[R] Linear Regression with slope equals 0

[(Ted Harding)]

On 14-Aug-07 11:36:37, E.N.D.Grew at exeter.ac.uk wrote:
> 
>  Hi there, am trying to run a linear regression with a slope of 0.
> 
>  I have a dataset as follows
> 
>  t d
>  1 303
>  2 302
>  3 304
>  4 306
>  5 307
>  6 303
> 
>  I would like to test the significance that these points would lie on a
> horizontal straight line.
> 
>  The standard regression lm(d~t) doesn't seem to allow the slope to be
> set.

The model d~1 will fit a constant (the mean), i.e. a regressio with
slope = 0. The model d~t will fit the usual linear regression.
The two con be compared with anova(), as well as getting the details
of the individual fits with summary().

E.g. (with your example):

  d<-c(303,302,304,306,207,303)
  t<-c(1,2,3,4,5,6)
  lm0<-lm(u~1);lm1<-lm(u~t);anova(lm0,lm1)

##Analysis of Variance Table
##Model 1: u ~ 1
##Model 2: u ~ t
##  Res.Df    RSS Df Sum of Sq      F Pr(>F)
##1      5 7785.5                           
##2      4 6641.4  1    1144.1 0.6891 0.4531

summary(lm0)
## Call: lm(formula = u ~ 1)
## Residuals:
##    1     2     3     4     5     6 
## 15.5  14.5  16.5  18.5 -80.5  15.5 
## Coefficients:
##            Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   287.50      16.11   17.85 1.01e-05 ***
##Residual standard error: 39.46 on 5 degrees of freedom

mean(d)
## [1] 287.5

summary(lm1)
## Call: lm(formula = u ~ t)
## Residuals:
##      1       2       3       4       5       6 
## -4.714   2.371  12.457  22.543 -68.371  35.714 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)   
## (Intercept)  315.800     37.934   8.325  0.00114 **
## t             -8.086      9.740  -0.830  0.45314   
## Residual standard error: 40.75 on 4 degrees of freedom
## Multiple R-Squared: 0.147,      Adjusted R-squared: -0.0663 
## F-statistic: 0.6891 on 1 and 4 DF,  p-value: 0.4531 

The P-value for the slope in lm1 is the same as the P-value
returned by anova().

If you want to force a particular non-zero slope (e.g. s0)
for comparison with the data, you can use

  lm0 <- lm(d - s0*t ~ 1),

compared with

  lm1<- lm(d- s0*t ~ t)

for instance.

Hoping this helps,
Ted.

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E-Mail: (Ted Harding) <Ted.Harding at manchester.ac.uk>
Fax-to-email: +44 (0)870 094 0861
Date: 14-Aug-07                                       Time: 13:16:05
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