Dear Achim, I suspect that the problem, involving a fifth-degree raw polynomial, is very ill-conditioned, and that the computation in linear.hypothesis() fails because it is not as stable as lm() and summary.lm(). (BTW, one would not normally call summary.lm() directly, but rather use the generic summary() function instead.) A possible solution would be to use a fifth-degree orthogonal polynomial, with the formula energyshare ~ poly(x.log, 5). (You don't need the 1 for the constant, since it's implied.)
That said, it's hard for me to understand why it's interesting to have standard errors for the individual coefficients of a high-degree polynomial, and I'd also be concerned about the sensibleness of fitting a fifth-degree polynomial in the first place. I hope this helps, John -------------------------------- John Fox, Professor Department of Sociology McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox/ ------------ original message ---------- Achim Voß achim.voss at uni-muenster.de Tue Dec 30 14:15:37 CET 2008 Hi! I am trying to estimate Engel curves using a big sample (>42,000) using lm and taking heteroskedasticity into account by using the summaryHCCM posted here by John Fox (Mon Dec 25 16:01:59 CET 2006). Having used the SIC (with MASS stepAIC) to determine how many powers to use I estimate the model: > # ========================================= > summary.lm(fit.lm.5) Call: lm(formula = energyshare ~ 1 + I(x.log) + I(x.log^2) + I(x.log^3) + I(x.log^4) + I(x.log^5), data = ev) Residuals: Min 1Q Median 3Q Max -0.098819 -0.023682 -0.007043 0.013924 0.486615 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -5.452e+01 1.234e+01 -4.418 9.97e-06 *** I(x.log) 3.177e+01 6.966e+00 4.560 5.13e-06 *** I(x.log^2) -7.330e+00 1.567e+00 -4.677 2.93e-06 *** I(x.log^3) 8.395e-01 1.757e-01 4.778 1.78e-06 *** I(x.log^4) -4.775e-02 9.814e-03 -4.865 1.15e-06 *** I(x.log^5) 1.079e-03 2.185e-04 4.939 7.90e-07 *** --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 0.03748 on 42738 degrees of freedom Multiple R-squared: 0.09236, Adjusted R-squared: 0.09226 F-statistic: 869.8 on 5 and 42738 DF, p-value: < 2.2e-16 > # ========================================= Now I use summaryHCCM(fit.lm.5): > # ========================================= > summaryHCCM(fit.lm.5) Fehler in solve.default(L %*% V %*% t(L)) : System ist für den Rechner singulär: reziproke Konditionszahl = 6.98689e-19 > # ========================================= ("Error in solve.default(L %*% V %*% t(L)) : System is singulary for the computer: reciprocal number of conditions = 6.98689e-19") This does not happen if I omit I(x.log^5). I do not know what it means and I'd be grateful if anyone could help. And I'd like to add a (more or less) related question: Can I still use AIC, SIC etc. if I know there's a heteroskedasticity problem? Thanks in advance, Achim ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
