confidence intervals for nls or nls2 model

Hi all

I have fitted a model usinf nls function to these data:
x
[1]   1   0   0   4   3   5  12  10  12 100 100 100
y
[1]  1.281055090  1.563609934  0.001570796  2.291579783  0.841891853
 [6]  6.553951324 14.243274230 14.519899320 15.066473610 21.728809880
[11] 18.553054450 23.722637370

The model fitted is:

modellogis<-nls(y~SSlogis(x,a,b,c))

It runs OK. Then I calculate confidence intervals for the actual data using:

dataci<-predict(as.lm(modellogis), interval = "confidence")

BUt I don?t get smooth curves when plotting it, so I want to get other "confidence 
vectors" based on a new x vector by defining a new data to do predictions:

x0 <-  seq(0,15,1)
dataci<-predict(as.lm(modellogis), newdata=data.frame(x=x0), interval = "confidence")

BUt it does not work: I get the same initial confidence interval

Any ideas on how to get tconfidence and prediction intervals using new X data on a 
previous model?

Thanks 

Francisco

----------------------
Francisco Mora Ardila
Estudiante de Doctorado
Centro de Investigaciones en Ecosistemas
Universidad Nacional Aut?noma de M?xico

Hi all

I have fitted a model usinf nls function to these data:

x
[1]   1   0   0   4   3   5  12  10  12 100 100 100

y
[1]  1.281055090  1.563609934  0.001570796  2.291579783  0.841891853
[6]  6.553951324 14.243274230 14.519899320 15.066473610 21.728809880
[11] 18.553054450 23.722637370

The model fitted is:

modellogis<-nls(y~SSlogis(x,a,b,c))

It runs OK. Then I calculate confidence intervals for the actual  
data using:

dataci<-predict(as.lm(modellogis), interval = "confidence")

BUt I don?t get smooth curves when plotting it, so I want to get  
other "confidence
vectors" based on a new x vector by defining a new data to do  
predictions:

x0 <-  seq(0,15,1)
dataci<-predict(as.lm(modellogis), newdata=data.frame(x=x0),  
interval = "confidence")

BUt it does not work:
I am really getting tired of seeing the phrase "doesn't work". And on  
my machine it throws an error saying there is no as.lm function> This  
seemes to give "smooth results:

dataci<-predict(as.lm(modellogis), newdata=data.frame(x=x0), interval  
= "confidence")

And notice that this message is in ?predict.nls

interval
"At present this argument is ignored.
I get the same initial confidence interval

Any ideas on how to get tconfidence and prediction intervals using  
new X data on a
previous model?

Thanks

Francisco

----------------------
Francisco Mora Ardila
Estudiante de Doctorado
Centro de Investigaciones en Ecosistemas
Universidad Nacional Aut?noma de M?xico

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
On Tue, 15 May 2012 20:33:02 -0400, David Winsemius wrote
On May 15, 2012, at 8:08 PM, Francisco Mora Ardila wrote:

Hi all

I have fitted a model usinf nls function to these data:

x
[1]   1   0   0   4   3   5  12  10  12 100 100 100

y
[1]  1.281055090  1.563609934  0.001570796  2.291579783  0.841891853
[6]  6.553951324 14.243274230 14.519899320 15.066473610 21.728809880
[11] 18.553054450 23.722637370

The model fitted is:

modellogis<-nls(y~SSlogis(x,a,b,c))

It runs OK. Then I calculate confidence intervals for the actual  
data using:

dataci<-predict(as.lm(modellogis), interval = "confidence")

BUt I don?t get smooth curves when plotting it, so I want to get  
other "confidence
vectors" based on a new x vector by defining a new data to do  
predictions:

x0 <-  seq(0,15,1)
dataci<-predict(as.lm(modellogis), newdata=data.frame(x=x0),  
interval = "confidence")

BUt it does not work:
I am really getting tired of seeing the phrase "doesn't work". And on  
my machine it throws an error saying there is no as.lm function> This  
seemes to give "smooth results:
Sorry for that "doesn't work". Also, I forgot to say function as.lm() is in tha package 
nls2, which also fit nls models.
dataci<-predict(as.lm(modellogis), newdata=data.frame(x=x0), interval  
= "confidence")

And notice that this message is in ?predict.nls

interval
"At present this argument is ignored.
Yeah, that?s why I?m using nls2 package

I get the same initial confidence interval

Any ideas on how to get tconfidence and prediction intervals using  
new X data on a
previous model?

