Rasch models

Karl Bang Christensen

Dept. of BiostatisticsUniv. of Copenhagen

http://publicifsv.sund.ku.dk/~kach/scaleval_IRT

Karl Bang Christensen Rasch models

IRT model for dichotomous item

specifies that the probability of response pattern xv = (xvi )i=1,...,I

is

P(xv ) =

∫ I∏i=1

(Pi (θ)xvi (1− Pi (θ))(1−xvi )

)φ(θ)dθ

where is Pi (θ) is the probability of a correct response on item i asa function of the trait θ, and φ is the population distribution for θ.

Last week

Pi (θ) 2PL, φ(θ) standard normal distribution (mean=zero, SD=1).

This week

Pi (θ) 1PL, no assumption distributional assumptions are needed.

Karl Bang Christensen Rasch models

Rasch model: Probabilities for item i

person location θ, location βi and scale αi of item i .

P(Xvi = 1|θv = θ) =exp(θ − βi )

1 + exp(θ − βi )

The ’trick’

P(Xvi = xi ,Xvi ′ = xi ′ |Rv = xi + xi ′ , θ)

where Rv = Xvi + Xvi ′ is independent of θ

Karl Bang Christensen Rasch models

Model framework, Assumptions

Unidimensional latent variable θv responsible for all correlationbetween the observed items items X = (Xi )i∈I , covariatesY = (Y1,Y2, ..)

(i) Θ unidimensional

(ii) Monotonous relationship between Θ and Xi

(iii) No differential item functioning (DIF) Xi ⊥ Yj |Θ(iv) Local independence: Xi ⊥ Xj |ΘRasch model (=1PL) further requirements

sufficiency, specific objectivity, exchangeability (invariance)

Karl Bang Christensen Rasch models

Invariance

P(X1 = 1|θ) P(X1 = 0|θ)

P(X2 = 1|θ) P(X2 = 0|θ)

=

exp(θ−β1)

1+exp(θ−β1)1

1+exp(θ−β1)

exp(θ−β2)1+exp(θ−β2)

11+exp(θ−β2)

Odds

P(X1=1|θ)P(X1=0|θ)

P(X2=1|θ)P(X2=0|θ)

=

exp(θ − β1)

exp(θ − β2)

Comparison OR = exp(β2 − β1) independent of θ

Karl Bang Christensen Rasch models

Sufficiency

Local independence gives us

P(X = x|θ) =I∏

i=1

exp(xi (θ − βi ))1 + exp(θ − βi )

=exp(

∑Ii=1 xi (θ − βi ))∏I

i=1(1 + exp(θ − βi ))

=exp(tθ −

∑Ii=1 xiβi )∏k

i=1(1 + exp(θ − βi ))=

exp(tθ −∑I

i=1 xiβi )

K(θ,β)

where t =∑k

i=1 xi is the sum score.

Karl Bang Christensen Rasch models

Sufficiency

The sum score t =∑

i xi is sufficient for θ.

This means that

P(X1 = h1, . . . ,XI = hI |X1 + · · ·+ XI = t)

does not depend on θ.

We can use this to estimate item parameters without makingassumptions about the distribution of the latent variable.

Karl Bang Christensen Rasch models

Estimation

Marginal likelihood

normal distribution (mean zero, SD σ)

LM(η) =N∏

v=1

∫P(Xv = xv |θv = θ)ϕσ(θ) (1)

consistent estimates if normal distribution is correct.

Conditional likelihood

no distributional assumptions

LC (η) =N∏

v=1

P(Xv = xv |Rv = rv ) (2)

conditionally consistent estimates1

1Andersen. Journal of the Royal Statistical Society B, 1970, 32:283-301.Karl Bang Christensen Rasch models

Estimation in SAS

Marginal likelihood

PROC IRT

macro %rasch mml.sas2

Graphics: item characteristic curves (ICC’s), person-item locationmaps, item and test information functions. Goodness-of-fit plots.

2Christensen, Olsbjerg. Marginal maximum likelihood estimation inpolytomous Rasch models using SAS. Pub. Inst. Stat. Univ. Paris, vol. 57,fasc. 1-2, 69-84, 2013.

