eba (version 1.10-0)

trineq: Trinary Inequality

Description

Checks if binary choice probabilities fulfill the trinary inequality.

Usage

trineq(M, A = 1:I)

Arguments

M

a square matrix or a data frame consisting of absolute choice frequencies; row stimuli are chosen over column stimuli

A

a list of vectors consisting of the stimulus aspects; the default is 1:I, where I is the number of stimuli

Value

Results checking the trinary inequality.

n

number of tests of the trinary inequality

prop

proportion of triples confirming the trinary inequality

quant

quantiles of \(R(x, y, z)\)

n.tests

number of transitivity tests performed

chkdf

data frame reporting \(R(x, y, z)\) for each triple where \(P(x, y) > 1/2\) and \((xy)z\)

Details

For any triple of stimuli \(x, y, z\), the trinary inequality states that, if \(P(x, y) > 1/2\) and \((xy)z\), then $$R(x, y, z) > 1,$$ where \(R(x, y, z) = R(x, y) R(y, z) R(z, x)\), \(R(x, y) = P(x, y)/P(y, x)\), and \((xy)z\) denotes that \(x\) and \(y\) share at least one aspect that \(z\) does not have (Tversky and Sattath, 1979, p. 554).

inclusion.rule checks if a family of aspect sets is representable by a tree.

References

Tversky, A., & Sattath, S. (1979). Preference trees. Psychological Review, 86, 542--573. 10.1037/0033-295X.86.6.542

See Also

eba, inclusion.rule, strans.

Examples

# NOT RUN {
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
trineq(celebrities, A)        # check trinary inequality for tree A
trineq(celebrities, A)$chkdf  # trinary inequality for each triple
# }