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author | ranke <ranke@5fad18fb-23f0-0310-ab10-e59a3bee62b4> | 2007-03-22 16:44:33 +0000 |
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committer | ranke <ranke@5fad18fb-23f0-0310-ab10-e59a3bee62b4> | 2007-03-22 16:44:33 +0000 |
commit | 49679639ded05e7ef954c23f50d3d94d3d6dc1dd (patch) | |
tree | b6dc323d45f296f9a296fb5586873a7818344be1 /inst/doc | |
parent | f4443942f10740ecc62b928181a1911ef14eeb04 (diff) |
Start of the integration of nonlinear calibration models
git-svn-id: http://kriemhild.uft.uni-bremen.de/svn/chemCal@19 5fad18fb-23f0-0310-ab10-e59a3bee62b4
Diffstat (limited to 'inst/doc')
-rw-r--r-- | inst/doc/chemCal.Rnw | 36 |
1 files changed, 34 insertions, 2 deletions
diff --git a/inst/doc/chemCal.Rnw b/inst/doc/chemCal.Rnw index 8cdc97c..956c664 100644 --- a/inst/doc/chemCal.Rnw +++ b/inst/doc/chemCal.Rnw @@ -29,8 +29,8 @@ on this subject. However, I did not encounter any proof or explanation of the formula cited below yet, so I can't really confirm that Massart's method is correct. When calibrating an analytical method, the first task is to generate a suitable -model. If we want to use the \texttt{chemCal} functions, we will have to restrict -ourselves to univariate, possibly weighted, linear regression so far. +model. If we want to use the \texttt{chemCal} functions, we will have to +restrict ourselves to univariate, possibly weighted, linear regression so far. Once such a model has been created, the calibration can be graphically shown by using the \texttt{calplot} function: @@ -119,6 +119,38 @@ where I interpret $\frac{{s_s}^2}{w_s}$ as an estimator of the variance at locat $\hat{x_s}$, which can be replaced by a user-specified value using the argument \texttt{var.s} of the function \texttt{inverse.predict}. +\section*{Fitting and using a variance function} + +In the R package \texttt{nlme} variance functions are used for weighted +regressions. But since the \texttt{predict.nlme} method does not calculate +prediction intervals, this is not useful for the \texttt{calplot} function. + +Two approaches could be used for fitting variance functions, one based on +residuals from an unweighted fit, and one based on just the variances +of the different samples along the x axis. If we used the residuals for +fitting, a bias of the model in a certain area would result in a higher +variance, so it seems preferable to choose the second approach. Of course, +a prerequisite is to have sufficient repetitions for each sample in any +case. + +Let's take the above example and estimate a variance function + +<<>>= +massart97ex3 +massart97ex3$x <- factor(massart97ex3$x) +summary <- summaryBy(y~x, data = massart97ex3,FUN=c(mean,sd,var)) +summary$x <- as.numeric(as.vector((summary$x))) +plot(summary$x, summary$y.var, + xlim=c(0,50), + ylim=c(0,max(summary$y.var))) +varModel <- lm(y.var ~ I(x^2) + x, data=summary) +varCurve <- predict(varModel, newdata=data.frame(x=0:5000/100)) +lines(0:5000/100,varCurve) + + + + + \begin{thebibliography}{1} \bibitem{massart97} Massart, L.M, Vandenginste, B.G.M., Buydens, L.M.C., De Jong, S., Lewi, P.J., |