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simple lm question

On Dec 1, 2011, at 10:50 PM, Worik R wrote:

            
Use `lm` the way it is designed to be used, with a data argument:

 > l2 <- lm(e~. , data=as.data.frame(M))
 > summary(l2)

Call:
lm(formula = e ~ ., data = as.data.frame(M))

Residuals:
     Min      1Q  Median      3Q     Max
-0.5558 -0.2396  0.1257  0.2213  0.4586

Coefficients:
             Estimate Std. Error t value Pr(>|t|)
(Intercept)  0.41110    0.25484   1.613    0.128
a           -0.03258    0.28375  -0.115    0.910
b            0.09088    0.25971   0.350    0.731
c            0.09382    0.29555   0.317    0.755
d            0.14725    0.33956   0.434    0.671

Residual standard error: 0.3317 on 15 degrees of freedom
Multiple R-squared: 0.04667,	Adjusted R-squared: -0.2076
F-statistic: 0.1836 on 4 and 15 DF,  p-value: 0.9433