Karl Bang Christensen Rasch models

Estimation in SAS

Conditional likelihood

macro %rasch cml.sas3

Graphics: item characteristic curves (ICC’s), person-item locationmaps, item and test information functions. Goodness-of-fit plots.

3Christensen. Conditional maximum likelihood estimation in polytomousRasch models using SAS, ISRN Computational Mathematics, vol. 2013, ArticleID 617475, 8 pages, 2013 http://doi.org/10.1155/2013/617475

Karl Bang Christensen Rasch models

ADL six months after breast cancer surgery

17 items (ADL=Activities of Daily Living)

dichotomized: (0=no problems, 1=problems)Karl Bang Christensen Rasch models

ADL six months after breast cancer surgery

%l e t i t ems=Q6m 1 Q6m 2 Q6m 3 Q6m 4 Q6m 5 Q6m 6 Q6m 7 Q6m 8 Q6m 9 Q6m 10 Q6m 11 Q6m 12 Q6m 13 Q6m 14 Q6m 15 Q6m 16 Q6m 17 ;∗ r ead ADL data ;data s a s u s e r .ADL;

f i l e n ame dat u r l ’ h t tp : // b i o s t a t . ku . dk/˜kach/ s c a l e v a l IRT/ADL. t x t ’ ;i n f i l e dat f i r s t o b s =2;i n pu t i d &i t ems ;

run ;

Karl Bang Christensen Rasch models

Dichotomous Rasch model for ADL items

Item i taking values 0,1:

P(Xvi = x |θv = θ) =exp(x(θ − βi ))

1 + exp(θ − βi )

proc i r t data=s a s u s e r .ADL p l o t s =( i c c i i c ) ;va r Q6m 1−Q6m 17 ;model Q6m 1−Q6m 17 / r e s f u n c=ra s ch ;

run ;

PROC IRT uses marginal likelihood (assumes θ ∼ N(0, σ2))

parameters

β1, . . . , β17, σ

Karl Bang Christensen Rasch models

Dichotomous Rasch model for ADL items

Item i taking values 0,1:

P(Xvi = x |θv = θ) =exp(xα(θ − βi ))

1 + exp(α(θ − βi ))

proc i r t data=s a s u s e r .ADL p l o t s =( i c c t i c ) ;va r Q6m 1−Q6m 17 ;model Q6m 1−Q6m 17 / r e s f u n c=onep ;

run ;

PROC IRT uses marginal likelihood (assumes θ ∼ N(0, 1))

parameters

β1, . . . , β17, α

Karl Bang Christensen Rasch models

Dichotomous Rasch model for ADL items

Item Parameter Estimate s.e.

Q6m 1 Difficulty 3.36981 0.21681Slope 1.00000

Q6m 2 Difficulty 1.54327 0.15923Slope 1.00000

Karl Bang Christensen Rasch models

Dichotomous Rasch model for ADL items

interpretation of βi is the usual (P = 12)

Karl Bang Christensen Rasch models

Dichotomous Rasch model for ADL items

Item i taking values 0,1:

P(Xvi = x |θv = θ, βi1) =exp(x(θ − β1))

1 + exp(θ − β1)

proc i r t data=s a s u s e r .ADL p l o t s =( i c c i i c ) ;va r Q6m 1−Q6m 17 ;model Q6m 1−Q6m 17 / r e s f u n c=ra s ch ;

run ;

Karl Bang Christensen Rasch models

Rasch model in SAS, CML

Include macros

%l e t u r l=h t t p s : // raw . g i t h u bu s e r c o n t e n t . com/Ka r lBangCh r i s t en s en /Rasch/master ;f i l e n ame r u r l ”&u r l / r a s ch i n c l u d e a l l . s a s ” ;%i n c l u d e r ;

