Friedman Test

Created:October 8, 2024

Last Updated:January 13, 2026

The Friedman Test Calculator helps you perform non-parametric analysis for comparing three or more matched or repeated measurements. 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 repeated measures ANOVA, such as normality or sphericity. Common applications include comparing multiple treatments across the same subjects, analyzing repeated measurements over time, or evaluating preferences across different conditions. for a quick example.

Important considerations before using Friedman Test:

The test requires at least 2 blocks/subjects and at least 3 treatments/conditions.
Data should be organized with subjects as blocks and treatments as conditions being compared.
The dependent variable should be at least ordinal (can be ranked).
Blocks should be independent of each other, but measurements within blocks are related.

Calculator

1. Load Your Data

2. Select Columns & Options

Select group/block column:

Select treatment column:

Select value (response) column:

Significance Level:

Related Calculators

One-Way ANOVA Calculator

Repeated Measures ANOVA Calculator

Wilcoxon Signed-Rank Test

Chi-Square Test of Independence Calculator

Learn More

Friedman Test

Definition

Friedman Test is a non-parametric alternative to the one-way repeated measures ANOVA. It tests for differences between three or more matched or paired groups, without requiring normality in the data distribution. The test statistic is based on the ranks of the data values within each block.

Test Statistic

Q = \frac{12}{nk(k+1)}\sum_{j=1}^k R_j^2 - 3n(k+1)

Where:

$n$ = number of blocks (subjects)
$k$ = number of treatments
$R_j$ = total rank of treatment $j$
$Q \sim \chi^2_{k-1}$ when $n > 15$ or $k > 4$
$Q \sim F_{k-1, (k-1)(n-1)}$ for better accuracy when $n \leq 15$ and $k \leq 4$

Key Assumptions

Matched Groups: Data must be paired or matched

Ordinal Data: Measurements must be ordinal or continuous

Independent Blocks: Subjects must be independent

Practical Example

Step 1: State the Data

Scores for three treatments across five subjects:

Subject	Treatment A	Treatment B	Treatment C
1	8	6	5
2	7	7	6
3	9	8	6
4	6	5	4
5	7	6	5

Step 2: State Hypotheses

$H_0$ : No difference between treatments
$H_a$ : At least two treatments differ
$\alpha = 0.05$

Step 3: Rank Within Blocks

Subject	A	B	C
1	3	2	1
2	2.5	2.5	1
3	3	2	1
4	3	2	1
5	3	2	1
$R_j$	14.5	10.5	5

Step 4: Calculate Test Statistic

1. Basic Friedman statistic with b = 5 subjects and k = 3 treatments:

Q = \frac{12}{5 \cdot 3(3+1)}(14.5^2 + 10.5^2 + 5^2) - 3 \cdot 5(3+1) = 9.1

2. Correction for ties:

In our data, we have one tie (7,7) in Subject 2's scores.

C = 1 - \frac{\sum T}{n k(k^2-1)}

where $T = (t^3 - t)$ for each group of $t$ tied ranks

One pair of ties gives $T = (2^3 - 2) = 6$

C = 1 - \frac{6}{5 \cdot 3(9-1)} = 0.95

3. Adjusted test statistic:

Q_{\text{adj}} = \frac{Q}{C} = \frac{9.1}{0.95} = 9.58

Step 5: Draw Conclusion

Critical value for $\chi^2$ with $df = 2$ at $\alpha = 0.05$ is $5.991$ . Since $Q = 9.58 \gt 5.991$ , we reject $H_0$ .

There is sufficient evidence to conclude that there are significant differences between at least two treatments.

With $p$ -value = $0.008317 \lt 0.05$ , there is strong evidence to conclude that there are significant differences between at least two treatments.

Effect Size

Kendall's W

Kendall's coefficient of concordance ( $W$ ) ranges from $0$ (no agreement) to $1$ (complete agreement):

W = \frac{Q}{n(k-1)}

For our example:

W = \frac{9.58}{5(3-1)} = 0.958

Interpretation guidelines:

$W \lt 0.3$ : Weak agreement
$0.3 \leq W \leq 0.5$ : Moderate agreement
$W \gt 0.5$ : Strong agreement

Code Examples

library(tidyverse)

