?offset
?formula
On Jun 4, 2010, at 4:53 PM, array chip wrote:
Hi, is it possible to specify a constant intercept (based on prior
knowledge) in linear regression using lm()?
Thanks
John
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R-help@r-project.org mailing list
- Original Message -
From: baptiste auguie ba...@exeter.ac.uk
To: Gundala Viswanath gunda...@gmail.com
Cc: r-h...@stat.math.ethz.ch R-help r-h...@stat.math.ethz.ch
Sent: Thursday, February 19, 2009 7:12:23 AM GMT -06:00 US/Canada Central
Subject: Re: [R] Insert value in a Vector
I would think this could be approached by segmenting the probability volume
using identities such as these:
P(Y1 Z1, Y2 Z2, Y3 Z3, Y4 Z4) + P(Y1 Z1, Y2 Z2, Y3 Z3, Y4 Z4) =
P(Y1 Z1, Y2 Z2, Y3 Z3, Y4 Inf)
and
P(Y1 Z1, Y2 Z2, Y3 Z3, Y4 Inf) + P(Y1 Z1, Y2 Z2, Y3 Z3, Y4 Inf)
It returns a chi-squared statistic with one degree of freedom.
--
David Winsemius
-- Original message --
From: Timthy Chang henchao.ch...@gmail.com
See the cpower() and spower() functions in Frank Harrell's Hmisc package
on CRAN.
HTH,
Marc Schwartz
-- Original message --
From: Dimitri Liakhovitski ld7...@gmail.com
Thank you for providing advice on this graphics question.
I am building an interaction.plot.
d=data.frame(xx=c(3,3,2,2,1,1),yy=c(4,3,4,3,4,3),zz=c(5.1,4.2,4.4,3.5,3.3,-1.1,-
1.3)
You have
Try setting exclude=NULL if you think that the NA's need to be considered as a
value.
The default is to exclude both NaN and NA.
--
David Winsemius
-- Original message --
From: Farley, Robert farl...@metro.net
I presume the behavior below (no Connector
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