Hello, Ok, got it, thanks.
Apparently I was right, the estimator is still x.pred = (y.new - b0)/b1 which can be obtained from predict.lm with the inverse regression model formula. According to the litterature, the main differences are in the confidence intervals for the true value of x.pred, for which there are several ways of computing the limits. So, I'll simply say, use lm() and predict.lm() but not the standard errors if you want CI's.
(The formulae for the CI's don't seem very hard to program, by the way. But there are several of them.)
Rui Barradas Em 29-08-2012 18:58, Peter Ehlers escreveu:
I think that what the OP is looking for comes under the heading of "inverse regression" or the "calibration" problem. One reference with a simple explanation including confidence intervals is "Applied regression analysis" by Draper and Smith. (It's in section 3.2 in my 3rd edition). Peter Ehlers On 2012-08-29 10:03, Rui Barradas wrote:Hello, Inline. Em 29-08-2012 16:06, John Thaden escreveu:Could it be that my newdata object needs to include a column for the concn term even though I'm asking for concn to be predicted? If so, what numbers would I fill it with? Or should my newdata object include the original data, too? Is there another mailing list I can ask?stackoverflow.com There's an R tag. Rui BarradasThanks, -JohnOn Wed, Aug 29, 2012 at 9:16 AM, John Thaden <johntha...@gmail.com> wrote:I think I may be misreading the help pages, too, but misreading how?I agree that inverting the fitted model is simpler, but I worry that I'm misusing ordinary least squares regression by treating my response, with its error distribution, as a predictor with no such error. In practice, with my real data that includes about six independent peak area measurements per known concentration level, the diagnostic plots from plot.lm(inv.model) lookvery strange and worrisome.Certainly predict.lm(..., type = "terms") must be able to do what I need.-JohnOn Wed, Aug 29, 2012 at 6:50 AM, Rui Barradas <ruipbarra...@sapo.pt> wrote:Hello,You seem to be misreading the help pages for lm and predict.lm, argument'terms'.A much simpler way of solving your problem should be to invert the fittedmodel using lm(): model <- lm(area ~ concn, data) # Your original model inv.model <- lm(concn ~ area, data = data) # Your problem's model. # predicts from original data pred1 <- predict(inv.model) # predict from new data pred2 <- predict(inv.model, newdata = new) # Let's see it. plot(concn ~ area, data = data) abline(inv.model) points(data$area, pred1, col="blue", pch="+") points(new$area, pred2, col="red", pch=16)Also, 'data' is a really bad variable name, it's already an R function.Hope this helps, Rui Barradas Em 28-08-2012 23:30, John Thaden escreveu:Hello all, How do I actually use the output of predict.lm(..., type="terms") to predict new term values from new response values? I'm a chromatographer trying to use R (2.15.1) for one of the most common calculations in that business:- Given several chromatographic peak areas measured for controlsamples containing a molecule at known (increasing) concentrations, first derive a linear regression model relating the known concentration (predictor) to the observed peak area (response) - Then, given peak areas from new (real) samples containing unknown amounts of the molecule, use the model to predict concentrations of the molecule in the unknowns.In other words, given y = mx +b, I need to solve x' = (y'-b)/m for newdata y' and in R, I'm trying something like this require(stats) data <- data.frame(area = c(4875, 8172, 18065, 34555), concn = c(25, 50, 125, 250)) new <- data.frame(area = c(8172, 10220, 11570, 24150)) model <- lm(area ~ concn, data) pred <- predict(model, type = "terms") #predicts from original data pred <- predict(model, type = "terms", newdata = new) #error pred <- predict(model, type = "terms", newdata = new, se.fit = TRUE) #error pred <- predict(model, type = "terms", newdata = new, interval = "prediction") #error new2 <- data.frame(area = c(8172, 10220, 11570, 24150), concn = 0) new2 pred <- predict(model, type = "terms", newdata = new2) #wrong results Can someone please show me what I'm doing wrong? ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
