lm(y ~ x1) vs. lm(y ~ x0 + x1 - 1) with x0 <- rep(1, length(y))
Thank you all (including Dennis), this was elucidating. I would have (maybe naively) anticipated that in this somewhat pathological case of fitting without an intercept and re-introducing it via constant x1, R might check whether the design matrix includes a column of ones, and adjust the degrees of freedom accordingly. But now I can see that by explicitly requesting via the formula interface not to fit a constant, I am implicitly stating my hypothesis that y==0, even if I re-introduce my suspicion that y==mu via x1 <- 1. If I understood correctly, x1 is treated as a variable in the latter case, right?
On Jan 22, 2011, at 5:18 , Bert Gunter wrote:
Well ... as x1 is continuous(numeric), it has no levels. So ...?? Note that the fits are identical for both models. The issue is only what is the Null that you are testing in the two cases. In the first case, it is just y = constant, so you are testing the 1 df for x1. In the second, it is y = 0 (which rarely makes any sense) and you are testing the 2 df for the two terms (x0 and x1). Etc. etc. -- Bert On Fri, Jan 21, 2011 at 7:48 PM, David Winsemius <dwinsemius at comcast.net> wrote:
On Jan 21, 2011, at 9:03 PM, jochen laubrock wrote:
Dear list, the following came up in an introductory class. Please help me understand the -1 (or 0+) syntax in formulae: Why do the enumerator dfs, F-statisics etc. differ between the models lm(y ~ x1) and lm(y ~ x0 + x1 - 1), if x0 is a vector containing simply ones?
You are testing something different. In the first case you are testing the difference between the baseline and the second level of x1 (so there is only one d.f.), while in the second case you are testing for both of the coefficients being zero (so the numerator has 2 d.f.). It would be easier to see if you did print() on the fit object. The first model would give you an estimate for an "Intercept", which is really an estimate for the first level of x1. Having been taught to think of anova as just a special case of regression is helpful here. Look at the model first and only then look at the anova table.
Example: N <- 40 x0 <- rep(1,N) x1 <- 1:N vare <- N/8 set.seed(4) e <- rnorm(N, 0, vare^2) X <- cbind(x0, x1) beta <- c(.4, 1) y <- X %*% beta + e summary(lm(y ~ x1)) # [...] # Residual standard error: 20.92 on 38 degrees of freedom # Multiple R-squared: 0.1151, Adjusted R-squared: 0.09182 # F-statistic: 4.943 on 1 and 38 DF, p-value: 0.03222 summary(lm(y ~ x0 + x1 - 1)) # or summary(lm(y ~ 0 + x0 + x1)) # [...] # Residual standard error: 20.92 on 38 degrees of freedom # Multiple R-squared: 0.6888, Adjusted R-squared: 0.6724 # F-statistic: 42.05 on 2 and 38 DF, p-value: 2.338e-10 Thanks in advance, Jochen ---- Jochen Laubrock, Dept. of Psychology, University of Potsdam, Karl-Liebknecht-Strasse 24-25, 14476 Potsdam, Germany phone: +49-331-977-2346, fax: +49-331-977-2793
______________________________________________ R-help at 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.
David Winsemius, MD West Hartford, CT
______________________________________________ R-help at 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.
-- Bert Gunter Genentech Nonclinical Biostatistics 467-7374 http://devo.gene.com/groups/devo/depts/ncb/home.shtml
---- Jochen Laubrock, Dept. of Psychology, University of Potsdam, Karl-Liebknecht-Strasse 24-25, 14476 Potsdam, Germany phone: +49-331-977-2346, fax: +49-331-977-2793