[R] Linear model

Bill.Venables at csiro.au Bill.Venables at csiro.au
Thu Feb 12 01:56:16 CET 2009


If you look carefully you will see that the p-value is actually 

< 2e-16

where the "<" sign is telling you this is an upper limit.  For many machines this is the double precision epsilon value:

> .Machine$double.eps
[1] 2.220446e-16

So there is no pretence to accuracy, other than "the value is postive, but smaller than the least positive value I can represent on this machine".

Bill Venables. 


Bill Venables
http://www.cmis.csiro.au/bill.venables/ 


-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of kayj
Sent: Thursday, 12 February 2009 6:36 AM
To: r-help at r-project.org
Subject: [R] Linear model


I want to know how accurate are the p-values when you do linear regression in
R?

I was looking at the variable x3 and the t=10.843 and the corresponding
p-value=2e-16 which is the same p-value for the intercept where the t-value
for the intercept is 48.402.

I tried to calculate the p-value in R and I got 0

x<-2*(1-pt(10.843,2838))
> x
[1] 0



> G<-lm(y~x1+x2+x3+x4+x5)
> summary(G)

Call:
lm(formula = y ~ x1 + x2 +x3 + x4 + x5)

Residuals:
     Min       1Q   Median       3Q      Max 
-14.3172  -3.2197  -0.2913   2.6938  23.3602 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  22.9461     0.4741  48.402  < 2e-16 ***
x1           -0.1139     0.3734  -0.305  0.76031    
x2           -0.0405     0.1936  -0.209  0.83437    
x3           2.0165     0.1860  10.843  < 2e-16 ***
x4           0.5313     0.1782   2.982  0.00289 ** 
x5           0.5879     0.1779   3.305  0.00096 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

Residual standard error: 4.724 on 2838 degrees of freedom
  (138 observations deleted due to missingness)
Multiple R-squared: 0.05279,    Adjusted R-squared: 0.05112 
F-statistic: 31.63 on 5 and 2838 DF,  p-value: < 2.2e-16


 Thanks for the help

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