Dear all,
does anyone know how I can run the histbackback function from R Commander? I am
going to teach this in my class and I hesitate to ask my students to write code
Thanks
Jason
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On Sun, 2010-02-14 at 12:42 +0500, Faiz Rasool wrote:
I am trying to perform goodness of fit test using R. I am using this website
http://wiener.math.csi.cuny.edu/Statistics/R/simpleR/stat013.html for help.
However, I am unable to carry out the test successfully. My code follows. It
is
On 14-Feb-10 07:42:12, Faiz Rasool wrote:
I am trying to perform goodness of fit test using R. I am using this
website
http://wiener.math.csi.cuny.edu/Statistics/R/simpleR/stat013.html
for help. However, I am unable to carry out the test successfully. My
code follows. It is taken from the
Dear Faiz,
On Sat, Feb 13, 2010 at 11:42 PM, Faiz Rasool fai...@gmail.com wrote:
The erorr I receive is erorr in chisq.test(freq,p=prob)/6 probabilities
must sum to 1
This error message seems a little off. It looks like the entire chi squared
test was divided by 6. Notice that the /6
Good that you got it to work somehow!
Faiz: can you report the result, exactly as returned by R, of
sum(c(1,1,1,1,1,1)/6) - 1
??
On my Linux with R version 2.10.0 (2009-10-26) this gives 0.
(And, by the way, did you really mean to say you were using
R 2.1.01? Was this a typing error for R
On 02/14/2010 01:27 AM, Orvalho Augusto wrote:
I am using pyramid.plot() from the plotrix package.
...
The problem is (1) I do not want plot agelabels on the center and (2)
I want plot different labels for each pair of the bars (one label for
masculine and the other feminine).
The data
Jim thanks for your great! I will try to use your source code.
Caveman
Ps:
Reconheco a minha trogolodice e por isso pedi ajuda. Lamento perturbar
a todos por isso.
On Sun, Feb 14, 2010 at 1:05 PM, Jim Lemon j...@bitwrit.com.au wrote:
On 02/14/2010 01:27 AM, Orvalho Augusto wrote:
I am using
Dear all,
I am trying to study the correlation between one independent variable (V1)
and several others dependent among them (V2,V3,V4 and V5). For doing
so, I would like to analyze my data by multiple-test (applying the Bonferroni´s
correction or other similar), but I do not find the proper
Pardon my ignorance here.
The Jim code works. But the problem the text comes over the plot.
Adjusting the laxlab and raxlab, to get more plot are, the text keeps
being cut.
And one more thing that might be added as a new functionality is an
option to show the value of the bar.
Thanks for your
On Sat, Feb 13, 2010 at 7:09 PM, Daniel Malter dan...@umd.edu wrote:
It seems to me that your question is more about the econometrics than about
R. Any introductory econometric textbook or compendium on econometrics will
cover this as it is a basic. See, for example, Greene 2006 or Wooldridge
Dear all,
I'm looking for a person who could help me to program a short code in R. The
code involves Bayesian analysis so some familiarity with WinBUGS or another
package/ software dealing with Bayesian estimation would be helpful.
I have an academic paper in which the code is described (Abe,
Bug fix:
first.day.of.quarter = function(date)
{
t = first.day.of.month(date)
l = month(date) %% 3
if (l == 0) return(t)
t = seq.Date(t, by = -1 month, length = l)
return(t[length(t)])
}
But the *apply part still does not work.
--
View this message in context:
Dataframe cust has Date-type column open.date. I wish to set up another
column, with (first day of) the quarter of open.date.
To be comprehensive (of course, improvement suggestions are welcome),
month = function(date)
{
return(as.numeric(format(date,%m)))
}
first.day.of.month =
Does anyone know, or know documentation that describes, how to declare
multiple values in R as missing that does not involve coding them as NA? I
wish to be able to treate values as missing, while still retaining codes
that describe the reason for the value being missing.
