inverse of 'predict' function

Question

Using predict() one can obtain the predicted value of the dependent variable (y) for a certain value of the independent variable (x) for a

Answer 1

Saw the previous answer is deleted. In your case, given n=50 and the model is binomial, you would calculate x given y using:

f <- function (y,m) {
  (logit(y/50) - coef(m)[["(Intercept)"]]) / coef(m)[["x"]]
}
> f(30,model)
[1] 48.59833

But when doing so, you better consult a statistician to show you how to calculate the inverse prediction interval. And please, take VitoshKa's considerations into account.

Answer 2

For just a quick view (without intervals and considering additional issues) you could use the TkPredict function in the TeachingDemos package. It does not do this directly, but allows you to dynamically change the x value(s) and see what the predicted y-value is, so it would be fairly simple to move x until the desired Y is found (for given values of additional x's), this will also show possibly problems with multiple x's that would work for the same y.

Answer 3

Came across this old thread but thought I would add some other info. Package MASS has function dose.p for logit/probit models. SE is via delta method.

> dose.p(model,p=.6)
             Dose       SE
p = 0.6: 48.59833 1.944772

Fitting the inverse model (x~y) would not makes sense here because, as @VitoshKa says, we assume x is fixed and y (the 0/1 response) is random. Besides, if the data weren’t grouped you’d have only 2 values of the explanatory variable: 0 and 1. But even though we assume x is fixed it still makes sense to calculate a confidence interval for the dose x for a given p, contrary to what @VitoshKa says. Just as we can reparameterize the model in terms of ED50, we can do so for ED60 or any other quantile. Parameters are fixed, but we still calculate CI's for them.

Answer 4

You just have to rearrange the regression equation, but as the comments above state this may prove tricky and not necessarily have a meaningful interpretation.

However, for the case you presented you can use:

(1/coef(model)[2])*(model$family$linkfun(30/50)-coef(model)[1])

Note I did the division by the x coefficient first to allow the name attribute to be correct.

Answer 5

The chemcal package has an inverse.predict() function, which works for fits of the form y ~ x and y ~ x - 1