Dear everybody,
I'm a new user of R 2.4.1 and I'm searching for information on improving
the output of regression tree graphs.
In the terminal nodes I am up to now able to indicate the number of
values (n) and the mean of all values in this terminal node by the command
text(tree, use.n=T,
On Thu, 16 Aug 2007, Juergen Kuehn wrote:
Dear everybody,
I'm a new user of R 2.4.1 and I'm searching for information on improving
the output of regression tree graphs.
In the terminal nodes I am up to now able to indicate the number of
values (n) and the mean of all values in this terminal
Hi all,
Can you please tell me what is the problem here.
My regression eq is y = B0 + B1X1 +B2X2 +e
And i am interested in coefficient B1
I am doing regression with two cases:
1) reg-lm(y ~ X1 + X2, sam) where sam is the data
2) reg-lm(y ~ X1 + X2, sam, na.action= na.exclude) . I have
na.exclude should give the same results as na.omit, which is the
default na.action. Is the number of complete cases the same in these
two regressions?
On 26/07/07, Vaibhav Gathibandhe [EMAIL PROTECTED] wrote:
Hi all,
Can you please tell me what is the problem here.
My regression eq is y =
Hi
I have two questions:
1)
I would like to know if there is a package in R that constructs a
regression tree using the 'goodness-of-split' algorithm for survival
analysis proposed by Le Blanc and Crowley (1993) (rather than the usual
CART algorithm that uses within-node difference and impurity
Ron,
you're right. It's not legitimate at all. I suggest you
to take a look at the HUGE bibliography on cointegration,
as a start up.
Rogerio
Dear all R user,
Please forgive me if my question is too simple.
My question is related to Statistics rather directly to R. Suppose I have
Dear all R user,
Please forgive me if my question is too simple.
My question is related to Statistics rather directly to R. Suppose I have two
time series of spot prices of two commodities X and Y for two years. Now I want
to see what percentage of spot price of X is explained by Y. Yes
On Fri, 2 Feb 2007, Henric Nilsson (Public) wrote:
Torsten, consider the following:
### ordinal regression
mammoct - ctree(ME ~ ., data = mammoexp)
Warning message:
no admissible split found
### estimated class probabilities
treeresponse(mammoct, newdata = mammoexp[1:5, ])
[[1]]
[1]
Den Fr, 2007-02-02, 06:03 skrev Stacey Buckelew:
Hi,
I am working on a regression tree in Rpart that uses a continuous response
variable that is ordered. I read a previous response by Pfr. Ripley to a
inquiry regarding the ability of rpart to handle ordinal responses in
2003. At that time
On Fri, 2 Feb 2007, Henric Nilsson (Public) wrote:
Den Fr, 2007-02-02, 06:03 skrev Stacey Buckelew:
Hi,
I am working on a regression tree in Rpart that uses a continuous response
variable that is ordered. I read a previous response by Pfr. Ripley to a
inquiry regarding the ability of
Hi,
I am working on a regression tree in Rpart that uses a continuous response
variable that is ordered. I read a previous response by Pfr. Ripley to a
inquiry regarding the ability of rpart to handle ordinal responses in
2003. At that time rpart was unable to implement an algorithm to handle
Sir I am not finding the function to plot least square regression line on
type=o plot of two variables.
guid me in this regard.
--
AMINA SHAHZADI
Department of Statistics
GC University Lahore, Pakistan.
Email:
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[[alternative HTML
On Wed, 2007-01-31 at 09:25 -0800, amna khan wrote:
Sir I am not finding the function to plot least square regression line on
type=o plot of two variables.
guid me in this regard.
Did you want something like this:
x - 1:50
y - rnorm(50)
plot(x, y, type = o)
abline(lm(y ~ x))
See ?abline
My simpleminded understanding of simple regression is that when
plotting regression lines for x on y and y on x in the same plot, the
lines should cross each other at the respective means. But, given the
R function below, abline (lm(y~x)) works fine, but abline (lm(x~y))
does not. Why?
Try this version of your function and then think about it
tst - function () {
attach (attitude)
x - rating
y - learning
detach (attitude)
plot (x, y)
abline(v=mean(x))
abline(h=mean(y))
abline (lm(y~x))
cc - coef(lm(x ~ y))
abline (-cc[1]/cc[2], 1/cc[2])
}
My simpleminded understanding of
On Fri, 12 Jan 2007, Tom Backer Johnsen wrote:
My simpleminded understanding of simple regression is that when
plotting regression lines for x on y and y on x in the same plot, the
lines should cross each other at the respective means. But, given the
R function below, abline (lm(y~x))
On Fri, 12 Jan 2007, Tom Backer Johnsen wrote:
My simpleminded understanding of simple regression is that when
plotting regression lines for x on y and y on x in the same plot, the
lines should cross each other at the respective means. But, given the
R function below, abline (lm(y~x)) works
Tom Backer Johnsen wrote:
My simpleminded understanding of simple regression is that when
plotting regression lines for x on y and y on x in the same plot, the
lines should cross each other at the respective means. But, given the
R function below, abline (lm(y~x)) works fine, but abline
Prof Brian Ripley wrote:
Where did you tell it 'x' was the abscissa and 'y' the ordinate?
(Nowhere: R is lacking a mind_read() function!)
Please stop complaining about missing features. Patches will be considered.
Oh, it's you, Brian. Never mind then. You'll get to it, I'm sure.
;-)
--
This should do the trick:
mind_reader - function() {
ll - letters[round(runif(6, 1, 26))]
ff - ll[1]
for (ix in 2:length(ll)) {
ff - paste(ff, ll[ix], sep = )
}
if (exists(ff)) {
cat(The function that you were
ken knoblauch wrote:
This should do the trick:
mind_reader - function() {
ll - letters[round(runif(6, 1, 26))]
I see my paraNormal distribution package hasn't found its way to CRAN yet:
http://tolstoy.newcastle.edu.au/R/help/05/04/1701.html
Barry
Barry Rowlingson wrote:
ken knoblauch wrote:
This should do the trick:
mind_reader - function() {
ll - letters[round(runif(6, 1, 26))]
I see my paraNormal distribution package hasn't found its way to CRAN yet:
http://tolstoy.newcastle.edu.au/R/help/05/04/1701.html
LOL! Nice!
On 1/12/2007 5:56 AM, Barry Rowlingson wrote:
ken knoblauch wrote:
This should do the trick:
mind_reader - function() {
ll - letters[round(runif(6, 1, 26))]
I see my paraNormal distribution package hasn't found its way to CRAN yet:
Fortune?
On 1/12/07, Peter Dalgaard [EMAIL PROTECTED] wrote:
Prof Brian Ripley wrote:
Where did you tell it 'x' was the abscissa and 'y' the ordinate?
(Nowhere: R is lacking a mind_read() function!)
Please stop complaining about missing features. Patches will be considered.
Oh, it's
Dear Helpers,
I have a simple question. In statistic studies. I have lear to make inference
on sampling. I want to know what should be the strategy when I have the whole
population ? If a suppose that data are collecte without error, does inference
made are useful ?
sincerly !
Justin BEM
I need to run a regression analysis with a large number of samples. Each
sample (identified in the first file column) has its own x and y values. I
will use the same model in all samples. How can I run the model for each
sample? In SAS code I would use the BY SAMPLE statement.
Alvaro
R 2.2.0
windows XP
How can I perform a regression analyses using a vector of means, a
variance-covariance matrix? I looked at the help screen for lm and did
not see any option for using the afore mentioned structures as input to
lm.
Thanks,
John
John Sorkin M.D., Ph.D.
