3 facet Subset reconnection

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

Model = ?,?,?,R9 ; judges, examinees and items produce ratings on "Creativity".

Labels=   ; to name the components

1,Judges  ; name of first facet

1=Avogadro  ; names of elements within facet

2=Brahe   ; these must be named, or they are treated as missing

3=Cavendish

4=Davey

*

2,Examinees

2=Betty 

5=Edward

7=George

1=Anne

3=Chris 

4=David 

6=Fred 

*

3,Items

1=Attack

2=Basis

3=Clarity

4=Daring

5=Enthusiasm

*

Dvalues = 3, 1-5 ; put in constant value for the 3 facet entry in the data

 

Data=

1,1, 5,5,3,5,3 ; typical paired-rater design

1,2, 9,7,5,8,5 ; raters 1 and 2 rate examinees 1, 2, 3

1,3, 3,3,3,7,1

2,1, 6,5,4,6,3

2,2, 8,7,5,7,2

2,3, 4,5,3,6,6

3,4, 5,3,3,3,1 ; raters 3 and 4 rate examinees 4, 5, 6, 7

3,5, 9,7,7,7,7

3,6, 3,3,3,5,3

3,7, 7,7,7,5,7

4,4, 5,6,4,5,5

4,5, 2,4,3,2,3

4,6, 4,4,6,4,2

4,7, 3,3,5,5,4

 

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

 

Resolving the ambiguity:

 

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

*


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