Thanks
John MAcInnes
Try
library(zoo)
x - as.Date(seq(1, 500, 50)) # test data
as.Date(as.yearmon(x))
as.Date(as.yearqtr(x))
On Sun, Feb 14, 2010 at 8:59 AM, Dimitri Shvorob
dimitri.shvo...@gmail.com wrote:
Dataframe cust has Date-type column open.date. I wish to set up another
column, with (first day of) the
Dear All,
I am trying to generate a LaTeX table from a small data frame using
xtable. I have three variable and 10 records. However, the resulted
LaTeX table has four columns (instead of three), of which the first
column seems to be an automatically generated ID. Is there a way to
get rid of this
On 14.02.2010 08:19, Daniel Nordlund wrote:
-Original Message-
From: Frank E Harrell Jr [mailto:f.harr...@vanderbilt.edu]
Sent: Saturday, February 13, 2010 5:49 AM
To: Daniel Nordlund
Cc: r-help@r-project.org
Subject: Re: [R] SAS and RODBC
Daniel Nordlund wrote:
. . .
This is just
On 2/14/10, Shige Song shiges...@gmail.com wrote:
column seems to be an automatically generated ID. Is there a way to
get rid of this column?
Perhaps
?print.xtable
include.rownames=F
Liviu
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R-help@r-project.org mailing list
See ?print.xtable and its argument include.rownames.
Uwe Ligges
On 14.02.2010 16:06, Shige Song wrote:
Dear All,
I am trying to generate a LaTeX table from a small data frame using
xtable. I have three variable and 10 records. However, the resulted
LaTeX table has four columns (instead of
NA, Inf, -Inf, NaN would give you 4 possibilities and is.finite would
check if its any of them:
x - c(1, NA, 2, Inf, 3, -Inf, 4, NaN, 5)
is.finite(x)
[1] TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE
You might need to map them all to NA before using it with various
functions depending
Dear Livlu and Uwe,
This is exactly what I need, thanks.
Shige
On Sun, Feb 14, 2010 at 10:11 AM, Uwe Ligges
lig...@statistik.tu-dortmund.de wrote:
See ?print.xtable and its argument include.rownames.
Uwe Ligges
On 14.02.2010 16:06, Shige Song wrote:
Dear All,
I am trying to generate a
Hello,
When drawing barcharts, I find it not helpful if ylim[1] != 0 - bars for a
quantity of 0, that do not show a length of 0 are quite non-intuitive.
I have tried to study
library(lattice)
panel.barchart
but am unable to figure out where ylim is taken care of and how one
I can think of a few solutions, none perfect.
* You could have a master dataset that has the
missing value codes you want, and a dataset that
you use which is a copy of it with real NA's in it.
* You could add an attribute that gives the types
of missing values in the various positions. The
# I want to plot bold text. The text should depend on a variable
containing a character string.
plot.new()
text(.5, .5, expression(bold(Some text)))
# now I would like to do the same replacing some text by a variable.
plot.new()
myText - some text
text(.5, .5, expression(bold(myText)))
#
Hello,
I have data for an ANOVA where the between-subjects factor has three
levels. How do I run a test of normality (using shapiro.test) on each
of the levels of the factor for the dependent variable separately
without creating extra datasets?
Thanks,
Ravi
On Sun, 14 Feb 2010, Ravi Kulkarni wrote:
Hello,
I have data for an ANOVA where the between-subjects factor has three
levels. How do I run a test of normality (using shapiro.test) on each
of the levels of the factor for the dependent variable separately
without creating extra datasets?
You
Hi,
Try with bquote,
plot.new()
myText - some text
text(.5, .5, bquote(bold(.(myText
basically, bquote( .(myText) ) performs the substitution before
applying bold() (see ?bquote).
HTH,
baptiste
On 14 February 2010 17:36, Mark Heckmann mark.heckm...@gmx.de wrote:
# I want to plot bold
Try
text(.5, .5, bquote(bold(.(myText
-Peter Ehlers
Mark Heckmann wrote:
# I want to plot bold text. The text should depend on a variable
containing a character string.
plot.new()
text(.5, .5, expression(bold(Some text)))
# now I would like to do the same replacing some text by a
Patrick Burns wrote:
I can think of a few solutions, none perfect.
* You could have a master dataset that has the
missing value codes you want, and a dataset that
you use which is a copy of it with real NA's in it.
* You could add an attribute that gives the types
of missing values in the
On Feb 14, 2010, at 10:33 AM, Johannes Graumann wrote:
Hello,
When drawing barcharts, I find it not helpful if ylim[1] != 0 -
bars for a
quantity of 0, that do not show a length of 0 are quite non-intuitive.
