Advanced Regression - Categorical X variables and Interaction terms 

Hey I know it's been awhile, but I just came across your comment and was curious. How'd the rest of the class go?

How'd the rest of your class go?

You are absolutely right in pointing this omission

think of it in terms of marginal effect: consider that you have a simpler form of regression with only pink slip (X1) and agecat4( X2) + interaction term (X1*X2) with a coef of b3, if you re going to asses the marginal effect of the var1 ( pink slip ) , you ll proceed with the derivative of Y over deriv of X1 and the result il the following ( Dy/Dx1 = b1+ b3* X2 ) , thus considering the " boolean" nature of X2 if X2=1 , then Y will increase by b1+b3 else it will only increase by b1, same thing if you want to see the marginal effect of X2 ! => Conclusion : always think of it as the derivative over the variable and you ll get the answer

There is no omission: Ln(P) = constant (b0) - 0.181Ac2 - 0.800Ac3 - 0.390Ac4 -0.209 LnD + 0.123PS +1.371(Ps x Ac4) The term [- 0.390Ac4] is what you counted as omitted, but it is present. Model7 Coef Var (ind) LN Anti-Ln c 9.125 1 9.1250 ac2 -0.181 0 0.0000 ac3 -0.800 0 0.0000 ac4 -0.390 1 -0.3900 lnd -0.209 5.669880923 -1.1850 ps 0.123 1 0.1230 psac 1.371 1 1.3710 9.0440 8,467.54 (Apologies, the table does not show very well here)

I was wondering about this as well. Glad you caught it as well! Edit: He did do it while calculating price in model 7 at the very end.

That time slot is about the effect of attaining a 'pink slip' and the -0.390 item is only for the age of the car, hence not included, I think.

Think about the kids learning advanced regression! /s lol

On the time stamp you provided, I think it should be 60.64% higher rather than 47.4%.

Yes! I found the same issue in the last video as well, it should be: the median of the car price will be e^0.474=1.6064 times. Thus 60.64% increase

@@imglenngarcia I think the pink slip interaction term case also has the same problem..

I think it is because he is not doing a logistic regression but a linear regression.

I have the same question, would appreciate if anyone willing to share the answer for this one

@@malikakbar It does, even if the variable on itself is not statistically significant. Think about this scenario: There's a new AgeCat5 variable that is added for all cars that are over 70 years old; and also we are gonna add another variable called redCar, that takes 1 as a value if the car is red, and 0 otherwise. Now, if I were to ask you: "Does the color of the car being red have an impact in the price of the car?" Probably not, cars are customizable and the color on itself doesn't seem to be too relevant to determine the price of the car. For the sake of the example, let's stirr things up and claim that in the 40's(80years ago) almost all the cars were either grey or black, and only BMW was producing fancy red cars that were way more expensive than the grey/black cars; but since it's been so long ago it's really hard nowadays to find those special edition BMW cars. So, I believe it's pretty easy to tell that if you find a car that is over 80years old whose original paint is red, it's almost certain that it is one of those special edition BMW red cars that are way more expensive than the others. So, for all the other cars that were fabricated less than 70 years ago, the color is not a factor that affects the price of the car, so the variable "redCar" is probably not gonna be statistically significant on itself, but in the case that the car is 70years older, the color of the car is gonna play a huge role to predict the value of the car, because if it is red,. the price is gonna be way more expensive, therefore, it's gonna be significant in that scenario. In this case, intuiton tells me we should add another predictive variable to the equation that describes that relationship, which you would do like this "+ Beta6 * redCar*ageCat5

Actually I think it is when the p value of the interaction term is statistically significant. Then you see there's an effect modifying e.g. age modifying selling a car with a pink slip.

Same here, I thought it was (exp(coef) - 1) * 100 for non-transformed independent variables. Someone please correct me.

Yes I was thinking this as well. I have always learned that amount of dummies = category levels (age in this case 4) - 1 = amount of dummies that meet be created

Restoring the Ac1 (age 1 to 5 years) into the Model is okay. However, the value of Ac1 = 0, annulling the entire term and contributing nothing to the Model, just like the Ac2 and Ac3. Other interactions are feasible. (e.g., odometer vs. age).

You can also interpret the effect of agecat4, which works the same way. Models that are in agecat4, but have NO pink-slip, have on average a 39% lower price than the baseline, holding all else constant. If they do have a pink-slip, the price is on average 98,1% (1,371-0,39) higher than the baseline, hold all else constant.

statistics is funny

...to them that are statistically lost, it is indeed a comedy! 1CO1!18

I have the same exact question. Have you found the answer?

If an interaction variable is SIGNIFICANT, t]then its components must be admitted into the model.

Whatever you remove demands that you recalculate your Model Regression and obtain new coefficients and significances.

I think according to what he said is a bit subjective (domain knowledge)

Any combination can interact! But to what effect? If the interaction is INSIGNIFICANT, we save time by ignoring the crossing, except we are constrained, say in an exam setting.

It should be e^1.104 times higher

@@helloworld1537 We generally think in decimals.

Why not? Use a calculator and type in -0.209log(290)

@@EpicLaith By now, you have settled this matter. To use Excel, change 'log' to 'LN' and add = ahead of the minus sign. = - 0.209 * LN(290) and enter (voila!)

Hi! Thanks for watching! Regarding your observations: 1. They certainly can be interpreted as a % change on y. Remember that if you were to exponentiate the log, youd find that y=exp(B0+B1X). Or in other words y = exp(B0)*exp(B1X), so the constant term just becomes a multiplying constant (ie. won't affect the percentage change in y for a given absolute change in X). 2. The ageCat4 term is def there! Just on the second line of the equation. Hope that helps.

@@zedstatistics 1. Erm exactly my point. When B1 increases by 10%, B1X increases by 10%. But Exp(B1X) does not. 2. It was in the forumula but not summed

Sorry ignore 2.

@@thenineteennn Be careful here, B1 does not increase at all. B1 is the coefficient. It is X that increases. Also, if you're talking about a log-linear relationship (ie. ln(y)=B0 + B1X...) then a 1 unit increase in X has a constant % effect on Y. As per my first response. If you have a log-log relationship (ie. ln(y)=B0+B1(ln(X))), only THEN can you interpret it as "a 1% increase in X relates to a 1% increase in Y". But this is different to your example above. Hope that helps :)

@@zedstatistics how come at 20:00 you added the main effect of pink slip but didn't subtract the main effect of agecat4?

We're talking about the effect of pinkslip. So not a mistake, sweetie.

@@zedstatistics but without considering Cat4, that term is 0. You should combine the effect of having the pinkSlip on cat4, not over Cat1 (BASE category). Thus, the price increases 110.4%.

Could you elaborate? Please

@@OskarBienko he shoud have sum the coefficients when cat4=1, pinkslip=1 and pinkslipXcat4=1

The extra effect of pinkslip is zero without consdering cat4

Rock that volume button bruh!