Solved – How to interpret logistic regression coefficients with interactions between binary and continuous variables

logisticlogitrregression coefficients

I know this has been already asked, but I am quite confused about the interpretation of logit regression estimates if I have interacted variables (continuous and binary ones).

I run the following regression:

model <- glm(elected ~ treat + factor(School) + factor(Race) + 
               treat*Treat.City, data = subset(df, Year == 2016),
             family = binomial(link = 'logit'))

My dependent variable elected is equal to 1 if a political candidate got elected, 0 otherwise. treat equals 1 if the candidate belongs to a treatment group, 0 if belongs to the control group.

After controlling for schooling and race dummy variables, I have put the interaction treat*Treat.City, in which Treat.City is a continuous variables indicating the percentage of treatment candidates in relation to the total number of challengers inside candidate's i city.

Running the regression in R, I have the following results:

Call:
glm(formula = elected ~ treat + factor(School) + factor(Race) + 
    treat * Treat.City, family = binomial(link = "logit"), data = subset(df, 
    Year == 2016))

Deviance Residuals: 
   Min      1Q  Median      3Q     Max  
-1.875  -1.321   1.000   1.039   1.262  

Coefficients:
                                  Estimate Std. Error z value Pr(>|z|)    
(Intercept)                        0.42387    0.15196   2.789 0.005281 ** 
treat                             -0.22397    0.03879  -5.775 7.71e-09 ***
factor(School)MÉDIO_INCOMPLETO     0.04055    0.03452   1.174 0.240227    
factor(School)SUPERIOR_COMPLETO    0.11976    0.03221   3.718 0.000201 ***
factor(School)SUPERIOR_INCOMPLETO  0.11576    0.02947   3.929 8.55e-05 ***
factor(Race)BRANCA                -0.12757    0.15054  -0.847 0.396742    
factor(Race)INDÍGENA              -0.57795    0.26393  -2.190 0.028542 *  
factor(Race)Preta_Parda           -0.20933    0.15073  -1.389 0.164897    
Treat.City                         1.50083    0.61352   2.446 0.014435 *  
treat:Treat.City                   2.80625    0.95484   2.939 0.003293 ** 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 54123  on 39893  degrees of freedom
Residual deviance: 54033  on 39884  degrees of freedom
AIC: 54053

Number of Fisher Scoring iterations: 4

How can I interpret such coefficients? More specifically, how can I numerically make a statement about the effect of the treatment on the probability in getting elected?

Can I make any clear interpretation about this 'Intensity of treatment' variable that Treat.City is?

Best Answer

The odds of being elected when you are not treated increases by a factor $\exp(1.50083)\approx 4.49$ or $(4.49-1)\times100\%=349\%$ if you move from a city with no one treated to a city where everyone is treated.

This effect of Treat.City increases by a factor $\exp(2.80625\approx16.55)$ or $(16.55-1)\times100\%=1555\%$ if one is treated. For more see: http://maartenbuis.nl/publications/interactions.html

Given the large size of the effect I will assume that Treat.City is not a percentage but a proportion. The effects will be more realistic and easier to interpret when you turn Treat.City into percentages.

Related Solutions

Logistic Regression – How to Interpret Coefficients

If you're fitting a binomial GLM with a logit link (i.e. a logistic regression model), then your regression equation is the log odds that the response value is a '1' (or a 'success'), conditioned on the predictor values.

Exponentiating the log odds gives you the odds ratio for a one-unit increase in your variable. So for example, with "gender", if Female = 0 and Male = 1 and a logistic regression coefficient of 0.014, then you can assert that the odds of your outcome for men are exp(0.014) = 1.01 times that of the odds of your outcome in women.

Solved – Understanding over-dispersion as it relates to the Poisson and the Neg. Binomial

This answer comes in two parts. The first addresses the issue of the standard errors and why that implies the models are not identical, as @ndoogan observed in comments, and the second addresses, partially, the issue of when the coefficient estimates might be substantially different.

Consider, for example, a hypothesis test on the coefficient of log(Vmaj) where the null is that the coefficient equals 0.5. The two sets of model estimates would result in rejection of the null in the Poisson case, unless testing at a very high confidence level, and failure to reject the null in the NB case.

More generally, there is more to a collection of estimates than just the point estimates. In the presence of (Negative binomial-like) overdispersion, the standard errors produced by the Negative Binomial-based model will be more accurate estimates of the underlying standard deviation of the coefficient estimates. Thus, the NB model as a whole is more accurate.

For example, a simple model with $\mu_i = \exp\{1 + x_i\}$, and $y_i \sim NB$ with mean $\mu_i$ and variance $5\mu_i$:

N <- 1000
mu <- exp(1 + (x <- rnorm(N)))

p <- 0.2
r <- mu * p / (1-p)
y <- rnbinom(N, r, p)

poisson_summary = summary(glm(y~x, family="poisson"))
nb_summary = summary(glm.nb(y~x))

# Parametric bootstrap calculation of the s.e. of the coefficient of x

coef_x <- rep(0,1000)
for (i in 1:length(coef_x)) {
   mu <- exp(1 + (x <- rnorm(N)))
   r <- mu * p / (1-p)
   y <- rnbinom(N, r, p)
   coef_x[i] <- summary(glm(y~x, family="poisson"))$coefficients["x","Estimate"]
}
data.frame("Sim. SE" = sd(coef_x - 1), 
           "Poisson SE" = poisson_summary$coefficients["x", "Std. Error"],
       "N.B. SE" = nb_summary$coefficients["x", "Std. Error"])

     Sim..SE Poisson.SE    N.B..SE
1 0.03492467 0.01273354 0.03984743

where you can see that the simulated std. deviation of the coefficient estimate is roughly 3x the Poisson-based estimate of same, and much better estimated by the NB-based estimate of same.

The coefficient estimates are pretty much the same, as one might expect with this sample size and the std. errors above, so I won't take up space by displaying them.

Additionally, although this is often a minor effect when overdispersion is low, by weighting the observations with more accurate weights (i.e., better estimates of observation-specific variances), the parameter estimates themselves will be more accurate, well, asymptotically at any rate. The rule of thumb I learned long ago was that heteroskedasticity adjustments (for that is what they are, in essence) don't buy you much unless the differences between weights are on the order of 5x or more.

Note, however, that in small samples you may well get more accurate (in MSE terms) point estimates with the Poisson model if there isn't much overdispersion, because you are reducing the variability induced by estimating the dispersion parameter. Of course, this is almost certainly more than offset by the loss of accuracy in the standard errors of the coefficient estimates.

Best Answer

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Logistic Regression – How to Interpret Coefficients

Solved – Understanding over-dispersion as it relates to the Poisson and the Neg. Binomial

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