I have tried to study
library(lattice)
panel.barchart
but am
Hi Mark,
Try also:
plot(1:10)
text(2, 5, Some text, font = 2)
HTH,
Jorge
On Sun, Feb 14, 2010 at 11:36 AM, Mark Heckmann wrote:
# I want to plot bold text. The text should depend on a variable
containing a character string.
plot.new()
text(.5, .5, expression(bold(Some text)))
# now
On Sun, Feb 14, 2010 at 7:33 AM, Johannes Graumann
johannes_graum...@web.de wrote:
Hello,
When drawing barcharts, I find it not helpful if ylim[1] != 0 - bars for a
quantity of 0, that do not show a length of 0 are quite non-intuitive.
I have tried to study
library(lattice)
I'm trying to fit a multinominal logistic model using package mlogit. I have
15 independent variables. The code looks like this:
m-mlogit(score~0|f1+f2+f3+f4+f5+f6+f7+f8+f9+f10+f11+f12+f13+f14+f15, data,
reflevel=1)
And it gives the following error message:
Error in parse(text = x) :
Gary King's Amelia package for R and a stand alone version does EM algorithm
multiple imputation.
Joe King
206-913-2912
j...@joepking.com
Never throughout history has a man who lived a life of ease left a name
worth remembering. --Theodore Roosevelt
-Original Message-
From:
Many thanks, but my focus is actually on *apply usage.
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Sent from the R help mailing list archive at Nabble.com.
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Hey R list,
Can someone help me with following question?
I can create a listing of the unique observations in a dataframe by using:
df - df [order(df $start),]
df $first_ob[!duplicated(df $ID)] - 1
df $first_ob[duplicated(df $ID)] - 0
Now I want to hold specific attributes of the first
Dataframe closed contains balances of closed accounts: each row has month of
closure (Date-type column month) and latest balance. I would like to plot
by-month distributions of balances. A qplot call below produces several
warnings and no output.
Can anyone help?
Thank you.
PS. A really basic
Hi,
it's hard to tell what's wrong without a reproducible example, but I
noted two things:
- AFAIK there is no plot method for ggplot2. You probably meant print(p) instead
- if you map x to factor(month), I think it will be incompatible with
your xlim values range(month).
HTH,
baptiste
On 14
Your function is not receiving what you may think its receiving. Try this:
x - Sys.Date() + 1:3
junk - lapply(x, print)
[1] 14655
[1] 14656
[1] 14657
So add this as the first statement in the body of each of your functions:
date - as.Date(date, 1970-01-01)
and then using x from above try
... Unfortunately, a problem remains: I cannot label x ticks a la 'names.arg
= '.
month has values like '2009-01-01', '2009-02-01', etc., while I would prefer
'Jan', 'Feb'. Using
closed$month = format(closed$month, %b)
disrupts the order of plot's panels, which now follows the alphabetic
I want to use lapply and a function returning a function in order to build a
list of functions.
genr1 - function(k) {function() {k}}
l1 - lapply(1:2,genr1)
l1[[1]]()
[1] 2
This was unexpected. I had expected the answer to be 1, since that is the
value k should be bound to when genr1 is
My bad: once I ran dev.off(), I did get a plot, albeit a blank one. Then I
removed xlim - which I put in after qplot's complain about xlim - and voila!
Thanks a lot.
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View this message in context:
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Sent from
Hi,
I believe it's lazy evaluation. See ?force
HTH,
baptiste
On 14 February 2010 20:32, Jyotirmoy Bhattacharya
jyotir...@jyotirmoy.net wrote:
I want to use lapply and a function returning a function in order to build a
list of functions.
genr1 - function(k) {function() {k}}
l1 -
Hi all,
I've been struggling with trying to specify a diagnoal matrix for linear mixed
effects model. I think I've got nearly everything correct, except the following
message appears:
In lme.formula(fixed = fwave ~ sex + sexXbulbar + visit + age + :
Fewer observations than random effects in
David Winsemius wrote:
On Feb 14, 2010, at 10:33 AM, Johannes Graumann wrote:
Hello,
When drawing barcharts, I find it not helpful if ylim[1] != 0 -
bars for a
quantity of 0, that do not show a length of 0 are quite non-intuitive.
