R help - Mar 2009 - plot confidence limits of a regression line - problem

hi,


I don't know what I am doing wrong,

but with that code;


x1 <- c(1.60, 0.27, 0.17, 1.63, 1.37, 2.00, 0.90, 1.07, 0.89, 0.43,  
0.37, 0.59,
0.47, 1.83, 1.79, 0.90, 0.72, 1.83, 0.23, 1.97, 2.03, 2.19, 2.03, 0.86)

x2 <- c(1.30, 0.24, 0.20, 0.50, 1.33, 1.87, 1.30, 0.75, 1.07, 0.43,  
0.37, 0.87,
1.40, 1.37, 1.63, 0.80, 0.57, 1.60, 0.39, 2.03, 1.90, 2.07, 1.93, 0.93)

model <- lm(x1 ~ x2)

predict(model, newdata=data.frame(x=seq(0,4)), ,interval =  
c("confidence"), level = 0.90,type="response")




I get an error message saying "newdate has four lines, but the  
variable found has 24"


what is wrong in the code?


thanks!
	[[alternative HTML version deleted]]

It is not an error but rather a warning. As you should have seen, R  
went ahead and returned estimates for 24 predicted values for x1 for  
arguments to the formula of x2. In R errors and warnings are very  
different. You are expected to post full console messages to prevent  
this sort of confusion. In this case the warning puts you on notice  
that your offer to newdata= was not acceptable. You did not construct  
it with names that match those in the formula which was used to create  
model.

Please follow the links offered and read the Posting Guide:

-- 
David Winsemius


On Mar 7, 2009, at 8:18 PM, Martin Batholdy wrote:
> hi,
>
>
> I don't know what I am doing wrong,
>
> but with that code;
>
>
> x1 <- c(1.60, 0.27, 0.17, 1.63, 1.37, 2.00, 0.90, 1.07, 0.89, 0.43,
> 0.37, 0.59,
> 0.47, 1.83, 1.79, 0.90, 0.72, 1.83, 0.23, 1.97, 2.03, 2.19, 2.03,  
> 0.86)
>
> x2 <- c(1.30, 0.24, 0.20, 0.50, 1.33, 1.87, 1.30, 0.75, 1.07, 0.43,
> 0.37, 0.87,
> 1.40, 1.37, 1.63, 0.80, 0.57, 1.60, 0.39, 2.03, 1.90, 2.07, 1.93,  
> 0.93)
>
> model <- lm(x1 ~ x2)
>
> predict(model, newdata=data.frame(x=seq(0,4)), ,interval >
c("confidence"), level = 0.90,type="response")
>
>
>
>
> I get an error message saying "newdate has four lines, but the
> variable found has 24"
>
>
> what is wrong in the code?
>
>
> thanks!
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> 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
Heritage Laboratories
West Hartford, CT

Hi,
the predict-function is not very good in guessing names from a given new 
dataset. So if you want to predict x1 from some new x2 values, your 
newdata argument should contain some x2 - values.

predict(model, newdata=data.frame(x2=seq(0,4)), interval =  
c("confidence"), level = 0.90,type="response")

without that, predict uses your old x2 values and issues a warning message.


Martin Batholdy schrieb:
> hi,
>
>
> I don't know what I am doing wrong,
>
> but with that code;
>
>
> x1 <- c(1.60, 0.27, 0.17, 1.63, 1.37, 2.00, 0.90, 1.07, 0.89, 0.43,  
> 0.37, 0.59,
> 0.47, 1.83, 1.79, 0.90, 0.72, 1.83, 0.23, 1.97, 2.03, 2.19, 2.03, 0.86)
>
> x2 <- c(1.30, 0.24, 0.20, 0.50, 1.33, 1.87, 1.30, 0.75, 1.07, 0.43,  
> 0.37, 0.87,
> 1.40, 1.37, 1.63, 0.80, 0.57, 1.60, 0.39, 2.03, 1.90, 2.07, 1.93, 0.93)
>
> model <- lm(x1 ~ x2)
>
> predict(model, newdata=data.frame(x=seq(0,4)), ,interval =  
> c("confidence"), level = 0.90,type="response")
>
>
>
>
> I get an error message saying "newdate has four lines, but the  
> variable found has 24"
>
>
> what is wrong in the code?
>
>
> thanks!
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> 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.
>   
-- 
Eik Vettorazzi
Institut f?r Medizinische Biometrie und Epidemiologie
Universit?tsklinikum Hamburg-Eppendorf

Martinistr. 52
20246 Hamburg

T ++49/40/42803-8243
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