The cor(mtcars$mpg, fitted(m0))^2 method works great - thanks so much Josh!
I've had another instance (also ex intercept) where it actually gave the
correct number, odd behavior.
Thanks for the other posts also. As an aside, in my application a zero
intercept makes economic sense and I'm using R
On Feb 18, 2011, at 14:20 , Jan wrote:
> One of the references you googled suggests that intercepts should never
> be omitted. Is this true even if I know that the physical reality behind
> the numbers suggests an intercept of zero?
No. That'll be a piece of pragmatic advice caused by the experie
On Feb 18, 2011, at 14:20 , Jan wrote:
> Hello Achim,
>
>> Not quite. Consult your statistics textbook for the correct interpretation
>> of p-values. Under the null hypothesis of a true intercept of zero, it is
>> very likely to observe an intercept as large as 13.52 or larger.
> thank you for
Hi,
R actually uses a different formula for calculating the R square
depending on whether the intercept is in the model or not.
You may also find this discussion helpful:
http://stats.stackexchange.com/questions/7948/when-is-it-ok-to-remove-the-intercept-in-lm/
If you conceptualize R^2 as the sq
I've just picked up R (been using Matlab, Eviews etc) and I'm having the same
issue. Running reg=lm(ticker1~ticker2) gives R^2=50% while running
reg=lm(ticker1~0+ticker2) gives R^2=99%!! The charts suggest the fit is
worse not better and indeed Eviews/Excel/Matlab all say R^2=15% with
intercept=0.
for example the same model with intercept R² = 0.6, without intercept R²
= 0.9 and higher. In my definition of R², R² has to be equal or less
without intercept
I do not know what R shows, but in the summary of the model without
intercept it does not show the R² of the regression line.
When I run
On Wed, Jun 27, 2012 at 4:36 AM, Christof Kluß wrote:
>
> for example the same model with intercept R² = 0.6, without intercept R²
> = 0.9 and higher. In my definition of R², R² has to be equal or less
> without intercept
>
> I do not know what R shows, but in the summary of the model without
> in
On Jun 27, 2012, at 13:15 , Uwe Ligges wrote:
>
> 1 - crossprod(residuals(model)) / crossprod(y - mean(y))
And the reason why that is not used in R:
> y<- rnorm(100,10,1)
> x <- 1:100
> model <- lm(y~x-1)
>
> 1 - crossprod(residuals(model)) / crossprod(y - mean(y))
[,1]
[1,] -27.600
On 27.06.2012 10:36, Christof Kluß wrote:
for example the same model with intercept R² = 0.6, without intercept R²
= 0.9 and higher. In my definition of R², R² has to be equal or less
without intercept
I do not know what R shows, but in the summary of the model without
intercept it does not s
for example the same model with intercept R² = 0.6, without intercept R²
= 0.9 and higher. In my definition of R², R² has to be equal or less
without intercept
I do not know what R shows, but in the summary of the model without
intercept it does not show the R² of the regression line.
When I run
On Jun 27, 2012, at 09:33 , Christof Kluß wrote:
> Hi
>
> is there a command that calculates the correct adjusted R-squared, when
> I work without intercept? (The R-squared from lm without intercept is
> false.)
When people say that, they are usually implying that a "correct" R-squared can
be
On 06/27/2012 10:33 AM, Christof Kluß wrote:
> Hi
>
> is there a command that calculates the correct adjusted R-squared, when
> I work without intercept? (The R-squared from lm without intercept is
> false.)
Hi! This answer in R-FAQ might help you:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Wh
On 27.06.2012 09:33, Christof Kluß wrote:
Hi
is there a command that calculates the correct adjusted R-squared, when
I work without intercept? (The R-squared from lm without intercept is
false.)
Then we need your definition of your version of "correct" - we know the
definition of your versi
Hi
is there a command that calculates the correct adjusted R-squared, when
I work without intercept? (The R-squared from lm without intercept is
false.)
Greetings
Chrsitof
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-hel
Hi,
thanks for your help. I'm beginning to understand things better.
> If you plotted your data, you would realize that whether you fit the
> 'best' least squares model or one with a zero intercept, the fit is
> not going to be very good
> Do the data cluster tightly around the dashed line?
No, a
Hello Achim,
> Not quite. Consult your statistics textbook for the correct interpretation
> of p-values. Under the null hypothesis of a true intercept of zero, it is
> very likely to observe an intercept as large as 13.52 or larger.
thank you for that help. I suppose the net doesn't have a detai
l post:
Message: 135
Date: Fri, 18 Feb 2011 11:49:41 +0100
From: Jan
To: "R-help@r-project.org list"
Subject: [R] lm without intercept
Message-ID: <1298026181.2847.19.camel@jan-laptop>
Content-Type: text/plain; charset="UTF-8"
Hi,
I am not a statistics expert, so I
Date: Fri, 18 Feb 2011 12:25:36 +0100
From: achim.zeil...@uibk.ac.at
To: jrheinlaen...@gmx.de
CC: r-help@r-project.org
Subject: Re: [R] lm without intercept
On Fri, 18 Feb 2011, Jan wrote:
> Hi,
>
> I am not a statistics expert, so I
Hi:
On Fri, Feb 18, 2011 at 2:49 AM, Jan wrote:
> Hi,
>
> I am not a statistics expert, so I have this question. A linear model
> gives me the following summary:
>
> Call:
> lm(formula = N ~ N_alt)
>
> Residuals:
>Min 1Q Median 3Q Max
> -110.30 -35.80 -22.77 38.07 122.76
On Fri, 18 Feb 2011, Jan wrote:
Hi,
I am not a statistics expert, so I have this question. A linear model
gives me the following summary:
Call:
lm(formula = N ~ N_alt)
Residuals:
Min 1Q Median 3Q Max
-110.30 -35.80 -22.77 38.07 122.76
Coefficients:
Estima
Hi,
I am not a statistics expert, so I have this question. A linear model
gives me the following summary:
Call:
lm(formula = N ~ N_alt)
Residuals:
Min 1Q Median 3Q Max
-110.30 -35.80 -22.77 38.07 122.76
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(
21 matches
Mail list logo
