Too many conditions?
Posted: Sat Mar 23, 2019 7:03 pm
Hi,
Thank you for all of the help so far. I am about to start a final experiment, following two rounds of piloting to get priors. We have a topic with a lot of conditions, because we are trying to make the experiment as true-to-life as possible, but I am concerned that this may be too much, as the list of conditions is very long compared to other designs I have seen.
My questions are:
1) Does the number of conditions look ok for this design? (and as an extension to that, does the code look ok in general?)
2) A minor second question - I have switched from a labelled to an unlabelled design - in that case, do I just need one alternative specific constant (b1) for all rather than the three included below?
Many thanks!
Thank you for all of the help so far. I am about to start a final experiment, following two rounds of piloting to get priors. We have a topic with a lot of conditions, because we are trying to make the experiment as true-to-life as possible, but I am concerned that this may be too much, as the list of conditions is very long compared to other designs I have seen.
My questions are:
1) Does the number of conditions look ok for this design? (and as an extension to that, does the code look ok in general?)
2) A minor second question - I have switched from a labelled to an unlabelled design - in that case, do I just need one alternative specific constant (b1) for all rather than the three included below?
Many thanks!
- Code: Select all
Design
;alts = alt1,alt2,alt3,alt4
;eff = (mnl,d,mean)
;bdraws = sobol(500)
;rows = 24
;block = 2
;cond:
if(alt1.att1=0, alt1.att3=[0,1]),
if(alt2.att1=0, alt2.att3=[0,1]),
if(alt3.att1=0, alt3.att3=[0,1]),
if(alt1.att1=1, alt1.att3=[1,2,3,4,5]),
if(alt2.att1=1, alt2.att3=[1,2,3,4,5]),
if(alt3.att1=1, alt3.att3=[1,2,3,4,5]),
if(alt1.att1=2, alt1.att3=[3,4]),
if(alt2.att1=2, alt2.att3=[3,4]),
if(alt3.att1=2, alt3.att3=[3,4]),
if(alt1.att1=0 AND alt1.att3=0, alt1.price=[1000,2500,5000]),
if(alt2.att1=0 AND alt2.att3=0, alt2.price=[1000,2500,5000]),
if(alt3.att1=0 AND alt3.att3=0, alt3.price=[1000,2500,5000]),
if(alt1.att1=0 AND alt1.att3=1, alt1.price=[1000,2500]),
if(alt2.att1=0 AND alt2.att3=1, alt2.price=[1000,2500]),
if(alt3.att1=0 AND alt3.att3=1, alt3.price=[1000,2500]),
if(alt1.att1=2, alt1.price=[250,500]),
if(alt2.att1=2, alt2.price=[250,500]),
if(alt3.att1=2, alt3.price=[250,500]),
if(alt1.att1=1 AND alt1.att3=1,alt1.price=[1000,2500]),
if(alt2.att1=1 AND alt2.att3=1,alt2.price=[1000,2500]),
if(alt3.att1=1 AND alt3.att3=1,alt3.price=[1000,2500]),
if(alt1.att1=1 AND alt1.att3=2,alt1.price=[500,1000]),
if(alt2.att1=1 AND alt2.att3=2,alt2.price=[500,1000]),
if(alt3.att1=1 AND alt3.att3=2,alt3.price=[500,1000]),
if(alt1.att1=1 AND alt1.att3=5,alt1.price=[25,250,500]),
if(alt2.att1=1 AND alt2.att3=5,alt2.price=[25,250,500]),
if(alt3.att1=1 AND alt3.att3=5,alt3.price=[25,250,500]),
if(alt1.att1=1 AND alt1.att3=3,alt1.price=[25,250,500]),
if(alt2.att1=1 AND alt2.att3=3,alt2.price=[25,250,500]),
if(alt3.att1=1 AND alt3.att3=3,alt3.price=[25,250,500]),
if(alt1.att1=1 AND alt1.att3=4,alt1.price=[25,250,500]),
if(alt2.att1=1 AND alt2.att3=4,alt2.price=[25,250,500]),
if(alt3.att1=1 AND alt3.att3=4,alt3.price=[25,250,500]),
if(alt1.att1=0, alt1.att2=[0,1,2,3]),
if(alt2.att1=0, alt2.att2=[0,1,2,3]),
if(alt3.att1=0, alt3.att2=[0,1,2,3]),
if(alt1.att1=1, alt1.att2=[0,1,2,3,4,5]),
if(alt2.att1=1, alt2.att2=[0,1,2,3,4,5]),
if(alt3.att1=1, alt3.att2=[0,1,2,3,4,5]),
if(alt1.att1=2, alt1.att2=[0,1,2,3,4]),
if(alt2.att1=2, alt2.att2=[0,1,2,3,4]),
if(alt3.att1=2, alt3.att2=[0,1,2,3,4])
;model:
U(alt1) = b1[-2.542] + att1.dummy[0.182|0.527]*att1[0,1,2] + place.dummy[1.386|-0.337|0.137|0.294|1.037]* place[0,1,2,3,4,5] + form.dummy[-0.924|-1.175|-0.612|0.094|0.66]* form[0,1,2,3,4,5] + price[-0.0001]*price[25,250,500,1000,2500,5000] + scenario.dummy[0.362|0.65]*scenario[0,1,2]/
U(alt2) = b2[-2.542] + att1* att1 + place* place + form* form + price*price + scenario*scenario[scenario]/
U(alt3) = b3[-2.542] + att1* att1 + place* place + form* form + price*price + scenario*scenario[scenario]/
U(alt4) = 0
$