Orthogonal design, attributes with more than 4 levels

[Sara]

Hi every one,

Actually I have small question on orthogonal design:
I want to use efficient design but since I have no information on priors I have to have some small pretest using orthogonal design. I know more levels lead to more complicated choice task and big number of scenarios. I have 3 alternatives, first one has 3 attributes each one 5 levels, two other alternatives have 5 attributes (3 of them five levels and 2 with 3 levels). All attributes are alternative specific. I tried to get orthogonal design which I couldn't, orthogonal design couldn't be found. When I eliminated levels all to 3 or 4 and 2 levels however I am getting results. The question is whether 5 levels consider as too much making finding an orthogonal design impossible?

another question is number of scenarios, the best situation is with 20. I wonder if I eliminate some of those scenarios (based on human brain logic) would it result in loosing orthogonality?

Thanks
Sara

[Michiel Bliemer]

Most orthogonal designs currently known consist of 2 and 3 levels, they are the easiest to find. An orthogonal column with 4 levels is also easy, as it can be created by combining 2 columns with 2 levels. Same for 6, 8, and 9 levels, which are created from 2x3, 2x2x2 and 3x3. Orthogonal columns with 5 and 7 levels are much harder to find. Hence, only few designs with this number of levels is known to date. If Ngene cannot find one, it may not exist. We have put all currently known orthogonal designs in Ngene. Instead of 5 levels, maybe you can choose 6?

There will not exist an orthogonal design in 20 with 3 and 5 levels, as 20 is not a multiple of 3. You could try finding a larger design and remove choice tasks, but this will indeed lead to the loss of orthogonality.

The best option for designs for choice models is to actually let go of orthogonality. Orthogonality is not of importance for these models, only for linear models like linear regression. So I would advise generating an efficient design with zero priors, which will yield in a much better design than trying to find an orthogonal design and manually remove rows.

Michiel

[Sara]

Thanks Michiel,
That 20 was after eliminating some attributes and declining number of levels. Many thanks! so I am free of orthogonal design now. such a relief!

Sara

[Michiel Bliemer]

One thing to bear into mind, using zero priors can lead to dominant alternatives in some choice tasks (with any kind of design, orthogonal or efficient), so that manual inspection may still be necessary. The only way to avoid this is to put in the sign of each parameter as a prior, for example b[-0.001] will indicate a negative parameter (but set very close to zero). In this way, Ngene can automatically exclude such dominant alternatives from choice tasks. Dominant alternatives are altenatives in which the attribute levels are 'better' for each attribute compared to all other alternatives, and should not appear in your design. If your choice model in unlabelled, you can use something like ;alts = alt1*, alt2* to exclude such dominant alternatives.

Michiel

[Sara]

Dear Michiel,

First of all many thanks to you and all Ngene team for kind help, guide and patience. It s really great to know you guys are there to help and we can count on it. As suggested earlier, I m using MNL design for pretest, using near zero priors. I am using actual levels as it is easier, and I don't know why my levels which are 0.5, 1.5 and 7.5 in formatted scenario change to 1, 2, and 8. Here is the syntax I'm using and I wonder whether I m making some mistake. However, I changed the levels to 1,2,3 and I know there should be no such a problem, but kind of curious to know the reason with design using actual levels. By the way those price are absolute deviations from references (e.g. CB(+1.5,+1,+0.5) and same for FT and so on) and will be different for each respondent!

Design
;alts = Car, train, bus
;rows = 12
;block=2
;eff = (mnl,d)
;model:
U(car) = b1[-0.001] + b2 [0.001] * HC[15,25,50] + b3[-0.001]* TVC[50,25,0]+ b4[-0.001]*CR[2,3,4]
+ b5[-0.001]*CP[5,10,15]+b6[-0.001]*GT[10,20,30]/
U(Train) = B20 [-0.001]+ b7[-0.001] * TVT[3,10,20] + b8[-0.001] *Ac[1,2,3]+ b9[-0.001] * FT[7.5,10,15]+
b10 [-0.001]* TVT [-30,-20,-10] +b11[-0.001]*CT[2,1.5,1] /
U(Bus) = b12[-0.001] * HB[5,15,30] +b13[0.001] * ACB[1,2,3]+ b9 *FT + b14[-0.001]* TVB [-25,0,25]
+b15[-0.001]*CB[1.5,1,0.5] $

Bests
Sara