I have tried to study
library(lattice)
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: Sunday, February 14, 2010 7:09 AM
To: Daniel Nordlund
Cc: r-help@r-project.org
Subject: Re: [R] SAS and RODBC
On 14.02.2010 08:19, Daniel Nordlund wrote:
-Original Message-
From:
All
I want to overlay two variables on the same plot following their appropriate
grouping. I have attempted to use subscripting in panel with panel.xyplot,
but I can't get the grouping to follow into the panel...here is an
example...
dat-data.frame(
y= log(1:10),
y2=10:19,
x=1:10,
grp =
Deepayan Sarkar wrote:
On Sun, Feb 14, 2010 at 7:33 AM, Johannes Graumann
johannes_graum...@web.de wrote:
Hello,
When drawing barcharts, I find it not helpful if ylim[1] != 0 - bars
for a quantity of 0, that do not show a length of 0 are quite
non-intuitive.
I have tried to study
At 07:35 AM 2/14/2010, Manuel Jesús López Rodríguez wrote:
Dear all,
I am trying to study the correlation between one
independent variable (V1) and several others
dependent among them (V2,V3,V4 and V5).
For doing so, I would like to analyze my data by
multiple-test (applying the Bonferroni´s
On Feb 14, 2010, at 3:40 PM, Johannes Graumann wrote:
David Winsemius wrote:
On Feb 14, 2010, at 10:33 AM, Johannes Graumann wrote:
Hello,
When drawing barcharts, I find it not helpful if ylim[1] != 0 -
bars for a
quantity of 0, that do not show a length of 0 are quite non-
intuitive.
Ok...I realized that if I don't wish to compare the groupings directly
i can plot the groups as separate panels, which is a much more simple
solution, but doesn't necessarily give the same effect.
xyplot(y+y2 ~ x | grp, dat)
On Sun, Feb 14, 2010 at 3:19 PM, Dennis Murphy djmu...@gmail.com
On Feb 14, 2010, at 3:44 PM, Pat Schmitz wrote:
All
I want to overlay two variables on the same plot following their
appropriate
grouping. I have attempted to use subscripting in panel with
panel.xyplot,
but I can't get the grouping to follow into the panel...here is an
example...
On 15/02/2010, at 9:40 AM, Johannes Graumann wrote
SNIP
(In response to some advice from David Winsemius):
I am quite certain that this is the most elaborately worded version of
RTFM I have ever come across.
I nominate this as a fortune. (Despite Prof. Winsemius's later
Hi Maura,
Is there any R package which implements non-linear dimensionality reduction
(LLE, ISOMAP, GTM, and so on) and/or intrinsic dimensionality estimation ?
I am not exactly sure what is meant by non-linear dimensionality
reduction but there is an isomap function in vegan. This is the
I would use all of the data. If you want to drop one, control for it in
the model sacrifice a degree of freedom.
Why the call to poly() by the way?
KeithC.
-Original Message-
From: Peter Ehlers [mailto:ehl...@ucalgary.ca]
Sent: Saturday, February 13, 2010 1:35 PM
To: David Winsemius
Dear Rhonda,
Consider curve().
KeithC.
-Original Message-
From: Rhonda Reidy [mailto:rre...@gmail.com]
Sent: Saturday, February 13, 2010 11:36 AM
To: r-help@r-project.org
Subject: [R] Plot different regression models on one graph
The following variables have the following significant
Hi R users
I am trying to add a regression line to a graph for c for factor 2
only. Any suggestions?
library(lattice)
a=(1:10)
b=c(1:5,16:20)
c=as.factor(c(rep(1,5),rep(2,5)))
d=data.frame(a,b,c)
xyplot(a~b, pch=c(6,8),data = tavg, groups=d$c,
reg.line=lm, smooth=FALSE,
Dear R Users,
When using zeroinfl() function to fit a Zero-Inflated Negative Binomial (ZINB)
model to a dataset, the summary() gives an estimate of log(theta) and its
standard error, z-value and Pr(|z|) for the count component. Additionally, it
also provided an estimate of Theta, which I
On Feb 14, 2010, at 6:33 PM, Andrew McFadden wrote:
Hi R users
I am trying to add a regression line to a graph for c for factor 2
only. Any suggestions?
library(lattice)
a=(1:10)
b=c(1:5,16:20)
c=as.factor(c(rep(1,5),rep(2,5)))
d=data.frame(a,b,c)
xyplot(a~b, pch=c(6,8),data = tavg,
On 15/02/2010, at 12:49 PM, Lam, Tzeng Yih wrote:
Dear R Users,
When using zeroinfl() function to fit a Zero-Inflated Negative Binomial
(ZINB) model to a dataset, the summary() gives an estimate of log(theta) and
its standard error, z-value and Pr(|z|) for the count component.