# Create the data frame
data <- tibble(
  Subject = c(
    "Subject 1", "Subject 1", "Subject 1",
    "Subject 2", "Subject 2", "Subject 2",
    "Subject 3", "Subject 3", "Subject 3",
    "Subject 4", "Subject 4", "Subject 4",
    "Subject 5", "Subject 5", "Subject 5"
  ),
  Treatment = c(
    "Treatment A", "Treatment B", "Treatment C",
    "Treatment A", "Treatment B", "Treatment C",
    "Treatment A", "Treatment B", "Treatment C",
    "Treatment A", "Treatment B", "Treatment C",
    "Treatment A", "Treatment B", "Treatment C"
  ),
  Score = c(8, 6, 5, 7, 7, 6, 9, 8, 6, 6, 5, 4, 7, 6, 5)
)

# Perform Friedman test
friedman_result <- friedman.test(Score ~ Treatment | Subject, data = data)
print(friedman_result)

Python

import pandas as pd
from scipy.stats import friedmanchisquare
import numpy as np

# Create the data frame
data = pd.DataFrame({
    'Subject': [
        'Subject 1', 'Subject 1', 'Subject 1',
        'Subject 2', 'Subject 2', 'Subject 2',
        'Subject 3', 'Subject 3', 'Subject 3',
        'Subject 4', 'Subject 4', 'Subject 4',
        'Subject 5', 'Subject 5', 'Subject 5'
    ],
    'Treatment': [
        'Treatment A', 'Treatment B', 'Treatment C',
        'Treatment A', 'Treatment B', 'Treatment C',
        'Treatment A', 'Treatment B', 'Treatment C',
        'Treatment A', 'Treatment B', 'Treatment C',
        'Treatment A', 'Treatment B', 'Treatment C'
    ],
    'Score': [8, 6, 5, 7, 7, 6, 9, 8, 6, 6, 5, 4, 7, 6, 5]
})

# Reshape data to wide format for Friedman test
# Each row represents a subject, each column a treatment
data_wide = data.pivot(index='Subject', columns='Treatment', values='Score')

# Perform Friedman test
statistic, p_value = friedmanchisquare(
    data_wide['Treatment A'],
    data_wide['Treatment B'],
    data_wide['Treatment C']
)

print(f'Friedman chi-squared = {statistic:.4f}')
print(f'p-value = {p_value:.6f}')

# Calculate Kendall's W (effect size)
n = len(data_wide)  # number of subjects
k = len(data_wide.columns)  # number of treatments
kendalls_w = statistic / (n * (k - 1))
print(f"Kendall's W = {kendalls_w:.7f}")

Verification

Friedman Test

Created:October 8, 2024

Last Updated:January 13, 2026

Important considerations before using Friedman Test:

The test requires at least 2 blocks/subjects and at least 3 treatments/conditions.
Data should be organized with subjects as blocks and treatments as conditions being compared.
The dependent variable should be at least ordinal (can be ranked).
Blocks should be independent of each other, but measurements within blocks are related.

Calculator

1. Load Your Data

2. Select Columns & Options

Select group/block column:

Select treatment column:

Select value (response) column:

Significance Level:

Related Calculators

One-Way ANOVA Calculator

Repeated Measures ANOVA Calculator

Wilcoxon Signed-Rank Test

Chi-Square Test of Independence Calculator

Learn More

Friedman Test

Definition

Test Statistic

Q = \frac{12}{nk(k+1)}\sum_{j=1}^k R_j^2 - 3n(k+1)

Where:

$n$ = number of blocks (subjects)
$k$ = number of treatments
$R_j$ = total rank of treatment $j$
$Q \sim \chi^2_{k-1}$ when $n > 15$ or $k > 4$
$Q \sim F_{k-1, (k-1)(n-1)}$ for better accuracy when $n \leq 15$ and $k \leq 4$

Key Assumptions

Matched Groups: Data must be paired or matched

Ordinal Data: Measurements must be ordinal or continuous

Independent Blocks: Subjects must be independent

Practical Example

Step 1: State the Data

Scores for three treatments across five subjects:

Subject	Treatment A	Treatment B	Treatment C
1	8	6	5
2	7	7	6
3	9	8	6
4	6	5	4
5	7	6	5

Step 2: State Hypotheses

$H_0$ : No difference between treatments
$H_a$ : At least two treatments differ
$\alpha = 0.05$

Step 3: Rank Within Blocks

Subject	A	B	C
1	3	2	1
2	2.5	2.5	1
3	3	2	1
4	3	2	1
5	3	2	1
$R_j$	14.5	10.5	5

Step 4: Calculate Test Statistic

1. Basic Friedman statistic with b = 5 subjects and k = 3 treatments:

Q = \frac{12}{5 \cdot 3(3+1)}(14.5^2 + 10.5^2 + 5^2) - 3 \cdot 5(3+1) = 9.1

2. Correction for ties:

In our data, we have one tie (7,7) in Subject 2's scores.

