Labeled DCE with constants

[Fabio91]

Hi, I have set this syntax to be run on NGENE with 3 alternatives and for each one of them there is a different constant coefficient. This is a labeled DCE and the third option is the preferred one in the pilot as it is possible to see from the coefficients. Unfortunately, the final design seems not to take into account these coefficients. How can I fix this problem?
Below I put the syntax and its result.

Design
;alts = SQ, AltB, AltC, AltD
;rows =4
;eff = (mnl,d)
;model:
U(AltB) = b4[-0.2563516] + b2[0.00000001] * Bt1[0,1] + b3[-0.013]* Ct1[50,90,160,300] /
U(AltC) = b5[1.87443] + b2 * Bt1 + b3 * Ct1 /
U(AltD) = b6[3.300526]] + b2 * Bt1 + b3 * Ct1
$

[johnr]

Ngene ignores the constants by default in calculating the (D-, A-, etc.) errors, but accounts for them in computing the probabilities, Hessian, etc. If you want to include them in the calculation of the optimisation errors, you need to add the command ;con. Note that the constants won't appear in the design output - they are just columns of ones and you wouldn't show the constants to respondents, so we elected not to show them in the design output. But they are there.

John

[Michiel Bliemer]

If you follow the recommendation of John and add ;con to your syntax, your output will show S-estimates for each parameter as shown below (note that S-estimates for priors with unrealistic values such as 0.000000001 are meaningless and you can ignore).

Prior b4 b2 b3 b5 b6
Fixed prior value -0.2563516 1E-08 -0.013 1.87443 3.300526
Sp estimates 345.824152 7.39087604142757E+16 2.732314 7.510864 2.547695
Sp t-ratios 0.105397 0 1.185744 0.715173 1.227955

[Fabio91]

Dear Michiel and JOhn
thank you very much for your help!
I modified the desing which now is:

?Sintassi da utilizzare
Design
;alts = SQ, AltB, AltC, AltD
;rows =4
;eff = (mnl,d)
;con
;model:
U(AltB) = b4[-0.2563516] + b2[0.00000001] * Bt1[0,1] + b3[-0.013]* Ct1[50,90,160,300] /
U(AltC) = b5[1.87443] + b2 * Bt1 + b3 * Ct1 /
U(AltD) = b6[3.300526]] + b2 * Bt1 + b3 * Ct1
$
Our choicse sets and efficiency measures are the ones reported below.. just one more quesiton: in our pilot the respondents really liked the third alternative (as it is possible to see from the constant) Are the choice sets reflecting this information (see for instance choice set 2 in which the third alternative has a lower price and ttribute equals 1) ? or I am wrong in interpreting them ?

[johnr]

Ngene allows you to interrogate these types of questions. If you look at the design output, you will see a tree under design properties, MNL. If you open this tree, you can ask to view the computed choice probabilities and utilities (computed using the priors you used). If you can look at the choice probabilities, altD has the highest probabilities of being selected.

[Michiel Bliemer]

This is a result of the fact that you are asking for only 4 rows and that you are using the default algorithm (swapping) that satisfies attribute level balance. Essentially, you are telling Ngene that it HAS to use cost = 50 for AltD at least once, even though this is likely not a good idea since it is already quite a dominant alternative as you say.

You can avoid this issue by using the modified federov algorithm, in which attribute level balance is not imposed. If you add ;alg = mfederov to your syntax, then you will see that low levels for price will no longer be used. You can impose some degree of attribute level balance by setting lower and upper bounds on the number of times each attribute level needs to appear across the design.

As an additional comment, I think you really want to use more than 4 rows. Even though you can estimate the model with only 4 different questions, you will likely want a bit more variation in your data. You could use for example 16 rows that you split into 4 blocks, such that each respondent only faces 4 choice tasks in total.

Example syntax:

Code: Select all

Design
;alts = SQ, AltB, AltC, AltD
;rows = 16
;block = 4
;eff = (mnl,d)
;alg = mfederov
;con
;model:
U(AltB) = b4[-0.2563516] + b2[0.00000001] * Bt1[0,1](6-10,6-10) + b3[-0.013]* Ct1[50,90,160,300](2-6,2-6,2-6,2-6) /
U(AltC) = b5[1.87443] + b2 * Bt1 + b3 * Ct1 /
U(AltD) = b6[3.300526]] + b2 * Bt1 + b3 * Ct1a[50,90,160,300](0-4,2-8,3-7,5-9)
$

Michiel