Thanks

Francisco

----------------------
Francisco Mora Ardila
Estudiante de Doctorado
Centro de Investigaciones en Ecosistemas
Universidad Nacional Aut?noma de M?xico

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
----------------------
Francisco Mora Ardila
Estudiante de Doctorado
Centro de Investigaciones en Ecosistemas
Universidad Nacional Aut?noma de M?xico
On Tue, May 15, 2012 at 8:08 PM, Francisco Mora Ardila
Hi all

I have fitted a model usinf nls function to these data:

x
?[1] ? 1 ? 0 ? 0 ? 4 ? 3 ? 5 ?12 ?10 ?12 100 100 100

y
?[1] ?1.281055090 ?1.563609934 ?0.001570796 ?2.291579783 ?0.841891853
?[6] ?6.553951324 14.243274230 14.519899320 15.066473610 21.728809880
[11] 18.553054450?23.722637370

The model fitted is:

modellogis<-nls(y~SSlogis(x,a,b,c))

It runs OK. Then I calculate confidence intervals for the actual data using:

dataci<-predict(as.lm(modellogis), interval = "confidence")

BUt I don?t get smooth curves when plotting it, so I want to get other "confidence
vectors" based on a new x vector by defining a new data to do predictions:

x0 <- ?seq(0,15,1)
dataci<-predict(as.lm(modellogis), newdata=data.frame(x=x0), interval = "confidence")

BUt it does not work: I get the same initial confidence interval

Any ideas on how to get tconfidence and prediction intervals using new X data on a
previous model?

as.lm is a linear model between the response variable and the gradient
of the nonlinear model and as we see below x is not part of that
linear model so x can't be in newdata when predicting from the tangent
model.  We can only make predictions at the original x points.   For
other x's we could use Interpolation. See ?approx  (?spline can also
work in smooth cases but in the example provided the function has a
kink and that won't work well with splines.)
as.lm(modellogis)$model
y          a             b             c  (offset)
1   1.281055090 0.06601796 -4.411829e-01  1.168928e+00  1.397153
2   1.563609934 0.04798815 -3.268846e-01  9.766080e-01  1.015584
3   0.001570796 0.04798815 -3.268846e-01  9.766080e-01  1.015584
4   2.291579783 0.16311227 -9.767241e-01  1.597189e+00  3.451981
5   0.841891853 0.12203013 -7.665928e-01  1.512752e+00  2.582551
6   6.553951324 0.21464369 -1.206154e+00  1.564573e+00  4.542552
7  14.243274230 0.74450055 -1.361047e+00 -1.455630e+00 15.756031
8  14.519899320 0.59707858 -1.721353e+00 -6.770205e-01 12.636107
9  15.066473610 0.74450055 -1.361047e+00 -1.455630e+00 15.756031
10 21.728809880 1.00000000 -2.943955e-13 -9.073765e-12 21.163223
11 18.553054450 1.00000000 -2.943955e-13 -9.073765e-12 21.163223
12 23.722637370 1.00000000 -2.943955e-13 -9.073765e-12 21.163223
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
On Tue, May 15, 2012 at 11:20 PM, Gabor Grothendieck
On Tue, May 15, 2012 at 8:08 PM, Francisco Mora Ardila
<fmora at oikos.unam.mx> wrote:
Hi all

I have fitted a model usinf nls function to these data:

x
?[1] ? 1 ? 0 ? 0 ? 4 ? 3 ? 5 ?12 ?10 ?12 100 100 100

y
?[1] ?1.281055090 ?1.563609934 ?0.001570796 ?2.291579783 ?0.841891853
?[6] ?6.553951324 14.243274230 14.519899320 15.066473610 21.728809880
[11] 18.553054450?23.722637370

The model fitted is:

modellogis<-nls(y~SSlogis(x,a,b,c))

It runs OK. Then I calculate confidence intervals for the actual data using:

dataci<-predict(as.lm(modellogis), interval = "confidence")

BUt I don?t get smooth curves when plotting it, so I want to get other "confidence
vectors" based on a new x vector by defining a new data to do predictions:

x0 <- ?seq(0,15,1)
dataci<-predict(as.lm(modellogis), newdata=data.frame(x=x0), interval = "confidence")

