The Mann-Whitney U Test Calculator helps you perform non-parametric analysis to compare two independent groups. It determines whether samples originate from the same distribution by analyzing the ranks of the data rather than the raw values. This makes it particularly useful when your data violates the assumptions of the independent t-test, such as normality or equal variances. Common applications include comparing treatment outcomes between two groups, analyzing differences in measurements, or evaluating performance metrics across two departments. Click here to populate the sample data for a quick example.
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Mann-Whitney U Test
Definition
Mann-Whitney U Test (also known as Wilcoxon rank-sum test) is a non-parametric alternative to the independent t-test. It compares two independent groups by analyzing the rankings of the data rather than the raw values.
Formula
U Statistics:
Where:
- = sample sizes
- = sum of ranks for group 1 and group 2
- = the minimum of U₁ and U₂
Standardized Test Statistic:
Correction for Ties:
When ties occur in the data, a correction is applied to the standard deviation:
- = total sample size ()
- = number of tied groups
- = number of tied values in the group
Continuity Correction:
The 0.5 term is the continuity correction, which improves the approximation to the normal distribution.
Effect Size
Effect size r for Mann-Whitney U test:
Where:
- = standardized test statistic
- = total sample size
Interpretation:
- Small effect:
- Medium effect:
- Large effect:
Key Assumptions
Common Pitfalls
- Using with paired/dependent samples (use Wilcoxon signed-rank instead)
- Interpreting results as comparing means rather than distributions
- Not checking for ties in the data when using exact calculations
Example without Ties
Problem Statement
A researcher wants to test if a treatment affects test scores. Students were randomly assigned to either control or treatment group, and their test scores were recorded:
Control group: 45, 47, 43, 44
Treatment group: 52, 48, 54, 50
Sample sizes:
Step 1: State Hypotheses
: The distributions of scores are the same for both groups.
: The distributions of scores differ between the two groups.
Step 2: Combine and Rank Data
Group | Value | Rank |
---|---|---|
Control | 43 | 1 |
Control | 44 | 2 |
Control | 45 | 3 |
Control | 47 | 4 |
Treatment | 48 | 5 |
Treatment | 50 | 6 |
Treatment | 52 | 7 |
Treatment | 54 | 8 |
Step 3: Calculate Rank Sums
Control rank sum (R₁): 1 + 2 + 3 + 4 = 10
Treatment rank sum (R₂): 5 + 6 + 7 + 8 = 26
Step 4: Calculate U Statistics
Step 5: Calculate Z-Statistic
Step 6: Draw Conclusion
The -value for this test is . Since -value , we reject . There is sufficient evidence to conclude that the treatment and control groups have different distributions of scores.
Example with Ties
Problem Statement
A researcher wants to compare the effectiveness of two different teaching methods on student performance. Students were randomly assigned to either Method A or Method B, and their test scores were recorded:
Method A: 85, 92, 78, 90, 85, 76, 88
Method B: 79, 85, 81, 89, 84, 82, 85
Note that there are ties in the data: the score 85 appears three times (once in Method A and twice in Method B).
Step 1: State Hypotheses
: The distributions of scores are the same for both teaching methods.
: The distributions of scores differ between the two teaching methods.
Step 2: Combine and Rank Data
Group | Value | Rank |
---|---|---|
A | 76 | 1 |
A | 78 | 2 |
B | 79 | 3 |
B | 81 | 4 |
B | 82 | 5 |
B | 84 | 6 |
A | 85 | 8 |
B | 85 | 8 |
B | 85 | 8 |
A | 88 | 10 |
B | 89 | 11 |
A | 90 | 12 |
A | 92 | 13 |
Note: For the three tied values of 85, we assign the average rank of (7+8+9)/3 = 8 to each.
Step 3: Calculate Rank Sums
Method A rank sum (R₁): 1 + 2 + 8 + 10 + 12 + 13 = 46
Method B rank sum (R₂): 3 + 4 + 5 + 6 + 8 + 8 + 11 = 45
Step 4: Calculate U Statistics
Step 5: Apply Correction for Ties
Since we have ties (three values of 85), we need to apply the correction to the standard deviation:
Where N = 14, and we have one tied group with t = 3 (for the value 85).
Step 6: Calculate Z-Statistic (no continuity correction)
Step 7: Draw Conclusion
The -value for this test is (two-sided). Since -value , we fail to reject . There is insufficient evidence to conclude that the two teaching methods result in different distributions of test scores.
Step 8: Calculate Effect Size
This represents a small effect size, which aligns with our failure to reject the null hypothesis.