I have noticed that the inverse.predict function (chemCal package) does not take into account all the degrees of freedom of the model in order to calculate the confidence interval, and I am wondering why.
Let me explain a bit better. I was trying to figure out how much my calibration would change if I would use 5 sets of standards (6 levels) instead of 1 set. Building the two models I can see how each model has different degrees of freedom ($n-2$) where $n$ is the number of calibration standards. The model with 5 sets of standards has 28 df and the model with one set of standards has 4 df. (see below the code. It is actually a model with the logarithm both in $x$ and $y$)
However, when I apply the inverse.predict function to calculate the confidence interval, I see that it uses as $t$-value the one with 4 degrees of freedom (corresponding to one standard set). So I am wondering why is that?
I was not able to access the book that is give as a reference, but in most literature they consider $n$ the number of calibration points and not the number of levels (which I agree that in most cases is the same).
So the question is: Do you think it is correct to use the inverse.predict function with more than one set of standards? Or, should I make my own function considering $n$ as the number of calibration points instead of the calibration levels?
Would you recommend any other package? I am interested as well in applying this with models with weighted least squares regression. So it should be able to handle that as well.
logmod_all<-lm(log10(signal)~log10(conc), data=calibdata) #model with the whole dataset
logmod_1<-lm(log10(signal)~log10(conc), data=calibdata) #model with one set of standards
summary(logmod_all)
summary(logmod_1)
Up to here is all ok. I see that the two models have different degrees of freedom. And if I back calculate the $t$-value, I see that it is using different $t$-values with different df to calculate the confidence interval. But then when I do the inverse predict, I get:
library("chemCal")
inv_all<-inverse.predict(logmod_all, log(2500),0.05)
inv_all
$Prediction
[1] 3.600191
$`Standard Error`
[1] 0.006570478
$Confidence
[1] 0.01824257
$`Confidence Limits`
[1] 3.581948 3.618433
inv_1<-inverse.predict(logmod_1, log(2500),0.05)
inv_1
$Prediction
[1] 3.589216
$`Standard Error`
[1] 0.008305648
$Confidence
[1] 0.02306018
$`Confidence Limits`
[1] 3.566156 3.612277
inv_all$Confidence/inv_all$`Standard Error`
[1] 2.776445
inv_1$Confidence/inv_1$`Standard Error`
[1] 2.776445
So you can see how in both cases it calculated the $t$-value with 4 degrees of freedom.
qt(c(0.025,0.975), df=4)
[1] -2.776445 2.776445