BUt it does not work: I get the same initial confidence interval

Any ideas on how to get tconfidence and prediction intervals using new X data on a
previous model?

as.lm is a linear model between the response variable and the gradient
of the nonlinear model and as we see below x is not part of that
linear model so x can't be in newdata when predicting from the tangent
model. ?We can only make predictions at the original x points. ? For
other x's we could use Interpolation. See ?approx ?(?spline can also
work in smooth cases but in the example provided the function has a
kink and that won't work well with splines.)

as.lm(modellogis)$model
? ? ? ? ? ? ?y ? ? ? ? ?a ? ? ? ? ? ? b ? ? ? ? ? ? c ?(offset)
1 ? 1.281055090 0.06601796 -4.411829e-01 ?1.168928e+00 ?1.397153
2 ? 1.563609934 0.04798815 -3.268846e-01 ?9.766080e-01 ?1.015584
3 ? 0.001570796 0.04798815 -3.268846e-01 ?9.766080e-01 ?1.015584
4 ? 2.291579783 0.16311227 -9.767241e-01 ?1.597189e+00 ?3.451981
5 ? 0.841891853 0.12203013 -7.665928e-01 ?1.512752e+00 ?2.582551
6 ? 6.553951324 0.21464369 -1.206154e+00 ?1.564573e+00 ?4.542552
7 ?14.243274230 0.74450055 -1.361047e+00 -1.455630e+00 15.756031
8 ?14.519899320 0.59707858 -1.721353e+00 -6.770205e-01 12.636107
9 ?15.066473610 0.74450055 -1.361047e+00 -1.455630e+00 15.756031
10 21.728809880 1.00000000 -2.943955e-13 -9.073765e-12 21.163223
11 18.553054450 1.00000000 -2.943955e-13 -9.073765e-12 21.163223
12 23.722637370 1.00000000 -2.943955e-13 -9.073765e-12 21.163223

Also regarding your comment about not getting smooth curves be sure
that you order the x's (and permute the y's the same way) before
plotting.
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
On Tue, May 15, 2012 at 11:20 PM, Gabor Grothendieck
On Tue, May 15, 2012 at 8:08 PM, Francisco Mora Ardila
<fmora at oikos.unam.mx> wrote:
Hi all

I have fitted a model usinf nls function to these data:

x
?[1] ? 1 ? 0 ? 0 ? 4 ? 3 ? 5 ?12 ?10 ?12 100 100 100

y
?[1] ?1.281055090 ?1.563609934 ?0.001570796 ?2.291579783 ?0.841891853
?[6] ?6.553951324 14.243274230 14.519899320 15.066473610 21.728809880
[11] 18.553054450?23.722637370

The model fitted is:

modellogis<-nls(y~SSlogis(x,a,b,c))

It runs OK. Then I calculate confidence intervals for the actual data using:

dataci<-predict(as.lm(modellogis), interval = "confidence")

BUt I don?t get smooth curves when plotting it, so I want to get other "confidence
vectors" based on a new x vector by defining a new data to do predictions:

x0 <- ?seq(0,15,1)
dataci<-predict(as.lm(modellogis), newdata=data.frame(x=x0), interval = "confidence")

BUt it does not work: I get the same initial confidence interval

Any ideas on how to get tconfidence and prediction intervals using new X data on a
previous model?

as.lm is a linear model between the response variable and the gradient
of the nonlinear model and as we see below x is not part of that
linear model so x can't be in newdata when predicting from the tangent
model. ?We can only make predictions at the original x points. ? For
other x's we could use Interpolation. See ?approx ?(?spline can also
work in smooth cases but in the example provided the function has a
kink and that won't work well with splines.)

as.lm(modellogis)$model
? ? ? ? ? ? ?y ? ? ? ? ?a ? ? ? ? ? ? b ? ? ? ? ? ? c ?(offset)
1 ? 1.281055090 0.06601796 -4.411829e-01 ?1.168928e+00 ?1.397153
2 ? 1.563609934 0.04798815 -3.268846e-01 ?9.766080e-01 ?1.015584
3 ? 0.001570796 0.04798815 -3.268846e-01 ?9.766080e-01 ?1.015584
4 ? 2.291579783 0.16311227 -9.767241e-01 ?1.597189e+00 ?3.451981
5 ? 0.841891853 0.12203013 -7.665928e-01 ?1.512752e+00 ?2.582551
6 ? 6.553951324 0.21464369 -1.206154e+00 ?1.564573e+00 ?4.542552
7 ?14.243274230 0.74450055 -1.361047e+00 -1.455630e+00 15.756031
8 ?14.519899320 0.59707858 -1.721353e+00 -6.770205e-01 12.636107
9 ?15.066473610 0.74450055 -1.361047e+00 -1.455630e+00 15.756031
10 21.728809880 1.00000000 -2.943955e-13 -9.073765e-12 21.163223
11 18.553054450 1.00000000 -2.943955e-13 -9.073765e-12 21.163223
12 23.722637370 1.00000000 -2.943955e-13 -9.073765e-12 21.163223

