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So, I have a binomial glm function, with two predictors, the second (as factor) being with two levels (50, 250).

model <- glm(GetResp.RESP ~ speed + PadLen, family = binomial(link = "logit"), data = myData)

The plot for it looks like this:

enter image description here

My question: How can I find the covariate (ball speed) specific to the .5 probability for each level of the second predictor?

For example, I've tried using the function dose.p(), from the package 'MASS':

dose.p(model, p = 0.5)

and I get

p = 0.5: 36.9868

which, by just looking at the plot, it would be the value for the first (50) level. Now, how can I find it for the second (250) level as well?

Thank you.

dput(myData):

https://pastebin.com/7QYXwwa4

Benjamin
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1 Answers1

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Since this is a logistic regression, you're fitting the function:

log(p/(1-p)) = b0 + b1*speed + b2*PadLen

where p is the probability of GetResp.RESP being equal to 1, b0, b1, and b2 are the regression coefficients, and PadLen is a dummy variable equal to zero when myData$PadLen is 50 and equal to 1 when myData$PadLen is 250.

So you can solve for the speed at p = 0.5:

log(1) = b0 + b1*speed + b2*PadLen
b1*speed = log(1) - b0 - b2*PadLen
speed = (log(1) - b0 - b2*PadLen)/b1

Since log(1) = 0, this reduces to:

speed = (-b0 - b2*c(0,1))/b1

Or, putting in the actual coefficient values:

speed = (-coef(model)[1] - coef(model)[3]*c(0,1))/coef(model)[2]

To solve for speed at other probabilities, just keep the log-odds factor in the equation and enter whatever value of p you desire:

speed = (log(p/(1-p)) - coef(model)[1] - coef(model)[3]*c(0,1))/coef(model)[2]
eipi10
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  • Thank you very much! There is a small correction I would like to add, though: log(p/1-p) will give -Inf, because it will calculate p/1, then subtract p, which is not ideal, so log(p/(1-p)) would work. – Benjamin Apr 16 '18 at 17:38
  • Thanks. I forgot the parentheses! I've fixed it now. – eipi10 Apr 16 '18 at 17:54