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Friedman’s test

k$ 3 related samples

data must at least be ordinal

talking about a situation where we want to compare T objects matched in B attributes

H0: M1=M2=....MT

roughly equivalent to 2-way ANOVA with 1 observation per cell

where Mi is the median object effect

test statistic is

a large value of S calls for rejection of the null

for larger sample sizes (> 15) we calculate S and substitute into Q=12S/BT(T+1)

Q is approximately P2 distributed with df=T-1

example

suppose we have rankings of regions on 4 criteria on a scale of 1 to 10
 

employment

opportunity

Scenery

weather

hospitality

A

6

4

8

8.5

B

8

7

6

7

C

7

5

4

6

D

5

6

8

4

E

2

3

6

4

F

5

6

9

9

G

8

7

4

3

 

note this data is ordinal

if we want to compare the median overall rating of the 7 regions we should rank these ratings 1 to 7 for each attribute separately

this means the objects are the regions and the criteria are the attributes

 

so T=7 B=4

we get
 

employment

Opportunity

Scenery

weather

hospitality

total

A

4

6

2.5

2

14.5

B

1.5

1.5

4.5

3

10.5

C

3

5

6.5

4

18.5

D

5.5

3.5

2.5

5.5

17.0

E

7

7

4.5

5.5

24.0

F

5.5

3.5

1

1

11.0

G

1.5

1.5

6.5

7

16.5

 

S=G R2t - B2T(T+1)2/4=

14.52+10.52+18.52+172+242+112+16.52=42(7)(7+1)2)/4=

210.25+110.25+342.25+289+576+121+272.25=

1921-16(7)(64)/4=1921-1792=129

S=129

Q=12S/BT(T+1)=12(129)/(4)(7)(8)=1548/224=

Q=6.91

df=T-1=6

P2=11.07 "=0.05

Q<11.07 so we accept H0

the regions are not different on the basis of the rankings on the 4 criteria