1 / 3

Logits : Odds of 1’s vs. 0’s in each Combination of Age, Education, Membership

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:

imaran
Télécharger la présentation

Logits : Odds of 1’s vs. 0’s in each Combination of Age, Education, Membership

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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 green1is the same as green0except 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 = -0.35. 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).

More Related