Min number of choice tasks

[ASAL2433]

Hi,

For the following formula:

Min number of choice tasks:

S > K/ (J-1)

For J, the number of alternatives, do you include the opt out option? I.e. If included: TxA, TxB, and no Tx, j is equal to 3.

Thank you!

[Michiel Bliemer]

Yes J also includes the opt-out alternative.
Note that the resulting value is only a MINIMUM, larger number of recommended to allow more variation in the data and to be able to estimate more advanced models. My rule of thumb is to multiply the minimum with at least 3.

Michiel

[ASAL2433]

Thank you for your quick response!

I have two follow up questions:

I have 4 attributes with 3 levels, and then 3 attributes with 4 levels. However, one of the attributes is cost (with 4 levels) which I am assuming will be coded as a continuous variable, and I will be using it to estimate WTP. Firstly, in the equation for min choice tasks, can I assume that cost will only have one parameter. Secondly, in a mixed MNL model do I double the number of parameters to account for random parameters:

Mixed MNL model:
Number of parameters = 4* 3 + 2*4 + 1 = 21
Double the number of parameters to account for random parameters = 42
Min number of choice tasks: 42/2 = 21

Number of choice tasks = 21 * 3 = 62

Are there any mistakes in my steps?

Thank you!

[Michiel Bliemer]

That looks correct to me.

Note that some have pointed out that the degrees of freedom calculation is actually not affected by having random coefficients, so instead of assuming 42 parameters you could consider just 21 coefficients (each with a mu and sigma to estimate). With 21 coefficients it would be 21 / 2 * 3 = 32. So 32 rows would likely be sufficient in your case, but using a larger number is never a bad idea. Having 62 rows should definitely be sufficient.

Michiel

[ASAL2433]

Thank you again!!

I have some follow up questions to help me understand the following formula a bit more:

Min number of choice tasks:

S > K/ (J-1)

1. As mentioned above, if I have two choices and an opt out option then J=3. Therefore by adding the opt out option, we are decreasing the minimum number of choice sets required as we are diving by a bigger number (2 vs 1). What is the reason for this? Is it because the opt out option assumes levels of zero for each attribute so we are getting more information by adding this alternative?

2. If I am using a two-stepped approach in my DCE (begins with two choices and an opt out option, and then if the opt out is chosen then a forced choice is displayed), do I still have J=3?

Thanks,
Amber

[Michiel Bliemer]

1. Yes, there are still trade-offs being made between the opt out alternative and the other alternatives. If you have two alternatives, A and B, then making a choice yields one data point, i.e. either A is preferred over B or the other way around. If you have three alternatives, A, B, and C, then selecting A would yield two data points, namely A is preferred over B and A is preferred over C. So with J alternatives you get J-1 data points. Of course the information you capture with an opt-out alternative is less than with a status quo alternative or regular alternative that has attribute levels, but technically adding any alternative will give you more data points. But some people do not count the opt-out alternative in the calculation S > K/(J-1) because of the fact that the opt-out alternative does not increase information much (it mainly provides information about an alternative-specific constant), which is also fine. I use S > K/(J-1) merely as a quick rule of thumb to determine the minimum bound and then I multiply with 2 or 3 to ensure sufficient variation, I would never use the minimum bound anyway.

2. Yes then you technically have 3 alternatives and you actually have two choice sets, one with opt-out and one without opt-out, so you obtain more data.

Michiel

[ASAL2433]

Thank you!!