Calibration data from Massart et al. (1997), example 3

Usage

data(massart97ex3)

Examples

data(massart97ex3)
attach(massart97ex3)

Die folgenden Objekte sind maskiert von massart97ex3 (pos = 3):

    x, y

yx <- split(y, x)
ybar <- sapply(yx, mean)
s <- round(sapply(yx, sd), digits = 2)
w <- round(1 / (s^2), digits = 3)
weights <- w[factor(x)]
m <- lm(y ~ x, w = weights)
calplot(m)

Warning message:
Assuming constant prediction variance even though model fit is weighted



# The following concords with the book p. 200
inverse.predict(m, 15, ws = 1.67)  # 5.9 +- 2.5

$Prediction
[1] 5.865367

$`Standard Error`
[1] 0.8926109

$Confidence
[1] 2.478285

$`Confidence Limits`
[1] 3.387082 8.343652


inverse.predict(m, 90, ws = 0.145) # 44.1 +- 7.9

$Prediction
[1] 44.06025

$`Standard Error`
[1] 2.829162

$Confidence
[1] 7.855012

$`Confidence Limits`
[1] 36.20523 51.91526



# The LOD is only calculated for models from unweighted regression
# with this version of chemCal
m0 <- lm(y ~ x) 
lod(m0)

$x
[1] 5.407085

$y
       1 
13.63911 



# Limit of quantification from unweighted regression
loq(m0)

$x
[1] 13.97764

$y
      1 
30.6235 



# For calculating the limit of quantification from a model from weighted
# regression, we need to supply weights, internally used for inverse.predict
# If we are not using a variance function, we can use the weight from
# the above example as a first approximation (x = 15 is close to our
# loq approx 14 from above).
loq(m, w.loq = 1.67)

$x
[1] 7.346195

$y
       1 
17.90777 


# The weight for the loq should therefore be derived at x = 7.3 instead
# of 15, but the graphical procedure of Massart (p. 201) to derive the 
# variances on which the weights are based is quite inaccurate anyway.