Attribute level frequency

[Corinna]

Hi,

I am trying to run a design for an unlabeled choice experiment. Since level "2" of attribute B should occur more often than the other levels, I wrote the constraint as described in the manual on page 151:

;eff=(mnl,d)
;cond:
If(alt1.c=110, alt1.a=[0,100]),
If(alt1.c=469, alt1.a=[100,200]),
If(alt1.b=2, alt1.d=0),
If(alt2.c=110, alt2.a=[0,100]),
If(alt2.c=469, alt2.a=[100,200]),
If(alt2.b=2, alt2.d=0),
If(alt3.c=110, alt3.a=[0,100]),
If(alt3.c=469, alt3.a=[100,200]),
If(alt3.b=2, alt3.d=0)
;model:
U(alt1)= b0+ b1.effects [0.01|0,02]* A[100,200,0]+ b2.effects [0.005|0,01|0,01]*B[100,200,300,2] (1-4,1-4,1-4,4) + b3 [-0.004]* C[110,219,299,359,469]+ b4.effects [0.01 | 0.005 | 0.005]* D[100,200,300,0] /
U(alt2)= b1.effects*A + b2.effects*B(1-4,1-4,1-4,4) + b3*C + b4.effects*D /
U(alt3)= b1.effects*A + b2.effects*B (1-4,1-4,1-4,4) + b3*C + b4.effects*D
$

The requested occurance of attribute leves for attribute B is ignored by ngene. Level "2" occurs only once for each of the three alternatives and no Error message is given.
Out of curiosity I tried to run a code where B is not effects coded, but then ngene gives the error message

Error: Attribute 'b' has an incorrect number of attribute level frequencies specified: '4,1-4,1-4,1-4'

Can you help me find my mistake?

Thank you.

Corinna

[Corinna]

Hello,

any idea if there is a possibility to influence Attribute Level frequency with effects coded variables?

Corrina

[Michiel Bliemer]

Apologies for the delayed response, I was teaching a course in London last week.

You need to remove (1-4,1-4,1-4,4) since conditions are using the swapping algorithm, which already automatically aims to satisfy attribute level balance. Adding these attribute level frequency constraints are only required when you use the modified Federov algorithm, for example when you use ;require or ;reject constraints.

Michiel

[Corinna]

Dear Michiel,

thank you very much for your answer, it was very helpful.
I modified the design, it is now labelled and I am using the federov algorithm.
I am very uncertain about the code and I have further questions.

Code: Select all

Design
;alts=alt1,alt2,alt3,alt4
;rows=10
;eff=(mnl,s)
;alg = mfederov(candidates=9000)
;reject:
alt1.a=0 and alt1.d>0,      
alt2.h=0 and alt2.d>0,
alt3.g=0 and alt3.d>0

;model:
U(alt1)= b0[0.33] + b1.effects   [0.01|0.02]*   A[1,2,0] (1-10,1-10,3) +       
b2      [-0.3] *      C  [1.10,2.19,2.99] (1-10,2,1-10)  +       
b3.effects   [0.01 | 0.005 | 0.005]*   D[1,2,3,0] /

U(alt2)=     b4[0.66]  + b10.effects[0.01|0.02|0.03]*H[1,2,3,0](1-10,1-10,1-2,3) +     b5[-0.3] *E[2.19,2.99,3.59] (1-10,2,1-10)    +     b3.effects*D /
U(alt3)=     b6[1]  + b7.effects[0.02|0.03]*G[2,3,0](1-2,1-10,3)  +     b8[-0.3] *F [2.99,3.59,4.69] (1-10,2,1-10)   +     b3.effects*D
$

1) the priors were derived from a pretest, so I am using ;eff=(mnl,s) now instead of ;eff=(mnl,d). I am still uncertain about the priors, not all of them were significant.
2) my S-estimate is very high with >100 000, which conclusion can I draw from this? I know the design is very restricted. Can I use it although the S-estimate is so high?

Thanks a lot!
Corinna

[Michiel Bliemer]

The code seems fine to me. You get very high S-estimates because your priors for example for b1 are very small and therefore attribute A hardly contributes to utility, which means that it will be difficult to estimate these parameters. S-estimates are only meaningful if you believe that your priors are reasonably OK.

You may want to increase the number of rows in order to include more variation in your design (e.g. ;rows = 10 and ;block = 2) since you are estimating quite a few parameters.

Michiel

[Corinna]

Thank you a lot for your help!

[Michiel Bliemer]

I notice I made a small mistake in my response, I meant of course ;rows = 20 and ;block = 2.

Good luck.