Dear Sir/Madam
Could you pls help me at these subjects in R-project?
How could I construct a table for 5 or 10 records of data set?
How could I construct a bar chart for categorical variable with overlay?
How could I construct a histogram for a categorical variable?
pls inform me.
Thanks in
Sometimes when I try to fit a model to data using the gnls function,
it doesn't return optimized values of the model parameters. Instead
it just returns the exact same values I used as initial guesses, but
with standard errors calculated. Example is as follows:
two_site_mHb - gnls(y ~
Hello everyone,
Sorry if my question is not clear, my first language is not English, but
Portuguese.
I am building a model for my data, using non-binomial error. I am having a
bit of a problem when updating the model to remove parameters that I no do
no autocorrelate with other variables (I have
Hi there,
I am new to R and feel so bloody stupid. Abut in spite of a search of
several hrs I could not find an answer to my problem.
I have imported SPSS data to R, and apart from some warnings regarding
duplicate labels, everything looks fine and names lists all my variables.
Then I try to
Those sound like basic questions. You should start by reading the Intro to R
that can be found here http://cran.r-project.org/doc/manuals/R-intro.html
or at www.r-project.org
For your specific questions, searching for histogram or bar chart at
http://www.rseek.org would be very helpful.
If you
Thanks, yes that does the trick, sorry inadvertently left the tavg in
my example. I haven't used the panel function before looks like it maybe
useful. Really appreciate your help.
regards
Andrew McFadden MVS BVSc | Incursion Investigator (animals),
Investigation and Diagnostic Centre |
kMan wrote:
I would use all of the data. If you want to drop one, control for it in
the model sacrifice a degree of freedom.
You like to live dangerously.
-Peter Ehlers
Why the call to poly() by the way?
KeithC.
-Original Message-
From: Peter Ehlers [mailto:ehl...@ucalgary.ca]
Peter wrote:
You like to live dangerously.
Clue me in, Professor.
Sincerely,
KeithC.
-Original Message-
From: Peter Ehlers [mailto:ehl...@ucalgary.ca]
Sent: Sunday, February 14, 2010 6:49 PM
To: kMan
Cc: 'David Winsemius'; 'Rhonda Reidy'; r-help@r-project.org
Subject: Re: [R] Plot
John wrote:
...
Does anyone know, or know documentation that describes, how to declare
multiple values in R as missing that does not involve coding them as NA? I
wish to be able to treate values as missing, while still retaining codes
that describe the reason for the value being missing.
I
Thank you.
Sorry. My question is rather ambiguous. Imeant to aske for Dimensionality
Reduction methods and/or Intrinsic Dimensionality Estimation methods that can
deal with non-linear data. That is, data that do not lie on a hyper-plane.
ISOMAP is one of such methods.
LLE, AutoEncoder, GTM
Dear R users,
I need some advises on how to use R to optimize this function with the
following constraints.
f(x1,x2,x3,y1,y2,y3,)
= gamma(x1+x2-1)/{gamma(x1)*gamma(x2)} * y1^(x2-1) * y2^(x1-1)
+ gamma(x1+x3-1)/{gamma(x1)*gamma(x3)} * y1^(x3-1) * y3^(x1-1)
+
Thanks! Exactly what I wanted.
Ravi
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Hello,
I notice that when I do Levene's test to test equality of variances across
levels of a factor, I get different answers in R and SPSS 16.
e.g.: For the chickwts data, in R, levene.test(weight, feed) gives
F=0.7493, p=0.5896.
SPSS 16 gives F=0.987, p=0.432
Why this difference? Which
I have written in R this:
require(tcltk)
tt-tktoplevel()
Name - tclVar()
entry.Name -tkentry(tt,width=20,textvariable=Name)
tkgrid(tklabel(tt,text=Please enter site number.))
tkgrid(entry.Name)
OnOK - function()
{
NameVal - tclvalue(Name)
use.this=NameVal
tkdestroy(tt)
}
Ravi Kulkarni wrote:
Hello,
I notice that when I do Levene's test to test equality of variances across
levels of a factor, I get different answers in R and SPSS 16.
e.g.: For the chickwts data, in R, levene.test(weight, feed) gives
F=0.7493, p=0.5896.