C = 1 - \frac{\sum T}{n k(k^2-1)}

where $T = (t^3 - t)$ for each group of $t$ tied ranks

One pair of ties gives $T = (2^3 - 2) = 6$

C = 1 - \frac{6}{5 \cdot 3(9-1)} = 0.95

3. Adjusted test statistic:

Q_{\text{adj}} = \frac{Q}{C} = \frac{9.1}{0.95} = 9.58

Step 5: Draw Conclusion

Critical value for $\chi^2$ with $df = 2$ at $\alpha = 0.05$ is $5.991$ . Since $Q = 9.58 \gt 5.991$ , we reject $H_0$ .

There is sufficient evidence to conclude that there are significant differences between at least two treatments.

With $p$ -value = $0.008317 \lt 0.05$ , there is strong evidence to conclude that there are significant differences between at least two treatments.

Effect Size

Kendall's W

Kendall's coefficient of concordance ( $W$ ) ranges from $0$ (no agreement) to $1$ (complete agreement):

W = \frac{Q}{n(k-1)}

For our example:

W = \frac{9.58}{5(3-1)} = 0.958

Interpretation guidelines:

$W \lt 0.3$ : Weak agreement
$0.3 \leq W \leq 0.5$ : Moderate agreement
$W \gt 0.5$ : Strong agreement

Code Examples

library(tidyverse)

# Create the data frame
data <- tibble(
  Subject = c(
    "Subject 1", "Subject 1", "Subject 1",
    "Subject 2", "Subject 2", "Subject 2",
    "Subject 3", "Subject 3", "Subject 3",
    "Subject 4", "Subject 4", "Subject 4",
    "Subject 5", "Subject 5", "Subject 5"
  ),
  Treatment = c(
    "Treatment A", "Treatment B", "Treatment C",
    "Treatment A", "Treatment B", "Treatment C",
    "Treatment A", "Treatment B", "Treatment C",
    "Treatment A", "Treatment B", "Treatment C",
    "Treatment A", "Treatment B", "Treatment C"
  ),
  Score = c(8, 6, 5, 7, 7, 6, 9, 8, 6, 6, 5, 4, 7, 6, 5)
)

# Perform Friedman test
friedman_result <- friedman.test(Score ~ Treatment | Subject, data = data)
print(friedman_result)

Python

import pandas as pd
from scipy.stats import friedmanchisquare
import numpy as np

# Create the data frame
data = pd.DataFrame({
    'Subject': [
        'Subject 1', 'Subject 1', 'Subject 1',
        'Subject 2', 'Subject 2', 'Subject 2',
        'Subject 3', 'Subject 3', 'Subject 3',
        'Subject 4', 'Subject 4', 'Subject 4',
        'Subject 5', 'Subject 5', 'Subject 5'
    ],
    'Treatment': [
        'Treatment A', 'Treatment B', 'Treatment C',
        'Treatment A', 'Treatment B', 'Treatment C',
        'Treatment A', 'Treatment B', 'Treatment C',
        'Treatment A', 'Treatment B', 'Treatment C',
        'Treatment A', 'Treatment B', 'Treatment C'
    ],
    'Score': [8, 6, 5, 7, 7, 6, 9, 8, 6, 6, 5, 4, 7, 6, 5]
})

# Reshape data to wide format for Friedman test
# Each row represents a subject, each column a treatment
data_wide = data.pivot(index='Subject', columns='Treatment', values='Score')

# Perform Friedman test
statistic, p_value = friedmanchisquare(
    data_wide['Treatment A'],
    data_wide['Treatment B'],
    data_wide['Treatment C']
)

print(f'Friedman chi-squared = {statistic:.4f}')
print(f'p-value = {p_value:.6f}')

# Calculate Kendall's W (effect size)
n = len(data_wide)  # number of subjects
k = len(data_wide.columns)  # number of treatments
kendalls_w = statistic / (n * (k - 1))
print(f"Kendall's W = {kendalls_w:.7f}")

