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Questions about choice sets

PostPosted: Tue May 14, 2024 4:43 pm
by ginger
Hello, I am having trouble with this and would like your guidance.
I have six attributes and the number of levels is 3,3, 2,2, 2,3 respectively.
The command I am using is
Design
;alts = choiceA, choiceB
;rows = 24
;blocks = 3
;eff = (mnl,d)
;model:U(choiceA) = asc+ b1 * b1 [0,1,2]+ b2 * b2[0,1,2]+ b3 * b3[0,1]+ b4 * b4[0,1]+ b5 * b5[0,1]+b6 * b6[0,1,2]/
U(choiceB) = b1 * b1+ b2 * b2+ b3 * b3+ b4 * b4+ b5 * b5+ b6 * b6$
However, the generated choice sets are only selected at low and high levels (0 and 2), and at medium and medium levels (1 and 1).
How can I modify the design commands? Thanks.

Re: Questions about choice sets

PostPosted: Tue May 14, 2024 5:12 pm
by Michiel Bliemer
This is because it is statistically most efficient to use 0 vs 2 and 1 vs 1 when assuming numerical attributes with 0 priors.

If you use zero priors, it is best to use dummy coding (or effects coding) for all attributes, see script below.

Code: Select all
Design
;alts = choiceA, choiceB
;rows = 24
;block = 3
;eff = (mnl,d)
;model:
U(choiceA) = asc
           + b1.dummy[0|0] * x1[1,2,0] ? last level is base level
           + b2.dummy[0|0] * x2[1,2,0]
           + b3.dummy[0]   * x3[1,0]
           + b4.dummy[0]   * x4[1,0]
           + b5.dummy[0]   * x5[1,0]
           + b6.dummy[0|0] * x6[1,2,0]
           /
U(choiceB) = b1.dummy      * x1
           + b2.dummy      * x2
           + b3.dummy      * x3
           + b4.dummy      * x4
           + b5.dummy      * x5
           + b6.dummy      * x6
$


Alternatively, you can use an orthogonal design, e.g. if your alternatives are unlabelled then an optimal orthogonal design can be achieved with 36 rows:

Code: Select all
Design
;alts = choiceA, choiceB
;rows = 36
;block = 3
;orth = ood
;model:
U(choiceA) = asc
           + b1 * x1[0,1,2]
           + b2 * x2[0,1,2]
           + b3 * x3[0,1]
           + b4 * x4[0,1]
           + b5 * x5[0,1]
           + b6 * x6[0,1,2]
           /
U(choiceB) = b1 * x1
           + b2 * x2
           + b3 * x3
           + b4 * x4
           + b5 * x5
           + b6 * x6
$


Michiel