Testing Linear Hypotheses in Elimination-by-Aspects (EBA) Models

Tests linear hypotheses of the form \(Cp = 0\) in elimination-by-aspects (EBA) models using the Wald test.

wald.test(object, C, u.scale = TRUE)

Arguments

object

an object of class eba, typically the result of a call to eba

C

a matrix of contrasts, specifying the linear hypotheses

u.scale

logical, if TRUE the test is performed on the utility scale, if FALSE the test is performed on the EBA parameters directly

Details

The Wald test statistic, $$W = (Cp)' [C cov(p) C']^{-1} (Cp),$$ is approximately chi-square distributed with \(rk(C)\) degrees of freedom.

C is usually of full rank and must have as many columns as there are parameters in p.

Value

C

the matrix of contrasts, specifying the linear hypotheses

W

the Wald test statistic

df

the degrees of freedom (\(rk(C)\))

pval

the p-value of the test

See also

eba, group.test, uscale, cov.u.

Examples

data(celebrities)                     # absolute choice frequencies
A <- list(c(1,10), c(2,10), c(3,10),
          c(4,11), c(5,11), c(6,11),
          c(7,12), c(8,12), c(9,12))  # the structure of aspects
eba1 <- eba(celebrities, A)           # fit elimination-by-aspects model

## Test whether JU, CY, and AJF have equal utility scale values
C1 <- rbind(c(0,0,0,1,-1, 0,0,0,0),
            c(0,0,0,1, 0,-1,0,0,0))
wald.test(eba1, C1)
#> 
#> Wald Test: Cp = 0
#> 
#> C:
#>      LBJ HW CDG JU CY AJF BB ET SL
#> [1,]   0  0   0  1 -1   0  0  0  0
#> [2,]   0  0   0  1  0  -1  0  0  0
#> 
#> W = 19.16994, df = 2, p-value = 6.875427e-05
#> 

## Test whether the three branch parameters are different
C2 <- rbind(c(0,0,0,0,0,0,0,0,0,1,-1, 0),
            c(0,0,0,0,0,0,0,0,0,1, 0,-1))
wald.test(eba1, C2, u.scale = FALSE)
#> 
#> Wald Test: Cp = 0
#> 
#> C:
#>      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
#> [1,]    0    0    0    0    0    0    0    0    0     1    -1     0
#> [2,]    0    0    0    0    0    0    0    0    0     1     0    -1
#> 
#> W = 0.3946142, df = 2, p-value = 0.8209385
#>