On Jan 15, 2009, at 11:31 AM, e-letter wrote:
Perhaps a coding error on my part (or on your part). Perhaps
different
methods (none of which you describe)?
I suspect that my method only used the first two points (I just
checked by plotting and -2.7 is closer to the paper and pen result I
get than is -3.28. Perhaps you made an extrapolation from a linear
fit
of a dataset that is not co-linear?
lm(c(0,5) ~ c(16,45))
Call:
lm(formula = c(0, 5) ~ c(16, 45))
Coefficients:
(Intercept) c(16, 45)
-2.7586 0.1724
It not that "R is different", it is merely that I used it
differently
than you used your other tools.
Here's another method ( using all points and again reversing the
roles
of x and y) :
lm(c(0,5,10,15,20) ~ c(16,45,77,101,125))
Call:
lm(formula = c(0, 5, 10, 15, 20) ~ c(16, 45, 77, 101, 125))
Coefficients:
(Intercept) c(16, 45, 77, 101, 125)
-3.2332 0.1818
My understanding from gnuplot manual is that a marquart-levenberg
algorithm is used, which I applied to the data to perform a least
squares best fit linear curve. Gnuplot returns values for the
intercept and gradient which I then apply to solve the linear equation
y=mx+c. Similarly with scilab, where the regress(ion?) function was
applied. Qtiplot performed non-weighted linear regression to output
values similar to those from gnuplot.
Why reverse the roles of x and y in your method?
I accidentally switched x and y ad then realized I could get an
intercept value without the labor of solving by hand.
Although your revised value is closer to those from other programs,
how do I understand and
explain the discrepancy?
The regression line for x ~ y is *not* the same as the regression line
for y ~ x.
If you want to check that the numbers from R agree with your other
solutions, then take the regression equation from
> lmmod
Call:
lm(formula = c(16, 45, 77, 101, 125) ~ c(0, 5, 10, 15, 20))
Coefficients:
(Intercept) c(0, 5, 10, 15, 20)
18.00 5.48
... and then solve for x = 0 .... as you apparently did with the other
systems.
--
David Winsemius
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