Item information

data i n ;i n pu t i tem no i tem name $ i tem t e x t $ max group ;d a t a l i n e s ;1 Q6m 1 x 1 12 Q6m 2 x 1 23 Q6m 3 x 1 34 Q6m 4 x 1 45 Q6m 5 x 1 56 Q6m 6 x 1 67 Q6m 7 x 1 78 Q6m 8 x 1 89 Q6m 9 x 1 910 Q6m 10 x 1 1011 Q6m 11 x 1 1112 Q6m 12 x 1 1213 Q6m 13 x 1 1314 Q6m 14 x 1 1415 Q6m 15 x 1 1516 Q6m 16 x 1 1617 Q6m 17 x 1 17;run ;

Karl Bang Christensen Rasch models

Rasch model in SAS, CML

call macro

%ra s ch data (data=s a s u s e r .ADL,i tem names=i n ) ;

%ra s ch CML(data=s a s u s e r .ADL,i tem names=in ,out=CML) ;

takes a long time for big tables (contingency table has 217 cells).

Karl Bang Christensen Rasch models

Rasch model in SAS, MML

call macro

%ra s ch data (data=s a s u s e r .ADL,i tem names=i n ) ;

%ra s ch MML(data=s a s u s e r .ADL,i tem names=in ,out=MML) ;

takes a long time for big data sets.

Karl Bang Christensen Rasch models

Does the Rasch model fit data? Overall test of the model4

Divide sample into G groups: Lg likelihood in group g

CLR = −2 log

(L∏G

g=1 Lg

)

asymptotically χ2 on (G − 1) · (I − 1) degrees of freedom (fordichotomous items). Note that this uses the fact that we do nothave to assume anything about the distribution

4Andersen. Psychometrika, vol. 38, 123-140, 1973.Karl Bang Christensen Rasch models

Does the Rasch model fit data? Overall test of the model

Two data sets G = 1 (score 0-2), G = 2 (score 3-17)

data ADL;s e t s a s u s e r .ADL;s c o r e=sum( o f &i t ems ) ;nm=nmiss ( o f &i t ems ) ;

run ;data ADL;

s e t ADL;i f nm=0;

run ;data ADL1 ;

s e t ADL;i f s c o r e i n ( 0 , 1 , 2 , 3 ) ;

run ;data ADL2 ;

s e t ADL;i f s c o r e > 3 ;

run ;

Karl Bang Christensen Rasch models

Does the Rasch model fit data? Overall test of the model

Fit Rasch model in each data set

%ra s ch data (data=ADL,

i tem names=i n ) ;%ra s ch cml (

data=ADL,i tem names=in ,out=cml ) ;

%ra s ch data (data=ADL1 ,

i tem names=i n ) ;%ra s ch cml (

data=ADL1 ,i tem names=in ,out=cml1 ) ;

%ra s ch data (data=ADL2 ,

i tem names=i n ) ;%ra s ch cml (

data=ADL2 ,i tem names=in ,out=cml2 ) ;

Karl Bang Christensen Rasch models

Does the Rasch model fit data? Overall test of the model

Compare fit of original Rasch model L with combined fit in eachdata set L1 · L2proc s q l ;

s e l e c t v a l u e i n t o : l 1 from cml1 l o g l ;s e l e c t v a l u e i n t o : l 2 from cml2 l o g l ;s e l e c t v a l u e i n t o : l from cml l o g l ;

q u i t ;data l r t ;

l r t =(& l 1+& l2−& l ) ;d f =16;p=1−cd f ( ’ c h i s q u a r e d ’ , l r t , d f ) ;

run ;p roc p r i n t data= l r t round noobs ;run ;

by compare log likelihood values

χ2 = 48.2(df = 16)

model is rejected

Karl Bang Christensen Rasch models

Multiple groups analysis 5

G = 1 X1 X2

X3

:

Θ

KK EE

::

--

&&

:

X16

X17

G = 2 X1 X2

X3

:

Θ

KK EE

::

--

&&

:

X16

X17

5Andersen. Psychometrika, vol. 38, 123-140, 1973.Karl Bang Christensen Rasch models

Does the Rasch model fit data? Individual item fit

ResidualsRvi = Xvi − Evi

where Xvi is the response of person v to item i and Evi expectedresponse of person v to item i .

