df
.188
p
⚪ Fail to Reject H₀ — Pain levels do not differ significantly across clinics, H(2)=3.34, p=.188.
APA-7
A Kruskal-Wallis test indicated no significant difference in pain levels across the three clinics, H(2) = 3.34, p = .188, η² = .10.

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