# [R] residuals

Shawn Way sway at tanox.com
Fri Apr 16 16:49:22 CEST 2004

```I was going through all kinds of torture to figure this out.  An elegant
solution, indeed.

Thank you very much.

Shawn Way, PE
Engineering Manager
sway at tanox.com

-----Original Message-----
From: Liaw, Andy [mailto:andy_liaw at merck.com]
Sent: Friday, April 16, 2004 9:15 AM
To: 'Shawn Way'
Subject: RE: [R] residuals

Shawn,

This is what I'd do:

lof <- function(x, y) {
fit1 <- anova(lm(y ~ x))
fit2 <- summary(aov(y ~ factor(x)))[[1]]
SSpe <- fit2[["Sum Sq"]][2]
DFpe <- fit2[["Df"]][2]
SSresid <- fit1[["Sum Sq"]][2]
DFresid <- fit1[["Df"]][2]
SSlof <- SSresid - SSpe
DFlof <- DFresid - DFpe
Fstat <- (SSlof / DFlof) / (SSpe / DFpe)
pval <- pf(Fstat, DFlof, DFpe, lower.tail=FALSE)
out <- matrix(c(SSlof, SSpe, DFlof, DFpe, Fstat, NA, pval, NA), nrow=2,
dimnames=list(c("Lack-of-fit", "Pure Error"),
c("Sum Sq", "Df", "F value", "Pr(>F)")))
class(out) <- c("anova", class(out))
out
}

> lof(data\$ref, data\$actual)
Sum Sq Df F value Pr(>F)
Lack-of-fit 0.13441  1  0.7984 0.4374
Pure Error  0.50505  3

Best,
Andy

> From: Shawn Way
>
>
> I'm trying to determine the lack of fit for regression on the
> following:
>
> data <- data.frame(ref=c(0,50,100,0,50,100),
>                              actual=c(.01,50.9,100.2,.02,49.9,100.1),
>                              level=gl(3,1))
> fit <- lm(actual~ref,data)
> fit.aov <- aov(actual~ref+Error(level),data)
>
> According to the information I have, the lack of fit for this
> regression is
> the f-ratio between the residuals of the level contribution and the
> residuals within.  I'm trying to get the information from the
> fit to make
> this ratio of the residual mean squares.
>
> Any thoughts?
>
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>
>

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