Testing Transitivity with a True and Error Model

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Testing Transitivity with a True and Error Model Michael H. Birnbaum California State University, Fullerton

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Testing Transitivity with a True and Error Model. Michael H. Birnbaum California State University, Fullerton. Testing Algebraic Models with Error-Filled Data. Models assume or imply formal properties such as transitivity: If A > B and B > C then A > C - PowerPoint PPT Presentation

Transcript of Testing Transitivity with a True and Error Model

Page 1: Testing Transitivity with a True and Error Model

Testing Transitivity with a True and Error Model

Michael H. BirnbaumCalifornia State University,

Fullerton

Page 2: Testing Transitivity with a True and Error Model

Testing Algebraic Models with Error-Filled Data

• Models assume or imply formal properties such as transitivity:

If A > B and B > C then A > C• But such properties may not hold if

data contain “error.”• And different people might have

different “true” preferences.

Page 3: Testing Transitivity with a True and Error Model

Error Model Assumptions

• Each choice in an experiment has a true choice probability, p, and an error rate, e.

• The error rate is estimated from (and is the “reason” given for) inconsistency of response to the same choice by same person over repetitions

Page 4: Testing Transitivity with a True and Error Model

One Choice, Two Repetitions

A B

A

B€

pe2

+ ( 1 − p )( 1 − e )2

p ( 1 − e ) e + ( 1 − p )( 1 − e ) e

p ( 1 − e ) e + ( 1 − p )( 1 − e ) e

p ( 1 − e )2

+ ( 1 − p ) e2

Page 5: Testing Transitivity with a True and Error Model

Solution for e

• The proportion of preference reversals between repetitions allows an estimate of e.

• Both off-diagonal entries should be equal, and are equal to:

( 1 − e ) e

Page 6: Testing Transitivity with a True and Error Model

Ex: Stochastic Dominance

: 05 tickets to win $12

05 tickets to win $14

90 tickets to win $96

B: 10 tickets to win $12

05 tickets to win $90

85 tickets to win $96

122 Undergrads: 59% repeated viols (BB) 28% Preference Reversals (AB or BA) Estimates: e = 0.19; p = 0.85170 Experts: 35% show 2 violations (BB) 31% Reversals (AB or BA) Estimates: e = 0.196; p = 0.50

Page 7: Testing Transitivity with a True and Error Model

Testing Higher Properties• Extending this model to properties

relating 2, 3, or 4 choices:• Allow a different error rate on

each choice.• Estimate true probability for each

choice pattern. Different people can have different “true” patterns, which need not be transitive.

Page 8: Testing Transitivity with a True and Error Model

New Studies of Transitivity

• Work currently under way testing transitivity under same conditions as used in tests of other decision properties.

• Participants view choices via the WWW, click button beside the gamble they would prefer to play.

Page 9: Testing Transitivity with a True and Error Model

Some Recipes being Tested

• Tversky’s (1969) 5 gambles.• LS: Preds of Priority Heuristic• Starmer’s recipe• Additive Difference Model (regret;

majority rule)• Birnbaum, Patton, & Lott (1999) recipe.• Recipes based on Bleichrodt & Schmidt

context-dependent utility models.

Page 10: Testing Transitivity with a True and Error Model

Replications of Tversky (1969) with Roman

Gutierez• First two studies used Tversky’s 5

gambles, but formatted with tickets instead of pie charts.

• Two studies with n = 417 and n = 327 with small or large prizes ($4.50 or $450)

• No pre-selection of participants.• Participants served in other risky DM

studies, prior to testing (~1 hr).

Page 11: Testing Transitivity with a True and Error Model

Three of Tversky’s (1969) Gambles

• A = ($5.00, 0.29; $0, 0.79)• C = ($4.50, 0.38; $0, 0.62)• E = ($4.00, 0.46; $0, 0.54)Priority Heurisitc Predicts:A preferred to C; C preferred to E, and E preferred to A. Intransitive.

Tversky (1969) reported viol of WST

Page 12: Testing Transitivity with a True and Error Model

Response Combinations

Notation (A, B) (B, C) (C, A)

000 A B C *

001 A B A

010 A C C

011 A C A

100 B B C

101 B B A

110 B C C

111 B C A *

Page 13: Testing Transitivity with a True and Error Model

Weak Stochastic Transitivity

P ( A f B ) = P ( 000 ) + P ( 001 ) + P ( 010 ) + P ( 011 )

P ( B f C ) = P ( 000 ) + P ( 001 ) + P ( 100 ) + P ( 101 )

P ( C f A ) = P ( 000 ) + P ( 010 ) + P ( 100 ) + P ( 110 )

Page 14: Testing Transitivity with a True and Error Model

WST Can be Violated even when Everyone is Perfectly

Transitive

P ( 001 ) = P ( 010 ) = P ( 100 ) =1

3

P ( A f B ) = P ( B f C ) = P ( C f A ) =2

3

Page 15: Testing Transitivity with a True and Error Model

Triangle Inequality has similar problems:

• It is possible that everyone is transitive but WST is violated.

• It is possible that people are systematically intransitive and WST is satisfied.

• Possible that everyone is intransitive and triangle inequality is satisfied.

