Freedman's Test Demo

Friedman's test applies a non-parametric approach that provides a statistical measure for difference between groups. Consider the following data in wave1:

Group1Group2Group3Group4
Block17.3410.69.489.15
Block279.596.17
Block36.512.8110.711.35
Block44.7312.5513.245.7
Block56.5110.115.714.22

To run Friedman's test on this data, execute the command:

StatsFriedmanTest/T=1/Q /RW wave1

The results are stored in the wave M_FriedmanRanks and Friedman Test table:

RowsColsStat.Crit.appox. ChiI_D_FfI_D_Crit.Conclusion
549.247.8257.814734.81253.490290

and in the wave M_FriedmanRanks where each element represents the rank of the corresponding datum:

1432
2431
1423
1342
1243

In this case Friedman's statistic (9.24) exceeds the critical value (7.825) so the null hypothesis H0 (no difference between the groups) must be rejected. The Chi-square approximation and the Iman and Davenport Ff as well as the critical values all agree with this conclusion.

Although the operation does not provide the associated P-value, you can compute it using the command:

Print 1-statsFriedmanCDF(9.24,4,5,0,1)

The resulting P-value is: 0.0166709.

Applying Friedman's test to a second data set of the same dimensions in wave2:

Group1Group2Group3Group4
Block15.136.257.2814.17
Block26.1812.937.5212.08
Block312.939.398.8110.88
Block410.468.8813.084.4
Block59.2113.6414.3710.69

StatsFriedmanTest/T=1/Q /RW wave2

The results are found in the Friedman Test table:

RowsColsStat.Crit.appox. ChiI_D_FfI_D_Crit.Conclusion
541.087.8257.814730.2327593.490291

and in the wave M_FriedmanRanks:

1234
1423
4213
3241
1342

The statistic (1.08) is smaller than the critical value (7.825) and so H0 can't be rejected. Comparing the two distribution of ranks it is evident that in this case they appear to be distributed more randomly which is consistent with the test's conclusion.

You can compute associated P-value using the command:

Print 1-statsFriedmanCDF(1.08,4,5,0,1)
P = 0.85656.