[R] Rsquared in summary(lm)
apjaworski@mmm.com
apjaworski at mmm.com
Fri May 10 17:22:22 CEST 2002
I am not sure where the problem might be and I missed the original message
by Wouter, but here is a simple example I just ran (Win2000, 1.5.0) which
shows no apparent problems.
> x <- c(-3, -1.5, 2, 5, 7, 8)
> y <- c(2, 4, 5, 6, 6, 9)
> sxx <- sum((x-mean(x))^2)
> sxy <- sum((x-mean(x))*(y-mean(y)))
> bb <- sxy/sxx
> bb
[1] 0.4761517
> aa <- mean(y) - bb*mean(x)
> aa
[1] 3.944558
> ee <- y - aa - bb*x
> sst <- sum((y-mean(y))^2)
> sse <- sum(ee^2)
> ssr <- sst-sse
> ssr/sst
[1] 0.8477822
> summary(lm(y~x))
Call:
lm(formula = y ~ x)
Residuals:
1 2 3 4 5 6
-0.5161 0.7697 0.1031 -0.3253 -1.2776 1.2462
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 3.9446 0.5098 7.737 0.00150 **
x 0.4762 0.1009 4.720 0.00917 **
---
Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
Residual standard error: 1.02 on 4 degrees of freedom
Multiple R-Squared: 0.8478, Adjusted R-squared: 0.8097
F-statistic: 22.28 on 1 and 4 DF, p-value: 0.009172
Cheers,
Andy
__________________________________
Andy Jaworski
Engineering Systems Technology Center
3M Center, 518-1-01
St. Paul, MN 55144-1000
-----
E-mail: apjaworski at mmm.com
Tel: (651) 733-6092
Fax: (651) 736-3122
Wataru Shito
<shito at seinan-g To: wouter buytaert <wouter.buytaert at yucom.be>
u.ac.jp> cc: r-help at stat.math.ethz.ch
(bcc: Andrzej P. Jaworski/US-Corporate/3M/US)
05/09/2002 Subject: Re: [R] Rsquared in summary(lm)
20:24
Hi, Wouter
Actually, I have a similar problem too with a simple regression.
In my case, not only the R-square but also the estimates of intercept
and coefficient by lm() seem different from the calculation with the
well known formula for a simple regression.
What I used is the following code. (I have just started to use R last
week so don't blame my inmature code, please!)
Simply,
> ols1( y, x )
will give you the result of the simple regression.
Wouter, could you try the following code on your data and see whether
that's what you expect or not?
I will appreciate if anyone can give me some advice why this
differenct happens.
Thankk you.
Wataru Shito
-----------------------------------------
# Single Explanatory Variable Least Square Regression
#
library(methods)
# create ols class
setClass("ols", representation
( coefficients="list", standard.errors="list",
r.square="numeric" ))
setMethod("show", "ols",
function(object)
{
# create row names for data.frame
rownames <- c("(Intercept)", "X")
# create data.frame
z <- data.frame( row.names=rownames,
Estimate=object at coefficients,
Std.Error=object at standard.errors,
t.value=t.values )
cat("\n")
print(z)
cat( "\nR-Square:", object at r.square, "\n\n" )
}
)
ols1 <- function( y, x ){
size <- length(x) # number of ovservations
xbar <- mean(x)
ybar <- mean(y)
Sxx <- sum( (x-xbar)^2 )
b <- sum( (x-xbar)*(y-ybar) )/Sxx # coefficient
a <- ybar - b*xbar # interception
e <- y - a - b*x # residuals
# SSE (error sum of squares)
SSE <- sum( e^2 )
# SST (total sum of squares)
SST <- sum( (y-ybar)^2 )
# SSR (regression sum of squares)
SSR <- b^2 * Sxx
# Coefficient of determination
r2 <- SSR / SST
# unbiased estimator of sigma^2
s.square <- sum(e^2)/(size - 2)
# standard error for b
std.error.b <- sqrt( s.square/Sxx )
# standard error for intercept
std.error.a <- sqrt( s.square*(1/size + xbar^2/Sxx) )
standard.errors <- list( intercept=std.error.a, coeficient=std.error.b )
coefficients <- list( intercept=a, coefficient=b )
new("ols", coefficients=coefficients, standard.errors=standard.errors,
r.square=r2 )
}
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