Have you looked at results of str on a regression object? I would not
think that there would be a single p.value associated with such a
beast, but that there might be if you examined individual coefficients.
? coefficients
?coef
--
David Winsemius
On Nov 27, 2008, at 4:03 AM, ales grill wrote:
Dear all,
I have wrote a code for a linear regression. I want to
write a loop for so, that I can get estimate for pavlues for six
predictors.
But I am getting for estmate for only last one. How can I get
pvalues for
all my predictors in a loop??
Anticipating your help
Thanks
Ales
mat<-matrix(rnorm(36),nrow=6)
mat
[,1] [,2] [,3] [,4]
[,5] [,6]
[1,] 1.10536338 -0.7613770 -1.7100569 -1.8762241 -0.36579280
0.6465219
[2,] -1.34836804 -0.2174270 -0.1153477 -0.1727683 -1.88406206
1.7484955
[3,] 0.96814418 -2.1483727 0.5839668 -1.2361659 0.04592844
1.9937995
[4,] 0.01960219 -1.2339691 0.8290761 0.1002795 -0.15952881
0.3969251
[5,] 1.62343073 1.3741222 -1.2045854 0.4180127 -0.09898615
1.3575119
[6,] -0.95260509 -0.1522824 -1.4257526 1.0057412 -1.20068336
-0.4306761
res<-rnorm(6)
res
[1] 0.2045252 -0.9824761 0.7727004 0.6439993 1.8005737 1.0167214
pval<-NULL
for(i in c(1:6))
+ {
+ reg<-lm(res~mat[,i])
+ reg
+ pval[i]<-reg$p.value
+ }
pval
NULL
reg
Call:
lm(formula = res ~ mat[, i])
Coefficients:
(Intercept) mat[, i]
0.8195 -0.2557
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