Dear Something,
Ah, you need accessor functions. Set your lm() call to a variable name. If
my data frame or object is called dat, for example, I usually use dat.lm.
summary(dat.lm)
resid(dat.lm)
?summary.lm gives a host of information, including summary(dat.lm)$sigma and
$df for resolving SSE.
Alternatively, sum(c(resid(dat.lm)^2)).
See anova() as well.
Also, vif() [car] for the variance inflation factor (e.g. tolerance=1/vif)
Hope this helps.
Sincerely,
KeithC.
From: Something Something [mailto:mailinglists19@gmail.com]
Sent: Sunday, February 14, 2010 11:49 PM
To: kMan
Subject: Re: [R] lm function in R
Thank you so much, kMan. That makes sense. Only one question, how can I
see the value of 'error'? Here's what I see:
Call:
lm(formula = Y ~ X1 * X2 * X3, na.action = na.exclude)
Coefficients:
(Intercept) X1 X2 X3 X1:X2 X1:X3
X2:X3 X1:X2:X3
177.9245 0.2005 2.4482 3.1216 0.8127 -26.6166
-3.0398 29.6049
I don't see 'error' here. Please explain. Thanks again for your
help.
On Sun, Feb 14, 2010 at 3:26 PM, kMan <kchamberln@gmail.com> wrote:
Dear Something,
Tricky to differentiate subscripts vs. different variables in plain text, so
I'll just use 2, X and Z.
Y~X+Z yields the model Y=b0 + b1*X + b2*Z + error
Y~X*Z yields the model Y=b0 + b1*X + b2*Z + b3*X*Z + error |b3 is the
interaction term.
Y~X+Z+I(X*Z) is the same thing but with the interaction term spelled out.
See ?I
XZint<-X*Z
Y~X+Z+XZint same thing but avoids the call to I() in the model.
Note that:
Y~a*a or Y~a^2 does not evaluate an interaction, hence the use of I().
Sincerely,
KeithC.
-----Original Message-----
From: Something Something [mailto:mailinglists19@gmail.com]
Sent: Saturday, February 13, 2010 2:24 PM
To: Daniel Nordlund
Cc: r-help@r-project.org
Subject: Re: [R] lm function in R
Thanks Dan. Yes that was very helpful. I didn't see the change from
'*' to
'+'.
Seems like when I put * it means - interaction & when I put + it's not
an
interaction.
Is it correct to assume then that...
When I put + R evaluates the following equation:
Y-Hat = b0 + b1X1 + b2X2 + . . . bkXk + . . . + bkXk
But when I put * R evaluates the following equation; Y-Hat = b0 + b1X1 +
b2x2 + ... + bkXk + b12 X12+ b13 X13 +........ + c
Is this correct? If it is then can someone point me to any sources that
will explain how the coefficients (such as b0... bk, b12.. , b123..) are
calculated. I guess, one source is the R source code :) but is there any
other documentation anywhere?
Please let me know. Thanks.
On Fri, Feb 12, 2010 at 5:54 PM, Daniel Nordlund
<djnordlund@verizon.net>wrote:
> > -----Original Message-----
> > From: r-help-bounces@r-project.org
> > [mailto:r-help-bounces@r-project.org]
> > On Behalf Of Something Something
> > Sent: Friday, February 12, 2010 5:28 PM
> > To: Phil Spector; r-help@r-project.org
> > Subject: Re: [R] lm function in R
> >
> > Thanks for the replies everyone. Greatly appreciate it. Some
> > progress, but now I am getting the following values when I don't
use
> > "as.factor"
> >
> > 13.14167 25.11667 28.34167 49.14167 40.39167 66.86667
> >
> > Is that what you guys get?
>
>
> If you look at Phil's response below, no, that is not what he got.
> The difference is that you are specifying an interaction, whereas Phil
> did not (because the equation you initially specified did not include
> an interaction. Use Y ~ X1 + X2 instead of Y ~ X1*X2 for your formula.
>
> >
> >
> > On Fri, Feb 12, 2010 at 5:00 PM, Phil Spector
> > <spector@stat.berkeley.edu>wrote:
> >
> > > By converting the two variables to factors, you are fitting an
> > > entirely different model. Leave out the as.factor stuff and it
> > > will work exactly as you want it to.
> > >
> > > dat
> > >>
> > > V1 V2 V3 V4
> > > 1 s1 14 4 1
> > > 2 s2 23 4 2
> > > 3 s3 30 7 2
> > > 4 s4 50 7 4
> > > 5 s5 39 10 3
> > > 6 s6 67 10 6
> > >
> > >> names(dat) =
c('id','y','x1','x2') z = lm(y~x1+x2,dat)
> > >> predict(z)
> > >>
> > > 1 2 3 4 5 6 15.16667
24.66667
> > > 27.66667 46.66667 40.16667 68.66667
> > >
> > >
> > > - Phil Spector
> > > Statistical Computing
Facility
> > > Department of Statistics
> > > UC Berkeley
> > > spector@stat.berkeley.edu
> > >
> > >
> > >
> > > On Fri, 12 Feb 2010, Something Something wrote:
> > >
> > > Hello,
> > >>
> > >> I am trying to learn how to perform Multiple Regression
Analysis in
R.> > I
> > >> decided to take a simple example given in this PDF:
> > >> http://www.utdallas.edu/~herve/abdi-prc-pretty.pdf
> > >>
> > >> I created a small CSV called, students.csv that contains the
> > >> following
> > >> data:
> > >>
> > >> s1 14 4 1
> > >> s2 23 4 2
> > >> s3 30 7 2
> > >> s4 50 7 4
> > >> s5 39 10 3
> > >> s6 67 10 6
> > >>
> > >> Col headers: Student id, Memory span(Y), age(X1), speech
> > >> rate(X2)
> > >>
> > >> Now the expected results are:
> > >>
> > >> yHat[0]:15.166666666666668
> > >> yHat[1]:24.666666666666668
> > >> yHat[2]:27.666666666666664
> > >> yHat[3]:46.666666666666664
> > >> yHat[4]:40.166666666666664
> > >> yHat[5]:68.66666666666667
> > >>
> > >> This is based on the following equation (given in the PDF):
Y > > >> 1.67
> +
> > X1
> > >> +
> > >> 9.50 X2
> > >>
> > >> I ran the following commands in R:
> > >>
> > >> data = read.table("students.csv", head=F, as.is=T,
na.string=".",
> > >> row.nam=NULL)
> > >> X1 = as.factor(data[[3]])
> > >> X2 = as.factor(data[[4]])
> > >> Y = data[[2]]
> > >> mod = lm(Y ~ X1*X2, na.action = na.exclude) Y.hat =
fitted(mod)
> > >> Y.hat
> > >>
> > >> This gives me the following output:
> > >>
> > >> Y.hat
> > >>>
> > >> 1 2 3 4 5 6
> > >> 14 23 30 50 39 67
> > >>
> > >> Obviously I am doing something wrong. Please help. Thanks.
> > >>
>
> Hope this is helpful,
>
> Dan
>
> Daniel Nordlund
> Bothell, WA USA
>
>
> ______________________________________________
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> PLEASE do read the posting guide
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> and provide commented, minimal, self-contained, reproducible code.
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