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Dear Ngeners
I am trying to create an efficient design for my exp. My sample size will be approximately 120 respondents I will have two attributes one risk reduction (baseline risk 20% risk reduction 5% ,30%,50%,95%) and out of pocket cost. The main objective of the study is thus very simple to estimate the WTP for a health risk reduction.
The risk reduction is equal for all but the out of pocket four levels depend on the available budget. Following a previous survey I was able to divide the population in three budget categories (let’s call high medium, low) so the out of pocket payment are just a fixed percentage of the available budget (2%,7%,15% and 30%).
My questions are:
1)I still have to do a pilot study on a sample of 20-25 subjects. I am thinking of doing a dual response ( three choice with reference alternative included and two choices between two interventions) but there are no previous studies evaluating the WTP for this health risk so I have no priors for the pilot should I just go for an orthogonal design and use the coefficient obtained in the main study even if they are not stat significant? Is there a way to include status quo alternative in orthogonal design?
2) In the actual experiment I am using visual aids not numbers to report the risk reduction is there a way to integrate images in the Ngene formatted scenarios?
3)In the main study I would like to opt for a dual response choice with Bayesian design with three segments. I want to report the actual out of pocket payments (not the available budget after the payment) so for the cost attribute I will change both the reference and the levels in the three segments. With the following coding (reported below) and using as priors for cost and risk reductions the following values -0.7 for cost and -0.5 for risk reduction I made a preliminary design. Can I still optimize it for the C error (which for me is very important given that I am interested in the WTP?) By changing the distribution of priors can I check which perform best looking at the D error?
Doing this approach I don’t get the S coefficient so I am not able to estimate if my sample size will be OK is there another way I can get an estimate of S? My D error is low 0.016 but as you mention in previous posts it does not mean anything..
Design
;alts (fourth)= alt1, alt2,alt3
;alts (fifth)= alt1, alt2,alt3
;alts (inter)= alt1, alt2,alt3
;rows = 12
;block = 2
;eff = fish(mnl,d,mean)
;fisher(fish)= design1(fourth[0.33], fifth[0.33], inter[0.34])
;rep = 250
;bdraws = gauss(2)
;model(fourth):
U(alt1) = b1[(u,-0.7,-0.5)] * A.ref[0] + b2[(u,-0.5,0)] * B.ref[20] /
U(alt2) = b1* A.piv[1,3,6,11] + b2 * B.piv[-5%,-30%,-50%,-95%] /
U(alt3) = b1* A.piv[1,3,6,11] + b2 * B.piv[-5%,-30%,-50%,-95%]

;model(fifth):
U(alt1) = b1[(u,-0.7,-0.5)] * A.ref[0] + b2[(u,-0.5,0)] * B.ref[20] /
U(alt2) = b1* A.piv[1.5,5,11,22] + b2 * B.piv[-5%,-30%,-50%,-95%] /
U(alt3) = b1* A.piv[1.5,5,11,22] + b2 * B.piv[-5%,-30%,-50%,-95%]

;model(inter):
U(alt1) = b1[(u,-0.7,-0.5)] * A.ref[0] + b2[(u,-0.5,0)] * B.ref[20] /
U(alt2) = b1* A.piv%[3,11,25,50] + b2 * B.piv[-5%,-30%,-50%,-95%] /
U(alt3) = b1* A.piv[3,11,25,50] + b2 * B.piv[-5%,-30%,-50%,-95%] $
9
The prob estimates are divided as following (not exact numbs) :
Alt 1 Alt 2 Alt 3
0.73 0.15 0.11
0.01 0.01 0.97
0.12 0.79 0.08
0.15 0.97 0.01
.30 0.09 0.60
0.78 0.09 0.12
0.78 0.12 0.09
0.96 0.016 0.016
0.11 0.08 0.79
0.73 0.11 0.16
0.86 0.10 0.03
0.74 0.18 0.09

4) do I interpret the probabilities obtained from the model? Is it good when they are equally spread (among three alternatives 0.33,0.33,0.34 or when there are dominant alternatives 0.90 0.05,0.05?)
How you would judge the above probab distribution:
Thank you very very Much
Kind Regards
Carla

Dear Carla,

Only questions 2 and 3 are actually Ngene questions, the other two refer to experimental design in general, but I will try to answer all four.