Two issues to discuss

(i) How to calculate the expected values

(ii) How to summarize residuals

Karl Bang Christensen Rasch models

The conditional item characteristic curve

The item characteristic curve is the item response function

θ 7→ P(Xvi = x |θv = θ).

In the Rasch model: simple expression for a conditional version ofthis

r 7→ P(Xvi = x |Rv = r) =exp(−βi )γr−1(β1, . . . , βi−1, βi+1, . . . , βI )

β

Karl Bang Christensen Rasch models

(i) How to calculate the expected values

Expected response Evi :ideal situation: Known person locations

Evi = P(Xvi = 1|Θv = θv ) =exp(θv − βi )

1 + exp(θv − βi )

less than ideal situation: Person locations are estimated

Evi = P(Xvi = 1|Θv = θ̂v ) =exp(θ̂v − β̂i )

1 + exp(θ̂v − β̂i )

Two problems

Person locations are estimated with error

Formula is incorrect

Commercial Rasch model software ignores these problems ..

Karl Bang Christensen Rasch models

Standard evaluation of individual item fit

using

Ziv =Xiv − E (Xiv |Θv = θ̂v )√

V (Xiv |Θv = θ̂v )

Fit statistics like

OUTFITi =1

N

N∑v=1

Z 2iv

with no established null distribution are used. Early Raschliterature claims of χ2 distribution.6

6Wright, Stone (1979). Mesa Press, Chicago, USA.Karl Bang Christensen Rasch models

Wilson-Hilferty cube-root transformation

It has been suggested that the Wilson-Hilferty (1931) cube-roottransformation

ti = (OUTFIT1/3i − 1)

3

V (OUTFITi )+

V (OUTFITi )

3

has an approximate t distribution and that

FitResidi =f (log(N · OUTFITi )− log(f ))√

V (N · OUTFITi )

with f = (Nk − N − k + 1)/k, has a symmetrical distribution withmean zero and variance one.

Karl Bang Christensen Rasch models

Commercial software

WINSTEPS

OUTFIT, t-transformed OUTFIT

INFIT (weighted version of OUTFIT), t-transformed INFIT

RUMM

item χ2 fit statistic

item ANOVA fit statistic

item FitResidi

Karl Bang Christensen Rasch models

RUMM

groups respondents in G ’class intervals’ based on the estimatedperson locations θ̂v .

still relies on θ̂1, θ̂2, . . .

Item χ2 fit statistic

χ2(Xi ) =∑g

(∑v∈Vg

Xvi −∑

v∈VgE (Xvi )

)2∑v∈Vg

V (Xvi )

where Vg denotes the set of respondents in class interval g . ItemANOVA fit statistic uses ANOVA based on the grouping

Ziv = µg(v) + ERROR

Karl Bang Christensen Rasch models

Item fit statistics in SAS

Fit model

%ra s ch data (data=s a s u s e r .ADL,i tem names=i n ) ;

%ra s ch CML(data=s a s u s e r .ADL,i tem names=in ,out=CML) ;

creates output data sets with information.Estimate person locations

%ra s ch ppar (DATA=s a s u s e r .ADL,ITEM NAMES=in ,DATA IPAR=CML ipa r ,out=pp cml ) ;

also creates output data sets with information

Karl Bang Christensen Rasch models

Item fit statistics in SAS

Compute item fit statistics

%ra s ch i t e m f i t (DATA=s a s u s e r .ADL,ITEM NAMES=in ,DATA IPAR=cml i p a r ,DATA POPPAR=pp cml outdata ,NCLASS=3,OUT=f i t cm l ) ;

(takes a long time). We specify that we want three class intervals.Output data sets ’fitcml chisq’, ...

Karl Bang Christensen Rasch models

Exercise 10: Item fit statistics in SAS

Use the 11 symptom items from the colitis data set.

1 Evaluate item fit using PROC IRT

2 Evaluate item fit using SAS macros

use

http://publicifsv.sund.ku.dk/~kach/scaleval_IRT/exerc10.sas

Karl Bang Christensen Rasch models