Page 16: Testing Transitivity with a True and Error Model

Model for Transitivity

P ( 000 ) = p000

( 1 − e1

)( 1 − e2

)( 1 − e3

) + p001

( 1 − e1

)( 1 − e2

) e3

+

+ p010

( 1 − e1

) e2

( 1 − e3

) + p011

( 1 − e1

) e2e

3+

+ p100

e1

( 1 − e2

)( 1 − e3

) + p101

e1

( 1 − e2

) e3

+

+ p110

e1e

2( 1 − e

3) + p

111e

1e

2e

3

A similar expression is written for the other seven probabilities. These can in turn be expanded to predict the probabilities of showing each pattern repeatedly; i.e., up to six errors.

Page 17: Testing Transitivity with a True and Error Model

Expand and Simplify• There are 8 X 8 data patterns in an

experiment with 2 repetitions.• However, most of these have very small

frequencies.• Examine probabilities of each of 8

repeated patterns.• Frequencies of showing each of 8

patterns in one replicate OR the other, but NOT both. Mutually exclusive, exhaustive partition.

Page 18: Testing Transitivity with a True and Error Model

Tests of WSTPercentage Choosing Column >pr Row Gamble

Row A B C D E

A 73 77 80 85

B 30 68 79 79

C 16 29 74 78

D 11 16 24 63

E 13 17 15 33

Page 19: Testing Transitivity with a True and Error Model

Patterns for A, C, EPattern Rep. 1 Rep 2 Both

000 14 28 5

001 18 25 15

010 23 38 1

011 12 5 3

100 24 33 7

101 5 6 1

110 301 256 220

111 19 25 5

Sum 416 416 257

Page 20: Testing Transitivity with a True and Error Model

Pattern Both Rep 1 or 2Not both

Est p Pred Both

Pred 1 or 2 Not both

000 5 16 .03 8.1 8.6

001 15 6.5 .07 15.3 6.5

010 1 29.5 .00 4.7 37.2

011 3 5.5 .01 2.8 5.9

100 7 21.5 .03 7.8 26.0

101 1 4.5 .00 0.9 5.5

110 220 58.5 .85 196.6 67.6

111 5 17 .02 4.6 17.9

Sum 257 159 1 240.9 175.1

Page 21: Testing Transitivity with a True and Error Model

Comments• Results are surprisingly transitive, unlike

Tversky’s data (est. 95% transitive).• Of those 115 who were perfectly reliable,

93 perfectly consistent with EV (p), 8 with opposite ($), and only 1 intransitive.

• Differences: no pre-test, selection;• Probability represented by # of tickets

(100 per urn), rather than by pies.• Participants have practice with variety of

gambles, & choices;• Tested via Computer.

Page 22: Testing Transitivity with a True and Error Model

Pie Chart Format

Page 23: Testing Transitivity with a True and Error Model

Pies: with or without Numerical probabilities

• 321 participants randomly assigned to same study; except probabilities displayed as pies (spinner), either with numerical probabilities displayed or without.

• Of 105 who were perfectly reliable, 84 were perfectly consistent with EV (prob), 13 with the opposite order ($); 1 consistent with LS.

Page 24: Testing Transitivity with a True and Error Model

Lower Standards

• Look at AB,BC,CD,DE choices and EA choices only:

• 10 of 321 participants showed this pattern; all in the pies-only condition. Although the rate is low (6% of 160), association with condition is clear!

• By still lower standard used by Tversky: 75% agreement with above pattern: 37 people, 27 in pies-only condition.

Page 25: Testing Transitivity with a True and Error Model

Tests of Lexicographic Semi-order and Additive Difference• LS implies no integration of

contrasts (additive difference model allow integration)

• Both LS and additive difference models imply no interactions between probability and consequences.

Page 26: Testing Transitivity with a True and Error Model

Test of Interaction

R S Pies & p

Pies & No p

($7.25, .95; $1.25, .05)

($4.25, .95; $3.25, .05)

16 22

($7.25, .05; $1.25, .95)

($4.25, .05; $3.25, .95)

84 77

Page 27: Testing Transitivity with a True and Error Model

Among the 37 Leniently classified as Intransitive

• Are those 37 who are 75% consistent with the LS in the 5 choices also approx. consistent with LS in tests of Interaction?

• No. 26 of these have all four choices in the pattern of interaction predicted by TAX and other utility models.

Page 28: Testing Transitivity with a True and Error Model

Summary• Priority Heuristic model’s predicted

violations of transitivity are rare and rarely repeated when probability and prize information presented numerically.

• Violations of transitivity are still rare but more frequent when prob information presented only graphically.

• Evidence of Dimension Interaction violates PH and additive Difference models.

Page 29: Testing Transitivity with a True and Error Model

Conclusions

• Violations of transitivity are probably not due to intransitive strategy (LS or additive difference model), but rather to a configural assimilation of the probability values, which are then used in a numerical utility model.

• We are still unable to produce the higher rates of intransitivity reported by Tversky and others.

Page 30: Testing Transitivity with a True and Error Model

Transitivity Test: ADLTrial Choice % Response Pattern

RepsEst. parameters

00 01 10 11 p e

8 50 to win $10050 to win $20

50 to win $6050 to win $27

18 190 23 25 23 0.10

0.10

3 50 to win $6050 to win $27

50 to win $4550 to win $34

29 140 44 39 37 0.17

0.20

21 50 to win $4550 to win $34

50 to win $10050 to win $20

74 35 33 20 172

0.85

0.12

Page 31: Testing Transitivity with a True and Error Model

Results-ADLpattern Rep 1 Rep 2 Both

000 LPH 21 13 1

001 TAX 134 147 106

010 20 18 8

011 38 37 10

100 15 9 0

101 14 10 0

110 12 15 7

111 6 11 1

Sum 260 260 133