I have added a FAQ to the home page since this isn't the first time
this question has come up:

http://nls2.googlecode.com#FAQs
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
Thanks! Now it is clear.

Francisco

On Wed, 16 May 2012 07:32:56 -0400, Gabor Grothendieck wrote
On Tue, May 15, 2012 at 11:20 PM, Gabor Grothendieck
<ggrothendieck at gmail.com> wrote:
On Tue, May 15, 2012 at 8:08 PM, Francisco Mora Ardila
<fmora at oikos.unam.mx> wrote:
Hi all

I have fitted a model usinf nls function to these data:

x
?[1] ? 1 ? 0 ? 0 ? 4 ? 3 ? 5 ?12 ?10 ?12 100 100 100

y
?[1] ?1.281055090 ?1.563609934 ?0.001570796 ?2.291579783 ?0.841891853
?[6] ?6.553951324 14.243274230 14.519899320 15.066473610 21.728809880
[11] 18.553054450?23.722637370

The model fitted is:

modellogis<-nls(y~SSlogis(x,a,b,c))

It runs OK. Then I calculate confidence intervals for the actual data using:

dataci<-predict(as.lm(modellogis), interval = "confidence")

BUt I don?t get smooth curves when plotting it, so I want to get other "confidence
vectors" based on a new x vector by defining a new data to do predictions:

x0 <- ?seq(0,15,1)
dataci<-predict(as.lm(modellogis), newdata=data.frame(x=x0), interval = 
"confidence")
BUt it does not work: I get the same initial confidence interval

Any ideas on how to get tconfidence and prediction intervals using new X data on a
previous model?

as.lm is a linear model between the response variable and the gradient
of the nonlinear model and as we see below x is not part of that
linear model so x can't be in newdata when predicting from the tangent
model. ?We can only make predictions at the original x points. ? For
other x's we could use Interpolation. See ?approx ?(?spline can also
work in smooth cases but in the example provided the function has a
kink and that won't work well with splines.)

as.lm(modellogis)$model
? ? ? ? ? ? ?y ? ? ? ? ?a ? ? ? ? ? ? b ? ? ? ? ? ? c ?(offset)
1 ? 1.281055090 0.06601796 -4.411829e-01 ?1.168928e+00 ?1.397153
2 ? 1.563609934 0.04798815 -3.268846e-01 ?9.766080e-01 ?1.015584
3 ? 0.001570796 0.04798815 -3.268846e-01 ?9.766080e-01 ?1.015584
4 ? 2.291579783 0.16311227 -9.767241e-01 ?1.597189e+00 ?3.451981
5 ? 0.841891853 0.12203013 -7.665928e-01 ?1.512752e+00 ?2.582551
6 ? 6.553951324 0.21464369 -1.206154e+00 ?1.564573e+00 ?4.542552
7 ?14.243274230 0.74450055 -1.361047e+00 -1.455630e+00 15.756031
8 ?14.519899320 0.59707858 -1.721353e+00 -6.770205e-01 12.636107
9 ?15.066473610 0.74450055 -1.361047e+00 -1.455630e+00 15.756031
10 21.728809880 1.00000000 -2.943955e-13 -9.073765e-12 21.163223
11 18.553054450 1.00000000 -2.943955e-13 -9.073765e-12 21.163223
12 23.722637370 1.00000000 -2.943955e-13 -9.073765e-12 21.163223

I have added a FAQ to the home page since this isn't the first time
this question has come up:

http://nls2.googlecode.com#FAQs

-- 
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
----------------------
Francisco Mora Ardila
Estudiante de Doctorado
Centro de Investigaciones en Ecosistemas
Universidad Nacional Aut?noma de M?xico
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