choice-metrics.com

Dear everyone,

I am a new Ngene user and am designing a SP experiment which requires the respondents to make a binary choice between two options. We didn’t consider the feature of each option except their names (i.e., 1 and 2). Option=1 is the current service and option=2 is a new alternative under consideration. In each scenario, the respondent would be given an incentive (0= the current no incentive situation, 1=reward, and 2=fine) to motivate their choice. For reward, the levels are {2,5,10} while for fine, the levels are {20,30,50}. Since reward and fine will not exist at the same time, a level of 0 is added to reward and fine to be corresponding with incentive=0 level. As I would like to check whether option=2 would be chosen without any incentive, I added the condition when incentive=0, reward and penalty=0 and option=2.

With the following code, I can generate the experiments but I am not sure whether I coded correctly, or is there any better experiment design logic or coding approach to meet my research needs? And I would greatly appreciate it if you can point out my knowledge gap on experiment design and Ngene usage.

Code: Select all

Design
;alts = yes, no
;rows = 13
;eff=(mnl,fixed)
;cond:
if(yes.incentive=0, yes.reward=0 and yes.penalty=0 and yes.option=2),
if(yes.incentive=1,yes.reward>0 and yes.penalty=0),
if(yes.incentive=2,yes.reward=0 and yes.penalty>0)
;model:
U(yes) = cons + b_option * option[1,2] + b_incentive * incentive[0,1,2] + b_reward * reward[0,2,5,10] + b_penalty * penalty[0,20,30,50]$


Btw, I have noticed that we can use actually enumerate all 13 scenarios since the factors and their levels are not too many. But I still wonder how the program should be correctly coded.

Thanks in advance!

Dong

I think what you are looking for is interaction effects, which are multiplications of attribute levels.
If you create a dummy coded variable for incentive, where 0 is the base of no incentive, then you can multiply the respective dummy variables with reward and fine. If incentive is a reward, then incentive.dummy[1] = 1 and hence incentive.dummy[1'] * reward is equal to the reward. If incentive is a fine, then incentive.dummy[2] = 1 and hence incentive.dummy[2] * fine is equal to the fine. If there is no incentive, then incentive.dummy[1] = 0 and incentive.dummy[2] = 0.

I did not understand the yes/no altenatives and the options 1/2. It seems that you have two fixed alternatives that I describe by 'current' and 'new' that have no further attributes, is that correct? In that case, have a look at the syntax below.

Code: Select all

Design
;alts = new, current 
;rows = 12       
;eff=(mnl,fixed)
;alg = mfederov
;require:
current.incentive = 0
;model:
U(new)     = cons
           + b_incentive.dummy[0|0] * incentive[1,2,0]  ? 0 = none, 1 = reward, 2 = fine
           + b_reward               * incentive.dummy[1] * reward[2,5,10](4,4,4)
           + b_penalty              * incentive.dummy[2] * penalty[20,30,50](4,4,4)
           /
U(current) = b_incentive            * incentive
           + b_reward               * incentive.dummy[1] * reward
           + b_penalty              * incentive.dummy[2] * penalty
$

Michiel