SPSS 16 gives F=0.987, p=0.432
Why
kMan wrote:
Peter wrote:
You like to live dangerously.
Clue me in, Professor.
Sincerely,
KeithC.
Okay, Keith, here goes:
dat -
structure(list(x = c(62.5, 68.5, 0, 52, 0, 52, 0, 52, 23.5, 86,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0), y = c(0.054, 0.055, 0.017, 0.021,
0.02, 0.028, 0.032, 0.073,
Hi r-users,
I hope somebody can help me to understand the error message. Here is my code;
## Newton iteration
newton_gam - function(z)
{ n - length(z)
r - runif(n)
tol - 1E-6
cdf - vector(length=n, mode=numeric)
fprime - vector(length=n, mode=numeric)
f - vector(length=n,
Dear anyone who knows more about R than me (so everyone). I have been bashing
my head on the keyboard all day trying to do something with my table.
I have some data, like so:
-mm Rainfall(mm)
1 1977-0217.4
2 1977-0334.0
3 1977-0426.2
4 1977-05
the other alternative would be to edit my sql query so that that data is
brought in from the database and put in to the correct format initially.
sqlQuery(conn, select lsd,ttl_mo_prcp from mo_rains where stn_num=23090)
That is my very basic query. I have also been given this for use in orcale
Dear list,
I´m using the mgcv package to model the proportion by weight of certain prey
on the stomach content of a predator. This proportion is the ratio of two
weights (prey weight over stomach weight), and ranges between 0 and 1. The
variance is low when proportion is close to 0 and 1, and
Hi Raging Jim
may be this is a starting point.
myDF-read.table(stdin(),head=T,sep=,)
mm,Rainfall
1977-02,17.4
1977-03,34.0
1977-04,26.2
1977-05,42.6
1977-06,58.6
1977-07,23.2
1977-08,26.8
1977-09,48.4
1977-10,47.0
1977-11,37.2
1977-12,15.0
1978-01,2.6
1978-02,6.8
1978-03,9.0
1978-04,46.6
Dear R helpers
I am working on the credit risk default data and am referring to An
introduction to Credit Risk Modelling by Christian Bluhm. The literature
affirms that 'the default frequencies grow exponentially with decreasing credit
worthiness'.
I have a data of rating wise default
On Feb 15, 2010, at 1:22 AM, milton ruser wrote:
Hi Raging Jim
may be this is a starting point.
myDF-read.table(stdin(),head=T,sep=,)
Those mm entries will become factors, which can lead to
confusion for newbies. Might be more straightforward to always use
stringsAsFactors=FALSE in
Dear Peter,
Ah, I see your point, Professor. The point at x=23.5 is carrying the model.
Allow me to clarify. I was making a similar point.
I was suggesting that the cube term could be left in the model, as you did,
but rather than dropping the data-point, another model is fit with an
additional
Rolf Turner wrote:
On 15/02/2010, at 9:40 AM, Johannes Graumann wrote
SNIP
(In response to some advice from David Winsemius):
I am quite certain that this is the most elaborately worded version of
RTFM I have ever come across.
I nominate this as a fortune. (Despite Prof.
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
I have just got an idea which is using RExcel and coding with Excel VBA may
cope with it, its work!!!
Ryusuke
From: urishim...@optiver.com
To: ryusukeke...@hotmail.com
Date: Mon, 15 Feb 2010 08:31:49 +0100
Subject: RE: RE:Creating a new Access database with R
No idea, sorry.
Consider the SLR y = 50 + 10x + e where e is NID(0.16). n = 20 pairs of
observations. Generate 500 samples of 20 obersvations drawing 1 observations
drawing one observation from each level of x = .5,1,1.5 ... 10 for each
sample
A) For each sample computer the least squares regression estimates
it seems that this is a homework! Please read carefully the Posting
Guide: http://www.r-project.org/posting-guide.html -- the R-help mailing
list is not intended for Basic Statistics and classroom homework!
Best,
Dimitris
Schwonka wrote:
Consider the SLR y = 50 + 10x + e where e is
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