Regression (logistic) in R: Finding x value (predi

问题:

I've fitted a logistic regression model that predicts the a binary outcome vs from mpg (mtcars dataset). The plot is shown below. How can I determine the mpg value for any particular vs value? For example, I'm interested in finding out what the mpg value is when the probability of vs is 0.50. Appreciate any help anyone can provide!

model <- glm(vs ~ mpg, data = mtcars, family = binomial)

ggplot(mtcars, aes(mpg, vs)) + 
    geom_point() + 
    stat_smooth(method = "glm", method.args = list(family = "binomial"), se = FALSE)

回答1:

The easiest way to calculate predicted values from your model is with the predict() function. Then you can use a numerical solver to find particular intercepts. For example

findInt <- function(model, value) {
    function(x) {
        predict(model, data.frame(mpg=x), type="response") - value
     }
}

uniroot(findInt(model, .5), range(mtcars$mpg))$root
# [1] 20.52229

Here findInt just takes the model and a particular target value and returns a function that uniroot can solve for 0 to find your solution.

回答2:

You can solve for mpg directly as follows:

mpg = (log(p/(1-p)) - coef(model)[1])/coef(model)[2]

Detailed explanation:

When you fit the regression model, the equation you are fitting is the following:

log(p/(1-p)) = a + b*mpg

Where p is the probability that vs=1, a is the intercept and b is the coefficient of mpg. From the model fit results (just type model or summary(model)) we see that a = -8.8331 and b = 0.4304. We want to find mpg when p=0.5. So, the equation we need to solve is:

log(0.5/(1-0.5)) = -8.331 + 0.4304*mpg
log(1) = 0 = -8.331 + 0.4303*mpg

Rearranging,

mpg = 8.8331/0.4304 = 20.523

In general, to solve for mpg for any value of p:

mpg = (log(p/(1-p)) + 8.8331)/0.4304

Or, to make it more easily reproducible:

mpg = (log(p/(1-p)) - coef(model)[1])/coef(model)[2]