Chief, Biostatistics and
:[EMAIL PROTECTED] On Behalf Of John Sorkin
Sent: Saturday, October 14, 2006 3:27 PM
To: r-help@stat.math.ethz.ch
Subject: [R] regression analyses using a vector of means anda
variance-covariance matrix
R 2.2.0
windows XP
How can I perform a regression analyses using a vector of
means
Here is another approach using the same data as in
John Fox's reply. His is probably superior but this
does have the advantage that its very simple. Note
that it gives the same coefficients and R squared
to several decimal places. We just simulate a
data set with the given means and variance
There was a missing line:
On 10/14/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here is another approach using the same data as in
John Fox's reply. His is probably superior but this
does have the advantage that its very simple. Note
that it gives the same coefficients and R squared
to
Dear UseRs,
you can find on CRAN web site my last contribute about
R regression techniques:
http://cran.r-project.org/doc/contrib/Ricci-regression-it.pdf
It's in Italian language.
Regards.
Vito Ricci
Se non ora, quando?
Se non qui, dove?
Se non tu, chi
Andreas Svensson wrote:
So, how can I constrain the abline to the relevant region, i.e stop
abline from extrapolating beyond the actual range of data.
Or should I use a function line 'lines' to do this?
One elegant way of doing this is using 'xyplot' from 'lattice' and adding a
loess line
Hi
In R, using plot(x,y) followed by abline(lm(y~x)) produces a graph
with a regression line spanning the whole plot . This means that the
line extends beyond the swarm of data points to the defined of default
plot region. With par(xpd=T) it will span the entire figure region. But
how can
On Wed, 2006-05-24 at 18:51 +0200, Andreas Svensson wrote:
Hi
In R, using plot(x,y) followed by abline(lm(y~x)) produces a graph
with a regression line spanning the whole plot . This means that the
line extends beyond the swarm of data points to the defined of default
plot region.
Thankyou very much Marc for that nifty little script.
When I use it on my real dataset though, the lines are fat in the middle
and thinner towards the ends. I guess it's because lines draw one
fitted line for each x, and if you have hundreds of x, this turns into a
line that is thicker that it
On Wed, 2006-05-24 at 21:53 +0200, Andreas Svensson wrote:
Thankyou very much Marc for that nifty little script.
When I use it on my real dataset though, the lines are fat in the middle
and thinner towards the ends. I guess it's because lines draw one
fitted line for each x, and if you
Svensson
Sent: Wednesday, May 24, 2006 10:52 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Regression line limited by the rage of values
Hi
In R, using plot(x,y) followed by abline(lm(y~x)) produces a graph
with a regression line spanning the whole plot . This means that the
line extends beyond
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Dear R-users:
Sorry for the naiveness of my question but I have been trying in the
R-help and the CRAN website without any success. I am trying to perform
a regression through the origin (without intercept) and my main concern
is about its evaluative statistics. It is clear for me that R squared
this helps a bit.
Best,
Roland
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Trujillo L.
Sent: Tuesday, May 23, 2006 12:54 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Regression through the origin
Dear R-users:
Sorry for the naiveness of my
Trujillo L. skreiv:
Sorry for the naiveness of my question but I have been trying in the
R-help and the CRAN website without any success. I am trying to perform
a regression through the origin (without intercept) and my main concern
is about its evaluative statistics. It is clear for me that
@stat.math.ethz.ch
Subject: Re: [R] regression modeling
May I offer a perhaps contrary perspective on this.
Statistical **theory** tells us that the precision of estimates
improves as
sample size increases. However, in practice, this is not always the
case.
The reason is that it can take time
] On Behalf Of Weiwei Shi
Sent: Monday, April 24, 2006 12:45 PM
To: r-help
Subject: [R] regression modeling
Hi, there:
I am looking for a regression modeling (like regression
trees) approach for
a large-scale industry dataset. Any suggestion on a package
from R or from
other sources which
vary as the dataset gets
larger and larger?