Verification

Subject

Treatment A

Treatment B

Treatment C

Subject

2.5

R_j

14.5

10.5

library(tidyverse) # Create the data frame data <- tibble( Subject = c( "Subject 1", "Subject 1", "Subject 1", "Subject 2", "Subject 2", "Subject 2", "Subject 3", "Subject 3", "Subject 3", "Subject 4", "Subject 4", "Subject 4", "Subject 5", "Subject 5", "Subject 5" ), Treatment = c( "Treatment A", "Treatment B", "Treatment C", "Treatment A", "Treatment B", "Treatment C", "Treatment A", "Treatment B", "Treatment C", "Treatment A", "Treatment B", "Treatment C", "Treatment A", "Treatment B", "Treatment C" ), Score = c(8, 6, 5, 7, 7, 6, 9, 8, 6, 6, 5, 4, 7, 6, 5) ) # Perform Friedman test friedman_result <- friedman.test(Score ~ Treatment | Subject, data = data) print(friedman_result)

import pandas as pd from scipy.stats import friedmanchisquare import numpy as np # Create the data frame data = pd.DataFrame({ 'Subject': [ 'Subject 1', 'Subject 1', 'Subject 1', 'Subject 2', 'Subject 2', 'Subject 2', 'Subject 3', 'Subject 3', 'Subject 3', 'Subject 4', 'Subject 4', 'Subject 4', 'Subject 5', 'Subject 5', 'Subject 5' ], 'Treatment': [ 'Treatment A', 'Treatment B', 'Treatment C', 'Treatment A', 'Treatment B', 'Treatment C', 'Treatment A', 'Treatment B', 'Treatment C', 'Treatment A', 'Treatment B', 'Treatment C', 'Treatment A', 'Treatment B', 'Treatment C' ], 'Score': [8, 6, 5, 7, 7, 6, 9, 8, 6, 6, 5, 4, 7, 6, 5] }) # Reshape data to wide format for Friedman test # Each row represents a subject, each column a treatment data_wide = data.pivot(index='Subject', columns='Treatment', values='Score') # Perform Friedman test statistic, p_value = friedmanchisquare( data_wide['Treatment A'], data_wide['Treatment B'], data_wide['Treatment C'] ) print(f'Friedman chi-squared = {statistic:.4f}') print(f'p-value = {p_value:.6f}') # Calculate Kendall's W (effect size) n = len(data_wide) # number of subjects k = len(data_wide.columns) # number of treatments kendalls_w = statistic / (n * (k - 1)) print(f"Kendall's W = {kendalls_w:.7f}")

Subject

Treatment A

Treatment B

Treatment C

Subject

2.5

R_j

14.5

10.5

Friedman Test

Important considerations before using Friedman Test:

Calculator

1. Load Your Data

2. Select Columns & Options

Related Calculators

One-Way ANOVA Calculator

Repeated Measures ANOVA Calculator

Wilcoxon Signed-Rank Test

Chi-Square Test of Independence Calculator

Learn More

Friedman Test

Definition

Test Statistic

Key Assumptions

Practical Example

Step 1: State the Data

Step 2: State Hypotheses

Step 3: Rank Within Blocks

Step 4: Calculate Test Statistic

Step 5: Draw Conclusion

Effect Size

Kendall's W

Code Examples

Verification

View Verification Details

Friedman Test

Important considerations before using Friedman Test:

Calculator

1. Load Your Data

2. Select Columns & Options

Related Calculators

One-Way ANOVA Calculator

Repeated Measures ANOVA Calculator

Wilcoxon Signed-Rank Test

Chi-Square Test of Independence Calculator

Learn More

Friedman Test

Definition

Test Statistic

Key Assumptions

Practical Example

Step 1: State the Data

Step 2: State Hypotheses

Step 3: Rank Within Blocks

Step 4: Calculate Test Statistic

Step 5: Draw Conclusion

Effect Size

Kendall's W

Code Examples

Verification

View Verification Details

Friedman Test

Important considerations before using Friedman Test:

Calculator

1. Load Your Data

2. Select Columns & Options

Related Calculators

One-Way ANOVA Calculator

Repeated Measures ANOVA Calculator

Wilcoxon Signed-Rank Test

Chi-Square Test of Independence Calculator

Learn More

Friedman Test

Definition

Test Statistic

Key Assumptions

Practical Example

Step 1: State the Data

Step 2: State Hypotheses

Step 3: Rank Within Blocks

Step 4: Calculate Test Statistic

Step 5: Draw Conclusion

Effect Size

Kendall's W

Code Examples

Verification

View Verification Details

Friedman Test

Important considerations before using Friedman Test: