Stat 315c: Transposable Data Rasch model and friends

Transcription

1 Stat 315c: Transposable Data Rasch model and friends Art B. Owen Stanford Statistics Art B. Owen (Stanford Statistics) Rasch and friends 1 / 14

2 Categorical data analysis Anova has a problem with too much of the df being in the interaction. Generalizations of Anova for categorical data analysis face the same issue. They have some good solutions. We ll look at them. The Rasch model is a famous waypoint on our journey. Art B. Owen (Stanford Statistics) Rasch and friends 2 / 14

3 Contingency tables Setting Tables of counts 2 2 or I J or... I J K Z Art B. Owen (Stanford Statistics) Rasch and friends 3 / 14

4 Contingency tables Setting Notation Tables of counts 2 2 or I J or... I J K Z n ijkl = number of obs in cell (i,j,k,l) n ij = K k=1 L l=1 n ijkl etc. µ ijkl = E(n ijkl ) π ijkl = µ ijkl /µ π ij kl = π ijkl /π kl Art B. Owen (Stanford Statistics) Rasch and friends 3 / 14

5 Default sampling models The simplest probability model has independent Poisson random variables n ijkl Poi(µ ijkl ) Indep Art B. Owen (Stanford Statistics) Rasch and friends 4 / 14

6 Default sampling models The simplest probability model has independent Poisson random variables n ijkl Poi(µ ijkl ) Indep Conditioning N = n we get a multinomial n ijkl Mult(N; µ ijkl ) Art B. Owen (Stanford Statistics) Rasch and friends 4 / 14

7 Default sampling models The simplest probability model has independent Poisson random variables n ijkl Poi(µ ijkl ) Indep Conditioning N = n we get a multinomial n ijkl Mult(N; µ ijkl ) Conditioning (holding fixed) the K L margin we get a product of multinomials on I J: n ij kl Mult(n kl ; π ij kl ) Indep over k and l These usually lead to the same likelihoods and same inferences. There are also hypergeometric models and generalizations that hold fixed eg the I and J margins without fixing the I J margin. Art B. Owen (Stanford Statistics) Rasch and friends 4 / 14

8 Multinomial model Sum N independent observations from Mult(1; π ijkl ). Clearly π ijkl 0. Suppose π ijkl > 0. For I J, write log π ij = µ + α i + β j + (αβ) ij Suppose that (αβ) ij = 0 for all i and j Then π ij = exp(µ + α i + β j ) = exp(µ) exp(α i ) exp(β j ) So rows and columns are independent. And conversely. Art B. Owen (Stanford Statistics) Rasch and friends 5 / 14

9 Log linear models Anova expansion log π ijk = µ + α k + β j + γ k + (αβ) ij + (αγ) ik + (βγ) jk + (αβγ) ijk Hierarchical models Dropping interactions amounts to simplifying dependence EG I J indep of K, or, EG I and J indep given K, etc. Tests come from Poisson likelihood Downside Huge numbers of parameters that are hard to interpret response is a count, not one of the marginal variables rejecting independence is often uninteresting Art B. Owen (Stanford Statistics) Rasch and friends 6 / 14

10 Modeling the interaction Residence in 1966 and 1971 CC ULY WM GL CC ULY WM GL regions of UK Reject independence But so what? Mover-Stayer models (Quasi-Independence) Some people won t move, some might. Model via log π ij = µ + α i + β j + δ i 1 i=j using 3K params Might even use log π ij = µ + α i + β j + δ1 i=j Art B. Owen (Stanford Statistics) Rasch and friends 7 / 14

11 More models (from Agresti) Quasi-Symmetry log π ij = µ + α i + β j + (αβ) ij with (αβ) ij = (αβ) ji Symmetry log π ij = µ + α i + α j + (αβ) ij with (αβ) ij = (αβ) ji Equilibrium π i = π i exp(µ + α i + β j + (αβ) ij ) = j i exp(µ + α i + β j + (αβ) ij ) Art B. Owen (Stanford Statistics) Rasch and friends 8 / 14

12 Bradley-Terry models Preferences Customer prefers wine i to wine j Team i beats team j Models Basic model eα i α j Pr(i beats j) = 1 + e α i α j Fit by logistic regression (no intercept) Overidentified... so take α 1 = 0 or other constraint With covariates x eg home field advantage Pr(i beats j X = x) = exβ+α i α j 1 + e xβ+α i α j Art B. Owen (Stanford Statistics) Rasch and friends 9 / 14

13 Rasch model Educational testing Student i with ability θ i Question j with difficulty β j Probability correct e θ i β j 1+e θ i β j Art B. Owen (Stanford Statistics) Rasch and friends 10 / 14

14 Rasch model ctd Latent variables Threshold model with latent variables ε ij logistic Effect types Y ij = 1 θi β j +ε ij >0 Students random and questions fixed (But sometimes questions are drawn from a pool) Estimation Conditional maximum likelihood Huge number of student params Small number of item params Condition on estimates of ability so e θ i s cancel out See Agresti and web site for his book Art B. Owen (Stanford Statistics) Rasch and friends 11 / 14

15 Rasch problems Green question has steeper threshold Red question is hard, so they guess Art B. Owen (Stanford Statistics) Rasch and friends 12 / 14

16 Alternative models Two parameters for items Pr(Correct ij ) = e(θ i β j )δ j 1 + e (θ i β j )δ j...with guessing e (θ i β j )δ j Pr(Correct ij ) = γ i + (1 γ i ) 1 + e (θ i β j )δ j...and probit link Pr(Correct ij ) = γ i + (1 γ i )Φ((θ i β j )δ j ) Art B. Owen (Stanford Statistics) Rasch and friends 13 / 14

17 More item response theory Followups Local School of Ed: David Rogosa, Ingram Olkin, Ed Haertel Lord and Novick (1968) Statistical Theories of Mental Tests Thissen and Wainer (Eds) (2001) Test Scoring Check an item by plotting correctness vs ˆθ i (and smoothing) Interactions Can look at subtests Can t fit many params per student Can pool students e.g. CA vs NY vs... Art B. Owen (Stanford Statistics) Rasch and friends 14 / 14