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I am doing some basic fitting of data and exploring different fits. I understand that the residual is the difference between the sample and the estimated function value. The norm of the residuals is a measure of the deviation between the correlation and the data. A lower norm signifies a better fit.

Suppose a cubic fit has a norm of residuals of 0.85655 and a linear fit has a norm of residuals of 0.89182. What if the norms are 0.17113 and 0.24916? Is the difference between these two significant? Is a norm of residual less than 1 considered good? If not what is generally regarded as "acceptable" norm of residual.

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  • $\begingroup$ Is your question related to some extent to assessing model goodness of fit or the comparison of two nested models? $\endgroup$
    – chl
    Commented Sep 14, 2010 at 19:28
  • $\begingroup$ @chl, both, i have data and need to determine the proper fit and how well it fits $\endgroup$ Commented Sep 14, 2010 at 19:43

1 Answer 1

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So, I would recommend using standard method for comparing nested models. In your case, you consider two alternative models, the cubic fit being the more "complex" one. An F- or $\chi^2$-test tells you whether the residual sum of squares or deviance significantly decrease when you add further terms. It is very like comparing a model including only the intercept (in this case, you have residual variance only) vs. another one which include one meaningful predictor: does this added predictor account for a sufficient part of the variance in the response? In your case, it amounts to say: Modeling a cubic relationship between X and Y decreases the unexplained variance (equivalently, the $R^2$ will increase), and thus provide a better fit to the data compared to a linear fit.

It is often used as a test of linearity between the response variable and the predictor, and this is the reason why Frank Harrell advocates the use of restricted cubic spline instead of assuming a strict linear relationship between Y and the continuous Xs (e.g. age).

The following example comes from a book I was reading some months ago (High-dimensional data analysis in cancer research, Chap. 3, p. 45), but it may well serves as an illustration. The idea is just to fit different kind of models to a simulated data set, which clearly highlights a non-linear relationship between the response variable and the predictor. The true generative model is shown in black. The other colors are for different models (restricted cubic spline, B-spline close to yours, and CV smoothed spline).

library(rms)
library(splines)
set.seed(101)
f <- function(x) sin(sqrt(2*pi*x))
n <- 1000
x <- runif(n, 0, 2*pi)
sigma <- rnorm(n, 0, 0.25)
y <- f(x) + sigma
plot(x, y, cex=.4)
curve(f, 0, 6, lty=2, add=TRUE)
# linear fit
lm00 <- lm(y~x)
# restricted cubic spline, 3 knots (2 Df)
lm0 <- lm(y~rcs(x,3))
lines(seq(0,6,length=1000), 
      predict(lm0,data.frame(x=seq(0,6,length=1000))),
      col="red")
# use B-spline and a single knot at x=1.13 (4 Df)
lm1 <- lm(y~bs(x, knots=1.13))
lines(seq(0,6,length=1000),
      predict(lm1,data.frame(x=seq(0,6,length=1000))), 
      col="green")
# cross-validated smoothed spline (approx. 20 Df)
xy.spl <- smooth.spline(x, y, cv=TRUE)
lines(xy.spl, col="blue")
legend("bottomleft", c("f(x)","RCS {rms}","BS {splines}","SS {stats}"), 
       col=1:4, lty=c(2,rep(1,3)),bty="n", cex=.6)

exmaple

Now, suppose you want to compare the linear fit (lm00) and model relying on B-spline (lm1), you just have to do an F-test to see that the latter provides a better fit:

> anova(lm00, lm1)
Analysis of Variance Table

Model 1: y ~ x
Model 2: y ~ bs(x, knots = 1.13)
  Res.Df     RSS Df Sum of Sq      F    Pr(>F)    
1    998 309.248                                  
2    995  63.926  3    245.32 1272.8 < 2.2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Likewise, it is quite usual to compare GLM with GAM based on the results of a $\chi^2$-test.

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  • $\begingroup$ Your link at Frank Harrell's webpage, is broken, and it was sadly not crawled by the archive. By any chance, you summarize what was there or try finding another copy? (I understand the difficulty after more than a decade) $\endgroup$
    – user0
    Commented Oct 7, 2022 at 6:43

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