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Weiwei Shi
Sent: Monday, April 24, 2006 12:45 PM
To: r-help
Subject: [R] regression modeling
Hi, there:
I am looking for a regression modeling
]
[mailto:[EMAIL PROTECTED] On Behalf Of Weiwei Shi
Sent: Tuesday, April 25, 2006 12:10 PM
To: bogdan romocea
Cc: r-help
Subject: Re: [R] regression modeling
i believe it is not a question only related to regression
modeling. The
correlation between the sample size and confidence
Berton Gunter wrote:
May I offer a perhaps contrary perspective on this.
Statistical **theory** tells us that the precision of estimates improves as
sample size increases. However, in practice, this is not always the case.
The reason is that it can take time to collect that extra data, and
Hi, there:
I am looking for a regression modeling (like regression trees) approach for
a large-scale industry dataset. Any suggestion on a package from R or from
other sources which has a decent accuracy and scalability? Any
recommendation from experience is highly appreciated.
Thanks,
Weiwei
Hi list,
I'm looking for a way to easily extract regression p-values and export
them to one file for further evaluation.
Here is the problem.
I use lm() and step() to get my regression parameters/coefficients.
after that I can extract them with summary(lm-results)$coefficients[,4]
so far so
Have you considered lme in library(nlme)? The companion book
Pinheiro and Bates (2000) Mixed-Effects Models in S and S-Plus
(Springer) is my favorite reference for this kind of thing. From what I
understand of your question, you should be able to find excellent
answers in this
Dear R-users,
I set up an experiment where I put up bluebird boxes across an
urbanization gradient. I monitored these boxes and at some point I
pulled a feather from a chick and a friend used spectral properties
(rtot, a continuous var) to index chick health. There is an effect of
sex that I
Dear R users,
There is a method called style analysis where you make a regression being
Y=fund yield and X=benchmarks yield, where we have the restrictions to
calculatethe linear regression:
1. The regression must don have the intercept term.
2. The coefficient sum must be one.
3. All
Hi,
I have the following problem which I would appreciate some help on.
A variable y is to be modelled as a function of a set of variables
Vector(x).
The twist is that there is another variable z in the problem with the
property that y(i) = z(i).
So the data set is divided into three
Hi,
I would like to regress y (dependent variable) on x (independent variable) and
y(-1).
I have create the y(-1) variable in this way: ly-lag(y, -1)
Now if I do the following regression lm (y ~ x + ly) the results I obtain are
not correct.
Can someone tell me the code to use in R in order to
Create time series from your data and then use lm with
the dyn or dynlm package (as lm does not support time
series directly). With the dyn package you just preface
lm with dyn$ and then use lm as usual:
library(dyn)
yt - ts(y)
xt - ts(x)
dyn$lm(yt ~ xt + lag(yt, -1))
After loading dyn try this
On Sat, 15 Oct 2005, giacomo moro wrote:
Hi,
I would like to regress y (dependent variable) on x (independent variable)
and y(-1).
I have create the y(-1) variable in this way: ly-lag(y, -1)
Now if I do the following regression lm (y ~ x + ly) the results I obtain
are not correct.
The
On Thu, 29 Sep 2005, Christian Hennig wrote:
?confint
Thank you to all of you.
As far as I see this is not mentioned on the lm help page (though I
presumably don't have the recent version), which I would
suggest...
and I would suggest that you study a good book on the subject.
(confint
Hi list,
is there any direct way to obtain confidence intervals for the regression
slope from lm, predict.lm or the like?
(If not, is there any reason? This is also missing in some other statistics
softwares, and I thought this would be quite a standard application.)
I know that it's easy to
?confint
For example:
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group - gl(2,10,20, labels=c(Ctl,Trt))
weight - c(ctl, trt)
lm(weight ~ group)
Call:
lm(formula = weight ~ group)
Coefficients:
?confint
-Oprindelig meddelelse-
Fra: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] På vegne af Christian Hennig
Sendt: 29. september 2005 13:19
Til: r-help-request Mailing List
Emne: [R] Regression slope confidence interval
Hi list,
is there any direct way to obtain
On Thu, 29 Sep 2005, Christian Hennig wrote:
Hi list,
is there any direct way to obtain confidence intervals for the regression
slope from lm, predict.lm or the like?