(1) I would suggest using Bayesian Normal priors around zero if you do not know the design of the parameters, this is better than using an orthogonal design. If you known the sign, you can use Bayesian Uniform priors, e.g. (u,-0.5,0) for a negative parameter as you already did in your syntax. There is no problem adding a status quo. Note that the value of your priors need to be consistent with the attribute levels. So you cannot have a prior of -0.5 and a level of 20, as you have in your syntax, as -0.5 * 20 = -10, which is HUGE and will dominate the utility function. Also -0.7 * 11 is too large of a number, the priors do not make sense. You will have a lot of dominant choice tasks, as you can see from your probabilities. So please use priors that are consistent with your attribute levels. In your case, smaller prior values would make sense.

(2) Not yet, it is on the list.

(3) You would be able to see the S-estimates when you inspect in the design the separate designs, so go to 'fourth' for example, and inspect the MNL properties. The S-estimate will be given. Please note that 2 Gaussian draws is likely not enough, you may wish you increase this to 3, 4 or 5, depending on the range of your Bayesian prior.

(4) Utility balance (i.e. equal spread) is not a good property, but completely unbalanced is not good either because of dominant alternatives. Somewhere in between will be most efficient, e.g. (0.2, 0.5, 0.3). Probabilities of (0.33, 0.33, 0.33) and (1, 0, 0) are not very efficient.

Michiel

Dear Michiel
Thank you very much your reply was very helpful I will lower my prior coefficients!
C
Carla

Dear Ngeners
Sorry to bother you again ..
I changed my priors by making them small and as my model uses unlabeled alternatives I had to include the * to avoid dominant alternatives.
The problem is that if I do that let using this model:


Design
;alts (fourth)= alt1*, alt2*,alt3*
;alts (fifth)= alt1*, alt2*,alt3*
;alts (inter)= alt1*, alt2*,alt3*
;rows = 12
;block = 2
;eff = fish(mnl,d,mean)
;fisher(fish)= design1(fourth[0.33], fifth[0.33], inter[0.34])
;rep = 250
;bdraws = gauss(4)
;model(fourth):
U(alt1) = b1[(u,-0.07,-0.05)] * A.ref[0] + b2[(u,-0.05,0)] * B.ref[2] /
U(alt2) = b1* A.piv[1,3,6,11] + b2 * B.piv[-5%,-30%,-50%,-95%] /
U(alt3) = b1* A.piv[1,3,6,11] + b2 * B.piv[-5%,-30%,-50%,-95%]

;model(fifth):
U(alt1) = b1[(u,-0.07,-0.05)] * A.ref[0] + b2[(u,-0.05,0)] * B.ref[20] /
U(alt2) = b1* A.piv[1.5,5,11,22] + b2 * B.piv[-5%,-30%,-50%,-95%] /
U(alt3) = b1* A.piv[1.5,5,11,22] + b2 * B.piv[-5%,-30%,-50%,-95%]

;model(inter):
U(alt1) = b1[(u,-0.07,-0.05)] * A.ref[0] + b2[(u,-0.05,0)] * B.ref[20] /
U(alt2) = b1* A.piv%[3,11,25,50] + b2 * B.piv[-5%,-30%,-50%,-95%] /
U(alt3) = b1* A.piv[3,11,25,50] + b2 * B.piv[-5%,-30%,-50%,-95%] $

Ngene failed to identify random desings...
I also tried for the simple example provided in the textbook for Pivot and gain Ngene failed to identify choice sets without dominant alternative :
Design
;alts = alt1*, alt2*, alt3*
;rows = 12
;eff = (mnl,d)
;model:
U(alt1) = b1[0.6] * A.ref[2] + b2[-0.1] * B.ref[5] /
U(alt2) = b1 * A.piv[-1,0,1] + b2[-0.2] * B.piv[-25%,0%,25%] /
U(alt3) = b1 * A.piv[-1,0,1] + b2[-0.2] * B.piv[-25%,0%,25%] $

Is a problem with my coding or it is not possible to avoid dominant alternatives in Ngene when running a pivot design?
Looking forward to hearing from you
Regards
Carla

Dear Carla,

There exist in total 256 different choice tasks in your case, and you would like to have 12 out of these 256. However, most choice tasks will include a dominant alternative, as you only have 2 attributes while you have 3 alternatives. Since Ngene randomly selects 12 out of 256, it is quite likely that at least one choice task contains a dominant choice task. Using only 2 attributes is always a bit tricky.

To increase the likelihood of finding a design, you can decrease the number of rows, or increase the number of attribute levels.

Michiel

Dear Michel
hank you very much for your suggestion
Indeed I just deleted the block command from my model alowwing for only 6 choice sets in general and now is running perfectly !
Thank you very much
Kind Regards
Carla