R help - May 2013 - significantly different from one (not zero) using lm

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]]

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.tw@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]]

On Wed, 1 May 2013, Elaine Kuo 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)
In addition to the solutions suggested by Paul and Thomas, you could use 
linearHypothesis() from the "car" package. A (non-sensical) example
using
the cars dataset is:

m <- lm(dist ~ speed, data = cars)
linearHypothesis(m, c("(Intercept) = 0", "speed = 1"))

The output is equivalent to the anova() for the offset model that Thomas 
suggested:

m0 <- lm(dist ~ 0 + offset(1 * speed), data = cars)
anova(m0, m)
> 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 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.
>