﻿ LLTM - Linear Logistic Test Model

# LLTM - Linear Logistic Test Model

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.

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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