LLTM - Linear Logistic Test Model |
This is for 32-bit Facets 3.87. Here is Help for 64-bit Facets 4
Recommendation: In Facets, all facets are treated the same. The only difference between facets is that one facet can be specified as the inter-rater facet which reports a few more statistics. For LLTM for any facet, my recommendation is that you do a standard MFRM analysis. This estimates measures for the elements of the facet. Then decompose the element measures for that facet using linear regression in a standard statistics program.
LLTM is designed to deconstruct and then predict item difficulties and is based on measurement. Here is how LLTM works:
Imagine a complex arithmetic item. Solving it correctly requires numerical operations, such as addition, subtraction, multiplication, division. Perhaps several of each of these operations.
We model the difficulty of this item to be a linear combination of the difficulties of these operations.
We repeat this for all our items, producing a design matrix.
We then analyze our response data using the design matrix.
LLTM tells us the difficulty of each operation.
As we construct further items, we can use the difficulties of these operations to predict the item difficulties.
The most successful application of this approach is the Lexile system for readability. This can predict the difficulty of English prose with great accuracy.
The Linear Logistic Test Model, LLTM, with integer "basic parameter" weights can be modeled with Facets easily.
For non-integer weights, multiply all the weights for all the items by the same value until all the weights approximate integers. Then use the integer weights.
A standard LLTM model for two "basic parameters" (components of item difficulty) looks like
Bn - W1i * P1 - W2i *P2 -> Xni
where Bn is the ability of person n
P1 and P2 are the difficulties of basic parameters 1 and 2.
W1i and W2i are the pre-assigned weights of P1 and P2 in item i.
The same approach applies to any facet such as rater leniency, person ability, etc.
The standard LLTM can be estimated with Facets if W1i and W2i are integers. The Facets model is
Bn - P1 - P1 - ... - P2 - P2 - .... -> Xni
with W1i "P1" facets and W2i "P2" facets
Here is the procedure with Facets:
1. The "basic parameters", BP, the components of the items, are elements of a "basic parameter" facet.
2. The items are a dummy facet. The item labels include the element numbers of the BP elements.
3. There are as many BP facets in the data as the largest sum of all the integer BP weights for any item = BPL. For example, BPL= 7
4. The Facets specifications:
Facets = 9 ; Person+item+BPL facets in the data
Entered-in-data = 1, 2, 3, 3, 3, 3, 3, 3, 3 ; BPL=7 facets in the data for facet 3
Models = ?, ?, ?,?,?,?,?,?,?, D ; 2+BPL=9 facets
Labels=
1, Persons
1= ...
...
*
2, Items, D ; a dummy facet
1=1124000 ; the BP elements for item 1 are 2*BP1 + BP2 + BP4. 0 = no BP element.
2=1222344 ; the BP elements for item 2 are BP1+3*BP2+BP3+2*BP4
....
10=3440000
*
3, Basic Parameters (BP)
1 = basic parameter one
....
4 = basic parameter four
*
Dvalues=
3,2,1,1 ; the element number for facet 3 in the data (BP) is the first character in the label for the element in facet 2 (item)
4,2,2,1
5,2,3,1
6,2,4,1
7,2,5,1
8,2,6,1
9,2,7,1
*
; only facets 1 and 2 are needed in the data, Facets 3-9 are imputed by Dvalues=
Data=
1,1-10, 0,0,1,1,1,0,1,1,0,1 ; person 1, items 1-10, dichotomous scored responses
.....
5. In the output,
The BP estimates are the estimates for the "basic parameters".
The item displacement is the difference between the LLTM estimate and the item estimate from the data.
The sum of the weighted BP for an item is -"sum of the measures" in the Residualfile= for a Person with measure 0.00, or add in the Person measure to the -"sum of the measures".
Fischer, G. H., 'The linear logistic test model as an instrument in educational research', Acta Psychologica 37 (1973), 359-74.
Simple LLTM example:
LLTM is similar to a Facets-style model. It models the item difficulty as a combination of components. Each component has a difficulty and components are weighted to reflect their contribution to the item. In the simplest case, the weighting is 0 = does not contribute, 1 = does contribute.
For a Facets analysis, let's say there are 4 components. In the Facets "Labels=" specification, we set up a "component" facet with 4 elements. In the data, we have 4 component facets, all referencing the one Labels= component facet.
The Facets specification file:
Facets = 6 ; person, item, 4 components facets in data
Entered-in-data = 1, 2, 3, 3, 3, 3 ; all 4 components facets in the data reference the same "component" facet
Models= ?,?,?,?,?,?, D ; dichotomous right/wrong items
Labels=
1, Persons
1 = Jose
....
23 = Maria
...
*
2, items, D ; this is a dummy facet for specifying the components in each item
1 = 1300 item1 with active components 1, 3. Inactive components specified with 0.
...
20 = 2340 item 20 with active components 2,3,4
*
3, components of item difficulty
1= recognition
2= comprehension
3= computation
4= realization
*
Dvalues=
2, 1-20 ; facet 2 in the data: all persons respond to a 20 item test
3, 2, 1 ; facet 3 in the data: decode the components from the item labels
4, 2, 2
5, 2, 3
6, 2, 4
*
data=
23, 1, 0, 1, ... ; success by person 23, Maria, on item 1, failure on item 2, etc. of the 20 item test
...
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