Here is an example, using Guilford's data, of data that lack sufficient connection to yield unambiguous measure estimates. It reports subsets. In this example, pairs of judges rate subsets of examinees, but without crossing (overlap). Did Betty get the highest score because she was the most able or because her judges were more lenient? The data can't tell us. There is more about this in the discussion of subset connectedness.
Facets specifications and data (in file Subsets.txt):
; subsets.txt
Title = Ratings of Scientists (edited to illustrate ambiguity in measurement)
Facets = 3 ; three facets: judges, examinees, items
Umean = 50, 10 ; user-scaling = 50 +logit*10
Positive = 1,2,3 ; the examinees have greater creativity with greater score
Non-centered = 1 ; examinees and items are centered on 0, judges are allowed to float
pt-biserial = measure ; point-measure correlation
omitunobserved = No ; so we anchored elements which have no ratings
Arrange=N ; Table 7 in element-number order
Model = ?,?,?,R9 ; judges, examinees and items produce ratings on "Creativity".
Labels= ; to name the components
1,Judges, A ; name of first facet
1=Avogadro (1) ; name of element within facet (subset)
2=Brahe (1) ; element numbers must be in Labels=, or they are treated as missing
3=Cavendish (2)
4=Davey (2)
5=Anchored judge, 60 ; has no ratings: anchored at 60 units (umean=50)
10=Lone judge (3) ; rates only lone examinee
*
2,Examinees
1=Anne (1)
2=Betty (1)
3=Chris (1)
4=David (2)
5=Edward (2)
6=Fred (2)
7=George (2)
10=Lone examinee (3) ; rated only by lone judge
*
3,Items
1=Attack (1+2) ; Attack is in subsets 2 and 3
2=Basis (1+2)
3=Clarity (1+2)
4=Daring (1+2)
5=Enthusiasm (1+2)
10=Lone item (3)
*
Data=
; Subset 1
1,1, 1-5, 5,5,3,5,3 ; typical paired-rater design
1,2, 1-5, 9,7,5,8,5 ; raters 1 and 2 rate examinees 1, 2, 3
1,3, 1-5, 3,3,3,7,1 ; everyone is rated on items 1,2,3,4,5
2,1, 1-5, 6,5,4,6,3
2,2, 1-5, 8,7,5,7,2
2,3, 1-5, 4,5,3,6,6
; Subset 2
3,4, 1-5, 5,3,3,3,1 ; raters 3 and 4 rate examinees 4, 5, 6, 7
3,5, 1-5, 9,7,7,7,7 ; everyone is rated on items 1,2,3,4,5
3,6, 1-5, 3,3,3,5,3
3,7, 1-5, 7,7,7,5,7
4,4, 1-5, 5,6,4,5,5
4,5, 1-5, 2,4,3,2,3
4,6, 1-5, 4,4,6,4,2
4,7, 1-5, 3,3,5,5,4
; Subset 3
10,10,10, 5 ; lone rater rates lone examinee on lone item
In the Iteration Report to the screen:
Consolidating 2 subsets..
2 subsets remain
Warning (6)! There may be 2 disjoint subsets
In the Output file:
Table 6.0.0 Disjoint Subset Element Listing.
Subset number: 1
Facet: 1. Judges 2 Elements: 1 2
Facet: 2. Examinees 3 Elements: 1-3
Subset number: 2
Facet: 1. Judges 2 Elements: 3 4
Facet: 2. Examinees 4 Elements: 4-7
Investigating the ambiguity:
Click on the Output Files Menu pull-down menu
Click on Winsteps control & data file
This enables you to construct data files showing the pattern of responses of one facet against another. Subsetting will show as distinctive patterns.
Here is what it looks like with Examinees as rows, and Judges as columns:
1 Anne ; item-column labels
2 Betty
3 Chris
4 David
5 Edward
6 Fred
7 George
END LABELS
5 9 3 . . . . 1 Avogadro ; row data + label
6 8 4 . . . . 2 Brahe
. . . 5 9 3 7 3 Cavendish
. . . 5 2 4 3 4 Davey
Click on the Output Files Menu pull-down menu.
Click on Subset group-anchor file.
This produces:
To resolve subset problems, copy-and-paste after Labels=
Non-center= must reference a facet that is not anchored or group-anchored.
Group anchor this facet:
1,Judges, G ; group-anchoring at Umean = 50
1,Avogadro,50, 1
2,Brahe,50, 1
3,Cavendish,50, 2
4,Davey,50, 2
*
And/or group anchor this facet:
2,Examinees, G ; group-anchoring at Umean = 50
1,Anne,50, 1
2,Betty,50, 1
3,Chris,50, 1
4,David,50, 2
5,Edward,50, 2
6,Fred,50, 2
7,George,50, 2
*
Action:
To establish an unambiguous measurement structure,
(a) We can assert that the rater pairs have the same mean severity (i.e., are randomly equivalent) using group-anchoring:
Non-center=2 ; facet 2 is allowed to float, because facet 1 will be group-anchored.
1,Judges, G ; group-anchor
1=Avogadro,50, 1 ; rater 1 is in group 1 with a mean value of 0.
2=Brahe,50, 1
3=Cavendish,50, 2 ; rater 3 is in group 2 with a mean value of 0.
4=Davey,50, 2
*
or (b) We can assert that the examinee samples have the same mean ability (i.e., are randomly equivalent):
Non-center=1 ; facet 1 is allowed to float, because facet 2 will be group-anchored.
2,Examinees, G ; group-anchor
1=Anne,50, 1 ; examinee 1 is in group 1 with a mean value of 0
2=Betty,50, 1
3=Chris,50, 1
4=David,50, 2 ; examinee 2 is in group 1 with a mean value of 0
5=Edward,50, 2
6=Fred,50, 2
7=George,50, 2
*
