Or use an offset
lm( y ~ x+offset(x), data = dat)
The offset gives x a coefficient of 1, so the coefficient of x in this
model is the difference between the coefficient of x in the model without
an offset and 1 -- the thing you want.
-thomas
On Wed, May 1, 2013 at 2:54 PM, Paul Johnson <pauljoh...@gmail.com> wrote:
> It is easy to construct your own test. I test against null of 0 first so I
> can be sure I match the right result from summary.lm.
>
> ## get the standard error
> seofb <- sqrt(diag(vcov(lm1)))
> ## calculate t. Replace 0 by your null
> myt <- (coef(lm1) - 0)/seofb
> mypval <- 2*pt(abs(myt), lower.tail = FALSE, df = lm1$df.residual)
>
> ## Note you can pass a vector of different nulls for the coefficients
> myt <- (coef(lm1) - c(0,1))/seofb
>
> We could write this into a function if we wanted to get busy. Not a bad
> little homework exercise, I think.
>
>
>
>
> > dat <- data.frame(x = rnorm(100), y = rnorm(100))
> > lm1 <- lm(y ~ x, data = dat)
> > summary(lm1)
>
> Call:
> lm(formula = y ~ x, data = dat)
>
> Residuals:
> Min 1Q Median 3Q Max
> -3.0696 -0.5833 0.1351 0.7162 2.3229
>
> Coefficients:
> Estimate Std. Error t value Pr(>|t|)
> (Intercept) -0.001499 0.104865 -0.014 0.989
> x -0.039324 0.113486 -0.347 0.730
>
> Residual standard error: 1.024 on 98 degrees of freedom
> Multiple R-squared: 0.001224, Adjusted R-squared: -0.008968
> F-statistic: 0.1201 on 1 and 98 DF, p-value: 0.7297
>
> > seofb <- sqrt(diag(vcov(lm1)))
> > myt <- (coef(lm1) - 0)/seofb
> > mypval <- 2*pt(abs(myt), lower.tail = FALSE, df = lm1$df.residual)
> > myt
> (Intercept) x
> -0.01429604 -0.34650900
> > mypval
> (Intercept) x
> 0.9886229 0.7297031
> > myt <- (coef(lm1) - 1)/seofb
> > mypval <- 2*pt(abs(myt), lower.tail = FALSE, df = lm1$df.residual)
> > myt
> (Intercept) x
> -9.550359 -9.158166
> > mypval
> (Intercept) x
> 1.145542e-15 8.126553e-15
>
>
> On Tue, Apr 30, 2013 at 9:07 PM, Elaine Kuo <elaine.kuo...@gmail.com>
> wrote:
>
> > Hello,
> >
> >
> >
> > I am work with a linear regression model:
> >
> > y=ax+b with the function of lm.
> >
> > y= observed migration distance of butterflies
> >
> > x= predicted migration distance of butterflies
> >
> >
> >
> > Usually the result will show
> >
> > if the linear term a is significantly different from zero based on the
> > p-value.
> >
> > Now I would like to test if the linear term is significantly different
> from
> > one.
> >
> > (because I want to know if the regression line (y=ax+b) is significantly
> > from the line with the linear term =1 and the intercept =0)
> >
> >
> >
> > Please kindly advise if it is possible
> >
> > to adjust some default parameters in the function to achieve the goal.
> >
> > Thank you.
> >
> >
> > Elaine
> >
> > [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > 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.
> >
>
>
>
> --
> Paul E. Johnson
> Professor, Political Science Assoc. Director
> 1541 Lilac Lane, Room 504 Center for Research Methods
> University of Kansas University of Kansas
> http://pj.freefaculty.org http://quant.ku.edu
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> 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.
>
--
Thomas Lumley
Professor of Biostatistics
University of Auckland
[[alternative HTML version deleted]]
______________________________________________
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and provide commented, minimal, self-contained, reproducible code.
