1. Logits: Odds of 1’s vs. 0’s in each Combination of Age, Education, Membership
In the chapter, we shorted age to “A,” education to “E,” membership to “M,” and going green was “G.” Here it is with the variable names spelled out: In the category m=1 (want to buy a green car), the model would say: ℓn(eijk1) = intercept + agei + educationj + memberk + green1 + agei*educationj + agei*memberk + educationj*memberk + agei*educationj*memberk + agei*green1 + educationj*green1 + memberk*green1 + agei*educationk*green1 for the category m=0 (not interested in the green car), the model would say: ℓn(eijk0) = intercept + agei + educationj + memberk + green0 + agei*educationj + agei*memberk + educationj*memberk + agei*educationj*memberk + agei*green0 + educationj*green0 + memberk*green0 + agei*educationk*green0.

2. Logits: Odds of 1’s vs. 0’s in each Combination of Age, Education, Membership
In natural logs, ℓn(eijk1/eijk0) = ℓn(eijk1) - ℓn(eijk0), so to compare the odds of going green or not, in each cell in the table, we study the logit: ℓn(eijk1/eijk0) = ℓn(eijk1) - ℓn(eijk0): = ( intercept + agei + educationj + memberk + green1 + agei*educationj + agei*memberk + educationj*memberk + agei*educationj*memberk + agei*green1 + educationj*green1 + memberk*green1 + agei*educationk*green1) - ( intercept + agei + educationj + memberk + green0 + agei*educationj + agei*memberk + educationj*memberk + agei*educationj*memberk + agei*green0 + educationj*green0 + memberk*green0 + agei*educationk*green0). Lots of terms cancel, leaving only those terms with “m” in them (currently denoted as 1s and 0s): = (green1 + agei*green1 + educationj*green1 + memberk*green1 + agei*educationk*green1) - (green0 + agei*green0 + educationj*green0 + memberk*green0 + agei*educationk*green0)

3. Logits: Odds of 1’s vs. 0’s in each Combination of Age, Education, Membership
Rearranging terms gives us: = (green1 - green0) + (agei*green1 - agei*green0) + (educationj*green1 - educationj*green0) + (memberk*green1 - memberk*green0) + (agei*educationk*green1 - agei*educationk*green0). The response variable is binary (green cars: y/n), so parameter for green1 is the same as green0 except they’ll have opposite signs (there’s only 1 degree of freedom for a binary variable). So, if the estimate for “interest in green” is green1 = 0.35, then the estimate for “not interested in green” green0 = So the logit equation can be further simplified. For example, the first term (green1 - green0) can be written as (green1 - -green1) = 2green1 Thus, the final simplified model is: = 2 (green1 + agei*green1 + educationj*green1 + memberk*green1 + agei*educationk*green1).