R help - Feb 2000 - Follow-up: Inverse prediction with R?

(message from 22.2.2000 13:04 Uhr):
> 
> 
> Why don't you inverse the modelling instead:
> 
> t.m.i <- lm((x~y)
> 
Jan,

thanks for the tip, but it's not just the same. The coefficients come out
differently, since the squared y residuals are minimized. Orthogonal
regression would be symmetric, but least squares is not, I'm afraid. And,
what's more, I have to state my model that way since the x values I have got
are fixed. 

Perhaps I have to explain where my problem comes from: I want to measure the
age of trees. I cannot cut them and I can take core samples only from the
bigger ones. What I can measure is the number of bud rings and branchings,
which is proportional to the age, although the measurement is rather
inaccurate. So, I 'd like to model that number based on the core samples
I've got (N?40), which are accurate, and the inversely "predict core
samples" for the rest of my trees (N?300).

Maybe there is another way to do what I intend, but at the moment I've no
clue. I'm grateful for any hint.

 

Kaspar


-- 

Kaspar Pflugshaupt
Geobotanisches Institut
Zuerichbergstr. 38
CH-8044 Zuerich

Tel. ++41 1 632 43 19
Fax  ++41 1 632 12 15

mailto:pflugshaupt at geobot.umnw.ethz.ch
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You fit
y = a+b*x

Then you get the inverse regression estimate

x = (y-a)/b

I don't see what is hard about that.

As for the confidence interval for x, try bootstrapping:
- randomly sample your (x,y) data
- do the y = a+b*x fit
- get the estimate x = (y-a)/b
- use the bootstrapped x distribution to get your CI

Or use the formulas you saw in a book.

I don't see this as a hard programming project. (I am no ace R programmer
myself)

Bill

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