choice-sets per block

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choice-sets per block

Postby jctrujillo » Wed Apr 26, 2017 6:58 pm

Dear members of the Ngene team,

Thanks a lot for this opportunity to express our doubts in the use of this magnificent software.

I have been using Ngene 1 year, and I still have a doubt when it comes to the syntax. For instance, if I have the following syntax (whose priors were taken from a pilot test consisting of 3 blocks of 10 choice-sets):

Design
;alts = A*, B*, SQ
;rows = 39
;block = 3, minsum, total (10 mins), newblocking
;eff = (mnl, d)
;reject:
A.veh=0 and A.ped=0,
B.veh=0 and B.ped=0
;model:
U(A) = f[-0.00367645]*fatal[40,35,25,15,0] + v.dummy[0.06|0.2]*veh[3,2,1] + p.dummy[0.01|0.06]*ped[3,2,1] + pri[0.0015341]*price[2,5,10,15,18]/
U(B) = f*fatal + v.dummy*veh + p.dummy*ped + pri*price/
U(SQ)= asc[0.2]$

Should I be compelled to use the resulting 13 choice-sets per block or could I eliminate some of those choice-sets deliberately and reduce their quantity per block aiming to avoid the respondents’ fatigue?
jctrujillo
 
Posts: 1
Joined: Wed Apr 26, 2017 5:48 pm

Re: choice-sets per block

Postby Michiel Bliemer » Wed May 10, 2017 8:04 pm

I think that 13 may be a bit much, but you better how who your respondents are, and how you will interview them. You can consider lowering the number of experiments in your design, for example below I lowered the design to 30 so you show 10 to each respondent.

Note that you need to use the mfederov algorithm of you are adding constraints like ;reject and ;require, otherwise they are not taking into account. This algorithm lets go of attribute level balance, so in order to have some degree of balance in your design I have specified lower and upper limits for each attribute to appear within the design.


Design
;alts = A*, B*, SQ
;rows = 30
;block = 3
;eff = (mnl, d)
;alg = mfederov(candidates = 1000)
;reject:
A.veh=0 and A.ped=0,
B.veh=0 and B.ped=0
;model:
U(A) = f[-0.00367645]*fatal[40,35,25,15,0](4-8,4-8,4-8,4-8,4-8) + v.dummy[0.06|0.2]*veh[3,2,1](8-12,8-12,8-12) + p.dummy[0.01|0.06]*ped[3,2,1](8-12,8-12,8-12) + pri[0.0015341]*price[2,5,10,15,18](4-8,4-8,4-8,4-8,4-8)/
U(B) = f*fatal + v.dummy*veh + p.dummy*ped + pri*price/
U(SQ)= asc[0.2]$
Michiel Bliemer
 
Posts: 1705
Joined: Tue Mar 31, 2009 4:13 pm


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