There is a confint method: e.g.,
R fm - lm(dist ~ speed, data = cars)
R confint(fm, parm = speed)
2.5 % 97.5 %
?confint
Thank you to all of you.
As far as I see this is not mentioned on the lm help page (though I
presumably don't have the recent version), which I would
suggest...
Best,
Christian
On Thu, 29 Sep 2005, Chuck Cleland wrote:
?confint
For example:
ctl -
Why not use vcov() and the normal approximation ?
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group - gl(2,10,20, labels=c(Ctl,Trt))
weight - c(ctl, trt)
lm.D9 - lm(weight ~ group)
cbind(estimate =
Sorry, I forgot confint and I made a mistake in my suggestion which
should be:
cbind(estimate = coef(lm.D9),
lower = coef(lm.D9) - 1.96 * sqrt(diag(vcov(lm.D9))),
upper = coef(lm.D9) + 1.96 * sqrt(diag(vcov(lm.D9
Best,
Renaud
Christian Hennig a écrit :
Hi list,
is there
I retract the siggestion I proposed last night -- it was based
on a bad hunch! Sorry for wasting time.
Best wishes,
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 27-Sep-05
Dear R-users,
I have the following data
x - runif(300,min=1,max=230)
y - x*0.005 + 0.2
y - y+rnorm(100,mean=0,sd=0.1)
y - y%%1 # --- modulo operation
plot(x,y)
and would like to recapture the slope (0.005) and intercept(0.2). I wonder if
there are any clever algorithms to do this. I was
On 26-Sep-05 nwew wrote:
Dear R-users,
I have the following data
x - runif(300,min=1,max=230)
y - x*0.005 + 0.2
y - y+rnorm(100,mean=0,sd=0.1)
y - y%%1 # --- modulo operation
plot(x,y)
and would like to recapture the slope (0.005) and intercept(0.2).
I wonder if there are any
Hi,
I do not know the intercept and slope.
And you have to know them in order to do something like:
ix-(y 0.9*(x-50)/200
I am right?
cheers
(Ted Harding) wrote:
On 26-Sep-05 nwew wrote:
Dear R-users,
I have the following data
x - runif(300,min=1,max=230)
y - x*0.005 + 0.2
y -
On 26-Sep-05 Witold Eryk Wolski wrote:
Hi,
I do not know the intercept and slope.
And you have to know them in order to do something like:
ix-(y 0.9*(x-50)/200
I am right?
cheers
Although I really knew them from the way you generated the data,
I pretended I did not know them.
Read
Ted,
I agree with you that if you unwrap the data you can use lm.
And you can separate the data in the way you describe. However, if you
have thousands of such datasets I do not want to do it by looking at
the graph.
Yes the scatter may be larger as in the example and range(y) may be
larger
On 26-Sep-05 Witold Eryk Wolski wrote:
Ted,
I agree with you that if you unwrap the data you can use lm.
And you can separate the data in the way you describe. However, if you
have thousands of such datasets I do not want to do it by looking at
the graph.
Yes the scatter may be larger
I have not seen any replies, so I will offer a comment:
1. You speak of x1, x2, ..., x10, but your example includes only
x1+x2+x3+x4. I'm confused. If you could still use help with this,
could you please simplify your example further so there was only x1+x2,
say? Can
Dear WizaRds!
I am sorry to ask for some help, but I have come to a complete stop in
my efforts. I hope, though, that some of you might find the problem
quite interesting to look at.
