Ted
Thank you for giving me so much of your time to provide an explanation of
the likelihood ratio approach to the calibration problem. It has certainly
provided me with a new insight into this problem. I will check out the
references you mentioned to get a better understanding of the statistics.
Out of interest I have also tried the function provided by David Lucy back
in March 2002 with a correction to the code and addition of the t-value.
The results are very close to those obtained by the likelihood ratio method,
although the output needs tidying up.
####################
# R function to do classical calibration
# val is a data.frame of the new indep variables to predict dep
calib <- function(indep, dep, val, prob=0.95)
{
# generate the model y on x and get out predicted values for x
reg <- lm(dep~indep)
xpre <- (val - coef(reg)[1])/coef(reg)[2]
# generate a confidence
yyat <- ((dep - predict(reg))^2)
sigyyat <- ((sum(yyat)/(length(dep)-2))^0.5)
term1 <- sigyyat/coef(reg)[2]
sigxxbar <- sum((indep - mean(indep))^2)
denom <- sigxxbar * ((coef(reg)[2])^2)
t<-qt(p=(prob+(1-prob)/2), df=(length(dep)-2))
conf <- t*abs((((1+1/length(dep))+(((val -
mean(dep))^2)/denom))^0.5)*term1)
results<-list(Predicted=xpre, Conf.interval=c(xpre-conf, xpre+conf))
return(results)
}
conc<-seq(100, 280, 20) # mg/l
abs<-c(1.064, 1.177, 1.303, 1.414, 1.534, 1.642, 1.744, 1.852, 1.936, 2.046)
# absorbance units
new.abs<-data.frame(abs=c(1.251, 1.324, 1.452))
calib(conc, abs, new.abs, prob=0.95)
$Predicted
abs
1 131.2771
2 144.6649
3 168.1394
$Conf.interval
$Conf.interval$abs
[1] 124.4244 137.9334 161.5477
$Conf.interval$abs
[1] 138.1298 151.3964 174.7310
Thanks
Mike
----- Original Message -----
From: "Ted Harding" <[EMAIL PROTECTED]>
To: "Mike White" <[EMAIL PROTECTED]>
Cc: <R-help@stat.math.ethz.ch>
Sent: Wednesday, April 20, 2005 8:58 AM
Subject: RE: [R] Help with predict.lm
> Sorry, I was doing this too late last night!
> All stands as before except for the calculation at the end
> which is now corrected as follows:
>
> On 19-Apr-05 Ted Harding wrote:
> [code repeated for reference]
> > The following function implements the above expressions.
> > It is a very crude approach to solving the problem, and
> > I'm sure that a more thoughtful approach would lead more
> > directly to the answer, but at least it gets you there
> > eventually.
> >
> > ===========================================
> >
> > R.calib <- function(x,y,X,Y){
> > n<-length(x) ; mx<-mean(x) ; my<-mean(y) ;
> > x<-(x-mx) ; y<-(y-my) ; X<-(X-mx) ; Y<-(Y-my)
> >
> > ah<-0 ; bh<-(sum(x*y))/(sum(x^2)) ; Xh <- Y/bh
> > sh2 <- (sum((y-ah-bh*x)^2))/(n+1)
> >
> > D<-(n+1)*sum(x^2) + n*X^2
> > at<-(Y*sum(x^2) - X*sum(x*y))/D; bt<-((n+1)*sum(x*y) + n*X*Y)/D
> > st2<-(sum((y - at - bt*x)^2) + (Y - at - bt*X)^2)/(n+1)
> >
> > R<-(sh2/st2)
> >
> > F<-(n-2)*(1-R)/R
> >
> > x<-(x+mx) ; y<-(y+my) ;
> > X<-(X+mx) ; Y<-(Y+my) ; Xh<-(Xh+mx) ;
> > PF<-(pf(F,1,(n-2)))
> > list(x=x,y=y,X=X,Y=Y,R=R,F=F,PF=PF,
> > ahat=ah,bhat=bh,sh2=sh2,
> > atil=at,btil=bt,st2=st2,
> > Xhat=Xh)
> > }
> >
> > ============================================
> >
> > Now lets take your original data and the first Y-value
> > in your list (namely Y = 1.251), and suppose you want
> > a 95% confidence interval for X. The X-value corresponding
> > to Y which you would get by regressing x (conc) on y (abs)
> > is X = 131.3813 so use this as a "starting value".
> >
> > So now run this function with x<-conc, y<-abs, and these values
> > of X and Y:
> >
> > R.calib(x,y,131.3813,1.251)
> >
> > You get a long list of stuff, amongst which
> >
> > $PF
> > [1] 0.02711878
> >
> > and
> >
> > $Xhat
> > [1] 131.2771
> >
> > So now you know that Xhat (the MLE of X for that Y) = 131.2771
> > and the F-ratio probability is 0.027...
> >
> *****> You want to push $PF upwards till it reaches 0.05, so work
> *****> *outwards* in the X-value:
> WRONG!! Till it reaches ***0.95***
>
> R.calib(x,y,125.0000,1.251)$PF
> [1] 0.9301972
>
> ...
>
> R.calib(x,y,124.3510,1.251)$PF
> [1] 0.949989
>
>
> > and you're there in that direction. Now go back to the MLE
> > and work out in the other direction:
> >
> > R.calib(x,y,131.2771,1.251)$PF
> > [1] 1.987305e-06
>
> ...
>
> R.calib(x,y,138.0647,1.251)$PF
> [1] 0.95
>
> > and again you're there.
> >
> > So now you have the MLE Xhat = 131.2771, and the two
> ****> limits of a 95% confidence interval (131.0847, 131.4693)
> WRONG!!!
> limits of a confidence interval (124.3510, 138.0647)
>
> > for X, corresponding to the given value 1.251 of Y.
>
> As it happens, these seem to correspond very closely to
> what you would get by reading "predict" in reverse:
>
> > plot(x,y)
> > plm<-lm(y~x)
> > min(x)
> [1] 100
> > max(x)
> [1] 280
>
> > points((131.2771),(1.251),col="red",pch="+") #The MLE of X
> > lines(c(124.3506,138.0647),c(1.251,1.251),col="red") # The above CI
> > newx<-data.frame(x=(100:280))
> > predlm<-predict(plm,newdata=newx,interval="prediction")
> > lines(newx$x,predlm[,"fit"],col="green")
> > lines(newx$x,predlm[,"lwr"],col="blue")
> > lines(newx$x,predlm[,"upr"],col="blue")
>
> which is what I thought would happen in the first place, given
> the quality of your data.
>
> Sorry for any inconvenience due to the above error.
> Ted.
>
>
>
>
>
>
>
> --------------------------------------------------------------------
> E-Mail: (Ted Harding) <[EMAIL PROTECTED]>
> Fax-to-email: +44 (0)870 094 0861
> Date: 20-Apr-05 Time: 08:58:37
> ------------------------------ XFMail ------------------------------
>
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