**Friedman Test: **It is a non-parametric test alternative to the one way ANOVA with repeated measures. It tries to determine if subjects changed significantly across occasions/conditions. For example:- Problem-solving ability of a set of people is the same or different in Morning, Afternoon, Evening. It is used to test for **differences between groups **when the **dependent variable is ordinal. **This test is particularly useful when the sample size is very small.

**Elements of Friedman Test**

**One group**that is measured on**three or more****blocks**of**measures**overtime**/experimental conditions**.**One dependent variable**which can be Ordinal, Interval or Ratio.

**Assumptions of Friedman Test**

- The group is a random sample from the population.
- Samples are not normally distributed.

**Null and Alternate Hypothesis of Friedman Test**

**Null Hypothesis: **There is no significant difference between the given conditions of measurement OR the probability distributions for all the conditions are the same. (Medians are same)

**Alternate Hypothesis: **At least 2 of them differ from each other.

H_{0}: M_{1}= M_{2}= M_{3}= ..... M_{k}; M= Median H_{1}: At least two of them show significant difference.

**Test Statistic for Friedman Test**

n = total number of subjects/participants. k = total number of blocks to be measured. R_{i}= sum of ranks of all subjects for a block i

**Decision Rule for Friedman Test**

You can make the decision on the basis of the below-mentioned rules-

**Calculated Value vs Table Value:**If F_{R}is greater than the critical value limits reject the Null Hypothesis. Otherwise, accept the Null Hypothesis.**P-Value Approach:**Compare the P-Value with Alpha ( Level of Significance). If the p-value is less than or equal to alpha then reject the Null Hypothesis.

**Post Hoc Analysis: **You can find out whether there is a difference in any given pair of experimental conditions if the null hypothesis is rejected using the Post Hoc analysis which can be done using Wilcoxon signed-rank test, Conover’s test etc.. In Wilcoxon test, you can get the results for all pairs also but you will have to make a Bonferroni correction which will change the level of significance to **Given level of significance/total number of pairs.**

**Steps to perform Friedman Test: **

Let us take an example to understand how to perform this test.

**Example: **7 random people were given 3 different drugs and for each person, the reaction time corresponding to the drugs were noted. Test the claim at the 5% significance level that all the 3 drugs have the same probability distribution.

Drug A | Drug B | Drug C | |
---|---|---|---|

1 | 1.24 | 1.50 | 1.62 |

2 | 1.71 | 1.85 | 2.05 |

3 | 1.37 | 2.12 | 1.68 |

4 | 2.53 | 1.87 | 2.62 |

5 | 1.23 | 1.34 | 1.51 |

6 | 1.94 | 2.33 | 2.86 |

7 | 1.72 | 1.43 | 2.86 |

**Step 1: **Define NULL and Alternate Hypothesis

H_{0 }: All three drugs have the same probability distribution. M_{A}= M_{B}= M_{C}H_{1}: At least two of them differ from each other.

**Step 2: **State Alpha (Level of Significance)

Alpha = 0.05

**Step 3:** Calculate Degrees of Freedom

DF = K-1 ; K = number of blocks to be measured. Here ,DF = 3-1 =2.

**Step 4: ** Find out the Critical Chi-Square Value.

Use this table to find out the critical chi-square value for alpha = 0.05 and DF = 2.

X^{2}= 5.991

**Step 5: **State Decision Rule

You can check for any of the two rules –

1)If F_{R}is greater than 5.991 , reject the Null Hypothesis.

**Step 6: **Assign Ranks for the drugs corresponding to each person and find the sum.

Ranks will be in Ascending order.

Ranks | |||
---|---|---|---|

Drug A | Drug B | Drug C | |

1 | 1 | 2 | 3 |

2 | 1 | 2 | 3 |

3 | 1 | 3 | 2 |

4 | 2 | 1 | 3 |

5 | 1 | 2 | 3 |

6 | 1 | 2 | 3 |

7 | 2 | 1 | 3 |

∑ = 9 |
∑ = 13 |
∑ = 20 |

**Note:** If in the same row 2 or more columns have the same value then the rank assigned to them is the average of the ranks they get. For example: If a row has 2 columns with value x and the ranks which they get are 4 and 5. Then both the columns will be assigned with a rank of (4+5)/2 which is 4.5.

**Step 7: **Calculate Test Statistic

F_{R}= 8.857

**Step 8: **State Results

Since F_{R}is greater than 5.991 , We reject the Null Hypothesis.

**Step 9: **State Conclusion

All the three drugs do not have the same probability distribution.

You can apply the Post Hoc analysis with Wilcoxon Test to know which pairs have a significant difference between them.

Here,

Total number of pairs can be 3 (Drug A - Drug B , Drug B - Drug C , Drug A - Drug C). The new level of significance to be considered for each pair will be 0.05/3 = 0.0166.

**Implementation of Friedman Test using R**

## R

`# R program to illustrate ` `# Friedman Test ` ` ` `#input the data ` `y <- ` `matrix` `(` `c` `(1.24,1.50,1.62, ` ` ` `1.71,1.85,2.05, ` ` ` `1.37,2.12,1.68, ` ` ` `2.53,1.87,2.62, ` ` ` `1.23,1.34,1.51, ` ` ` `1.94,2.33,2.86, ` ` ` `1.72,1.43,2.86), ` `nrow = 7, byrow = ` `TRUE` `, ` `dimnames = ` `list` `(Person= ` `as.character` `(1:7),Drugs = ` `c` `(` `"Drug A"` `,` `"Drug B"` `,` `"Drug C"` `))) ` ` ` `#display the sample data ` `print` `(y) ` |

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**Output:**

## R

`#perform friedman test on the sample ` `result = ` `friedman.test` `(y) ` `print` `(result)` |

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**Output:**

As the p-value is less than the significance level (5%) it can be concluded that there are significant differences in the probability distribution.

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