I have been trying to estimate parameters for lotteries, the so called
utility of chance, i.e. the felt
Dear R users,
How can I do a regression analysis in R where there is more than one
observation per x value? I tried the example in SokalRohlf (3rd edn.,
1995), page 476 ff., but I somehow couldn´t find a way to partition the
sums of squares into linear regression, deviations from regression,
Sounds like you're looking for something like pure.error.anova in the `alr3'
package on CRAN...
Andy
From: Christoph Scherber
Dear R users,
How can I do a regression analysis in R where there is more than one
observation per x value? I tried the example in SokalRohlf
(3rd edn.,
Hi,
I suggest to give a look to:
Practical Regression and Anova using R by Julian
Faraway
http://cran.r-project.org/doc/contrib/Faraway-PRA.pdf
http://www.stat.lsa.umich.edu/~faraway/book/
see also package faraway for datasets:
hi all
i am busy teaching a regression analysis course to second year science
students. the course is fairly theoretical with all of the standard
theorems and proofs...
i would like to give the class a practical assignment as well. could you
suggest a good problem and the location of the data
Clark Allan wrote:
i would like to give the class a practical assignment as well. could you
suggest a good problem and the location of the data set/s?
it would be good if the data set has been analysed by a number of other
people so that students can see the different ways of tackling a
Sundar Dorai-Raj writes:
Hi, Laura,
Would ?predict.glm be better?
plot(logarea, hempresence,
xlab = Surface area of log (m2),
ylab=Probability of hemlock seedling presence,
type=n, font.lab=2, cex.lab=1.5, axes=TRUE)
lines(logarea, predict(hemhem, logreg, response),
Hi there,
I've looked through the very helpful advice about adding fitted lines to
plots in the r-help archive, and can't find a post where someone has offered
a solution for my specific problem. I need to plot logistic regression fits
from three differently-sized data subsets on a plot of
Laura M Marx wrote:
Hi there,
I've looked through the very helpful advice about adding fitted lines to
plots in the r-help archive, and can't find a post where someone has offered
a solution for my specific problem. I need to plot logistic regression fits
from three differently-sized
Can someone shed some light on this obscure portion of the help for lm?
Considerable care is needed when using 'lm' with time series.
Unless 'na.action = NULL', the time series attributes are stripped
from the variables before the regression is done. (This is
necessary as
Dr. Frank E. Harrell, Jr., Professor and Chair of the Department of
Biostatistics at Vanderbilt University is giving a one-day workshop on
Regression Modeling Strategies on Friday, April 29, 2005. Analyses of the
example datasets use R/S-Plus and make extensive use of the Hmisc library
I am running some models (for the first time) using rpart and am getting
results I don't know how to interpret. I'm using cross-validation to prune
the tree and the results look like:
Root node error: 172.71/292 = 0.59148
n= 292
CP nsplit rel error xerror xstd
1 0.124662 0
Sherri Miller wrote:
I am running some models (for the first time) using rpart and am getting
results I don't know how to interpret. I'm using cross-validation to prune
the tree and the results look like:
Root node error: 172.71/292 = 0.59148
n= 292
CP nsplit rel error xerror xstd
1
ReidH == Huntsinger, Reid [EMAIL PROTECTED]
on Thu, 3 Mar 2005 17:24:22 -0500 writes:
ReidH You might use lsfit instead and just do the whole Y
ReidH matrix at once. That saves all the recalculation of
ReidH things involving only X.
yes, but in these cases, we have been
From: Martin Maechler
ReidH == Huntsinger, Reid [EMAIL PROTECTED]
on Thu, 3 Mar 2005 17:24:22 -0500 writes:
ReidH You might use lsfit instead and just do the whole Y
ReidH matrix at once. That saves all the recalculation of
ReidH things involving only X.
yes, but in
Hi -
I am doing a monte carlo experiment that requires to do a linear
regression of a matrix of vectors of dependent variables on a fixed
set of covariates (one regression per vector). I am wondering if
anyone has any idea of how to speed up the computations in R. The code
follows:
#regression
-help@stat.math.ethz.ch
Subject: [R] regression on a matrix
Hi -
I am doing a monte carlo experiment that requires to do a linear
regression of a matrix of vectors of dependent variables on a fixed
set of covariates (one regression per vector). I am wondering if
anyone has any idea of how
see also the contributed document by John Verzani, Simple R, page 87f.
