How to interpret the intercept term in a GLM?

Question

I am using R and I have been analysing my data with GLM with Binomial link.

I want to know what is the meaning of the intercept in the output table. The intercept for one of my models is significantly different, however the variable is not. What does this mean?

What is the intercept. I don't know if I am just confusing myself but having searched the internet, there is nothing just saying, it is this, take notice of it...or don't.

Please help, a very frustrated student

---

glm(formula = attacked_excluding_app ~ treatment, family = binomial, 
    data = data)
Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-2.3548   0.3593   0.3593   0.3593   0.3593  
Coefficients:
                         Estimate Std. Error z value Pr(>|z|)   
(Intercept)                 2.708      1.033   2.622  0.00874 **
treatmentshiny_non-shiny    0.000      1.461   0.000  1.00000

(Dispersion parameter for binomial family taken to be 1)
Null deviance: 14.963  on 31  degrees of freedom
Residual deviance: 14.963  on 30  degrees of freedom
(15 observations deleted due to missingness)
AIC: 18.963
Number of Fisher Scoring iterations: 5

Answer 1

The intercept term is the intercept in the linear part of the GLM equation, so your model for the mean is $E[Y] = g^{-1}(\mathbf{X \beta})$, where $g$ is your link function and $\mathbf{X\beta}$ is your linear model. This linear model contains an "intercept term", i.e.:

$\mathbf{X\beta} = c + X_1\beta_1+X_2\beta_2+\cdots$

In your case the intercept is significantly non-zero, but the variable is not, so it is saying that

$\mathbf{X\beta} = c \neq 0$

Because your link function is binomial, then

$g(\mu) = \ln\left(\frac{\mu}{1-\mu}\right)$

And so with just the intercept term, your fitted model for the mean is:

$E[Y] = \frac{1}{1+e^{-c}}$

You can see that if $c=0$ then this corresponds to simply a 50:50 chance of getting Y=1 or 0, i.e. $E[Y] = \frac{1}{1+1} = 0.5$

So your result is saying that you can't predict the outcome, but one class (1's or 0's) is more likely than the other.

Answer 2

It looks to me like there may be some problem with the data. It is odd that the parameter estimate for the coefficient would be 0.000. It looks like both your DV and your IV are dichotomous and that the proportions of your DV do not vary at all with your IV. Is this right?

The intercept, as I noted in my comment (and as @corone 's answer implies) is the value of the DV when the IV is 0. How was your IV coded? As is, though, the fact that the estimate for the coefficient is 0.000 implies that the IV makes no difference.

Therefore, the intercept of 2.708 is the estimated logit of the DV: that is, $\text{log}(\frac{p}{1-p})$ at all levels of the IV.

Answer 3

In your case, the intercept is the grand mean of attacked_excluding_app, calculated for all data regardless of treatment. The significance test in the table of coefficients is testing whether it is significantly different from zero. Whether this is relevant depends on whether you have some a priori reason to expect it be zero or not.

For instance, imagine you had tested a drug and a placebo for their effect on blood pressure. For each subject, you record the change in their blood pressure by calculating (pressure after treatment - pressure before treatment) and treat this as the dependent variable in your analysis. You then find that the effect of treatment (drug vs. placebo) is non-significant but that the intercept is significantly > 0 - this would tell you that on average, your subjects' blood pressure increased between the two measurement times. This might be interesting and need further investigation.