Example of a Friedman Test

In a poll 10 subjects rated 4 different paintings on a scale from 0 (don’t like it at all) to 5 (like it very much). The following table shows the data and ranks for all Subjects and paintings:

	Painting 1	Rank	Painting 2	Rank	Paintin 3	Rank	Painting 4	Rank
1	0	1	5	4	1	2	4	3
2	3	2	4	3	2	1	5	4
3	1	1	4	3.5	3	2	4	3.5
4	4	4	2	1.5	2	1.5	3	3
5	2	1.5	2	1.5	4	4	3	3
6	0	1	3	2	5	3.5	5	3.5
7	3	2.5	1	1	3	2.5	4	4
8	5	3.5	3	2	1	1	5	3.5
9	1	1	5	4	2	2	4	3
10	2	2	4	4	0	1	3	3
Ti		19.5		26.5		20.5		33.5
		1.95		2.65		2.05		3.35

We then get for the total sum of ranks:

We have tied ranks so the Chi-square value is:

whereas

so the Chi-square value is:

The critical 5% Chi-square for 3 degrees of freedom is 7.81. The Chi-square test does reject H0. There must be at least one painting which is differentially rated compared to at least one other painting.

For Conovers’ F value we get:

and the 5% critical F value with 3 and 27 degrees of freedom is 2.96. The F Test does reject H0. There must be at least one painting which is differentially rated compared to at least one other painting.

The critical Difference of the Mean Ranks  is:

The following table shows the differences of the mean ranks:

Post hoc paired comparisons reveal that painting four is rated differentially compared to paintings 1 and 3.

BrightStat Output of this example