I have to compute and compare the least squares method on a model to the least trimmed method.

The LS model results were:

               Estimate Std. Error t value Pr(>|t|)    
(Intercept)    -39.9197    11.8960  -3.356  0.00375 ** 
Air.Flow         0.7156     0.1349   5.307  5.8e-05 ***
Water.Temp       1.2953     0.3680   3.520  0.00263 ** 
Acid.Conc.      -0.1521     0.1563  -0.973  0.34405    

The LTS results were:

  (Intercept)     Air.Flow      Water.Temp      Acid.Conc. 
-3.580556e+01  7.500000e-01    3.333333e-01    3.489094e-17 

How do I interpret the results for LTS? I know Air.Flow and Water.Temp are significant and ACid.conc is not. But don't know about LTS coefficients since there are no standard errors.

  • 2
    $\begingroup$ Am I correct that you're using R? What package / function are you using? Why do you need to compare these, is this a homework assignment? $\endgroup$ Nov 8, 2012 at 20:10
  • $\begingroup$ The standard errors for the LTS model can be displayed with: coef(summary(ltsReg(stack.loss ~ Air.Flow + Water.Temp + Acid.Conc., data=stackloss)))[, "Std. Error"] $\endgroup$
    – smillig
    Dec 9, 2012 at 10:13
  • $\begingroup$ Why do you say you don't know what the LTS s.e. are? It's in the R output: ltsreg <- ltsReg(y ~ x); summary(ltsreg) $\endgroup$
    – user603
    May 1, 2013 at 10:24

2 Answers 2


Why do you have to compare them? If this is a homework assignment then you should indicate that and you should give more detail on what the assignment means by compare.

One way to get standard errors/confidence intervals/significance tests for LTS coefficients would be to bootstrap the process.

A possibly more interesting comparison would be to compare in a scatterplot (or possibly a Bland-Altman plot) the predicted values from the 2 models with an $y=x$ line.

Then there is always the simple eye-ball comparison that the coefficient for Air.Flow did not change much, but the one for Water.Temp is a quarter of what it was under LS, which point(s) drive that? and Acid.Conc is much closer to 0, but was not sigificantly different before.

  • 1
    $\begingroup$ yes, it is homework. I will indicate homework next time. I am using R and using ltsreg compared to the least square model. $\endgroup$
    – stacy
    Nov 8, 2012 at 20:37

If you haven't been given a way of finding significance, it would be a fair bet that you're not supposed to compare the significance. (But maybe you're supposed to figure out how to do it.)

In any case, you can compare coefficients in size and sign, you can compare fitted values, you can compare residual plots and what they might tell you about the appropriateness of the various assumptions


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