Linear model
On Wed, Feb 11, 2009 at 1:36 PM, kayj <kjaja27 at yahoo.com> wrote:
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
Some comments: i) the printout says that the value is less than 2e-16 ii) It seems strange to interpret a p-value at that level of precision iii) you're confusing what is printed vs what is stored iv) see http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f v) rather than subtracting from 1, use the 'lower.tail' argument to pt:
2*pt(10.843,2838,lower=F)
[1] 7.185635e-27 hth, Kingsford Jones
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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