choice-metrics.com

Dear all,
I have attributes and levels for my DCEs and use N-gene to ran 14 choice sets but I dont' know if I use full factorial design How many choice sets in my DCEs ? I want to compare with pro of d-eff design.
PhT (2 years) PhT (4 years)
attributes1 they can do excellence they can do all PhT (2 years)and added excellence
competency of informatic technology (inf) competency of informatic technology (inf)

level 1 type1 type4
level 2 type2 type5
level 3 type3 type6
level 4 - type7
attributes2 they can do excellence competncy they can doall PhT (2 years)and added excellence competncy of drug product management (logis) of drug product management (logis)
level 1 type1 type5
level 2 type2 type6
level 3 type3 type7
level 4 type4 type8
attributes3 salary at entry
level 1 8,985 BAHT 11,000 BAHT
level 2 13,800 BAHT 15,000 BAHT
level 3 - 18,000 BAHT

Best Wishes,
Sukunta Muadthong

Hi Sukunta

The size of the full factorial is simply the product of the number of levels of each attribute. Taking your previous syntax

i1 = 3 levels (1,2,3), L1 = 4 levels (1,2,3,4), salarycer = 2 levels (8985,13800), i3= 4 levels (4,5,6,7), L2 = 4 levels (5,6,7,8), and salaryba = 3 levels (11000,15000,18000)

Hence, there are 3*4*2*4*4*3 = 1152 different combinations or sets in the full factorial.

John

Dear, Prof. John R.

thank you very much for your kindness. Then if I have 2 alternatives the number of choice sets is 1152*(1152-1)/2 =66,2976 or not?

Best Wishes,
Sukunta Muadthong

Hi Sukunta

No, I based my calculation on the attributes and levels of both alternatives as per your syntax.

i1 = 3 levels (1,2,3), L1 = 4 levels (1,2,3,4), salarycer = 2 levels (8985,13800) are from the first alternative.

and

i3= 4 levels (4,5,6,7), L2 = 4 levels (5,6,7,8), and salaryba = 3 levels (11000,15000,18000) are from the second.

Hence 3*4*2 (first alternative) *4*4*3 (second alternative) = 1152.

John

Dear Prof. John R.,
oh. thank you very much for your kindness.

Best Wishes,
Sukunta Muadthong