Adaikalavan Ramasamy wrote:
I would try to construct the confidence intervals and
compare them to
the value that you want
x - rnorm(20)
y - 2*x + rnorm(20)
summary( m1 - lm(y~x) )
snip
Coefficients:
On Tue, 2004-07-20 at 17:02, Avril Coghlan wrote:
Hello,
I'm a newcomer to R so please
forgive me if this is a silly question.
It's that I have a linear regression:
fm - lm (x ~ y)
and I want to test whether the
slope of the regression is significantly
less than 1. How can I do this in R?
Another
I will be giving a one-day short course related to my book Regression Modeling
Strategies in Toronto as part of the Joint Statistical Meetings on August 8.
For more information visit the American Statistical Association web site
amstat.org and biostat.mc.vanderbilt.edu/rms. The course applies
Hello,
I'm a newcomer to R so please
forgive me if this is a silly question.
It's that I have a linear regression:
fm - lm (x ~ y)
and I want to test whether the
slope of the regression is significantly
less than 1. How can I do this in R?
I'm also interested in comparing the
slopes of two
I would try to construct the confidence intervals and compare them to
the value that you want
x - rnorm(20)
y - 2*x + rnorm(20)
summary( m1 - lm(y~x) )
snip
Coefficients:
Estimate Std. Error t value Pr(|t|)
(Intercept) 0.1418 0.1294 1.0950.288
x 2.2058
At 06:44 PM 7/20/2004 +0100, Adaikalavan Ramasamy wrote:
I would try to construct the confidence intervals and compare them to
the value that you want
x - rnorm(20)
y - 2*x + rnorm(20)
summary( m1 - lm(y~x) )
snip
Coefficients:
Estimate Std. Error t value Pr(|t|)
(Intercept)
Adaikalavan Ramasamy wrote:
I would try to construct the confidence intervals and compare them to
the value that you want
x - rnorm(20)
y - 2*x + rnorm(20)
summary( m1 - lm(y~x) )
snip
Coefficients:
Estimate Std. Error t value Pr(|t|)
(Intercept) 0.1418 0.1294 1.0950.288
Hi All
I received a raw data set with one record per tennis player (both male and
female) and then i cured it by aggregation i.e by 4 age groups, 2 gender
levels and 6 income levels. Gender and Income are categorical variables.
Please advise me how to use 'R' to model this data set (Actually, i
where as in men's tennis it's fluctuating. Also, i
would like to which age group reflects the prime of a tennis player and
hence i've changed continuous variables to categorical.
Please advise
Thanks
-Dev
From: Peter Flom [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: Re: [R] Regression
From: Berton Gunter [EMAIL PROTECTED]
To: devshruti pahuja [EMAIL PROTECTED]
Subject: Re: [R] Regression Modeling query
Date: Tue, 22 Jun 2004 13:44:13 -0700
MIME-Version: 1.0
X-Sender: Berton Gunter [EMAIL PROTECTED]
Received: from compton.gene.com ([192.12.78.250]) by mc8-f14.hotmail.com
into books to find little useful. Any references that
you provide will
be pretty helpful
Please advise
-Dev
From: Berton Gunter [EMAIL PROTECTED]
To: devshruti pahuja [EMAIL PROTECTED]
Subject: Re: [R] Regression Modeling query
Date: Tue, 22 Jun 2004 13:44:13 -0700
MIME-Version
If variables are colinear, then looking at interactions among them
doesn't make much sense. High collinearity means that one variable is
nearly a linear combination of others. IOW, that variable is not adding
much information. So, if you look at the interaction, you are ALMOST
looking at a
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