R help - Jun 2003 - Residual plotting

Dear all,

So far i could do (in an informal way) to draw a Standardized Resisual plot 
in the following way-
---------------------
 >x <- c(104.1, 106.6, 105.5, 107.5, 109.6, 113.3, 115.5, 117.7, 119.9, 
122.1, 124.3, 126.5, 128.2)
 >y <- c(53732, 52912, 57005, 61354, 67682, 71602, 71961, 75309, 82931, 
93310, 102161, 103068, 108927)
 > beta1<-(n*sum(x*y)-sum(x)*sum(y))/(n*sum(x^2)-sum(x)^2)
 > beta0<-sum(y)/n-beta1*sum(x)/n
 > yhat<-beta0 + beta1*x
 > error<-(y-yhat)
 > se <- sqrt((sum(y^2)-beta0*sum(y)-beta1*sum(x*y))/(n-2))
 > SE<-sqrt(sum(error^2)/(n-2)) # just another way
 > plot(x,error)
 > plot(x,error/se)
---------------------
Or, in the formal way (using s-plus functions), in the following way, but i 
get stuck in the Resisual plot stage, don't know how to draw the 
"Standardized Resisual plot" in this way :

---------------------
 >x <- c(104.1, 106.6, 105.5, 107.5, 109.6, 113.3, 115.5, 117.7, 119.9, 
122.1, 124.3, 126.5, 128.2)
 >y <- c(53732, 52912, 57005, 61354, 67682, 71602, 71961, 75309, 82931, 
93310, 102161, 103068, 108927)
 > cripop<-rbind(x,y)
 > dimnames(cripop)<-NULL
 > columns <- c("1987", "1988", "1989",
"1990", "1991", "1992", "1993",
"1994", "1995", "1996", "1997",
"1998", "1999" )
 > rows<-c("Population","Crimes")
 > dimnames(cripop)<-list(rows,columns)
 > bd<-t(cripop)
 > bd.frame<-data.frame(bd)
 > attach(bd.frame)
 > regressions<-lm(Crimes~Population,data=bd.frame)
 > plot(Population,resid(regressions))
 > ...?...
---------------------
Can any one help me by telling me how can i draw "Standardized Resisual 
plot" from here ?

Also, is there any way i can construct 95% Confidence interval or 
Prediction interval for any value in R ?

_______________________

Mohammad Ehsanul Karim <appstat at HotPOP.com>
Institute of Statistical Research and Training
University of Dhaka, Dhaka- 1000, Bangladesh
_______________________


_______________________

Mohammad Ehsanul Karim <appstat at HotPOP.com>
Institute of Statistical Research and Training
University of Dhaka, Dhaka- 1000, Bangladesh

> -----Original Message-----
> From: wildscop [mailto:appstat at hotpop.com]
> Sent: Thursday, 26 June 2003 2:11 PM
> To: r-help at stat.math.ethz.ch
> Subject: [R] Residual plotting
> 
[snip]
> Or, in the formal way (using s-plus functions), in the 
> following way, but i 
> get stuck in the Resisual plot stage, don't know how to draw the 
> "Standardized Resisual plot" in this way :
> 
> ---------------------
>  >x <- c(104.1, 106.6, 105.5, 107.5, 109.6, 113.3, 115.5, 
> 117.7, 119.9, 
> 122.1, 124.3, 126.5, 128.2)
>  >y <- c(53732, 52912, 57005, 61354, 67682, 71602, 71961, 
> 75309, 82931, 
> 93310, 102161, 103068, 108927)
>  > cripop<-rbind(x,y)
>  > dimnames(cripop)<-NULL
>  > columns <- c("1987", "1988", "1989",
"1990", "1991",
> "1992", "1993", 
> "1994", "1995", "1996", "1997",
"1998", "1999" )
>  > rows<-c("Population","Crimes")
>  > dimnames(cripop)<-list(rows,columns)
>  > bd<-t(cripop)
>  > bd.frame<-data.frame(bd)
>  > attach(bd.frame)
>  > regressions<-lm(Crimes~Population,data=bd.frame)
>  > plot(Population,resid(regressions))
>  > ...?...
> ---------------------
> Can any one help me by telling me how can i draw 
> "Standardized Resisual 
> plot" from here ?
try plot(regressions). One of the plots produced is the standardized residuals v
quantiles. Another is the sqrt(standardized residuals) v fitted values. If you
just want to calculate the standardized residuals yourself, use stdres() in
package MASS.
> 
> Also, is there any way i can construct 95% Confidence interval or 
> Prediction interval for any value in R ?
see ?predict, or more specifically ?predict.lm for linear models.

Cheers,

Simon.

Simon Blomberg, PhD
Depression & Anxiety Consumer Research Unit
Centre for Mental Health Research
Australian National University
http://www.anu.edu.au/cmhr/
Simon.Blomberg at anu.edu.au  +61 (2) 6125 3379