significantly different from one (not zero) using lm

Hello,

Thanks.
But the parameter offset is new to me.
Please kindly explain why setting offset to x will give a significant
test
of whether the slope coefficient is different from one.
(I checked the ?lm but still do not understand it well)

Thanks again

Elaine


On Wed, May 1, 2013 at 11:12 AM, Thomas Lumley <tlumley at uw.edu> wrote:

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 <pauljohn32 at 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.tw at 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

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--
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

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--
Thomas Lumley
Professor of Biostatistics
University of Auckland

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