[R] problem to get coefficient from lm()

Jinsong Zhao jinsong_zh at yahoo.com
Fri Feb 6 17:50:36 CET 2004


Dear all,

The following is a example that I run and hope to get a linear model.
However, I find the lm() can not give correct coefficients for the
linear model.

I hope it's just my own mistake. Please help. TIA.

Regards,

Jinsong

> x
 [1] 3.760216 3.997288 3.208872 3.985417 3.265704 3.497505 2.923540
3.193937
 [9] 3.102787 3.419574 3.169374 2.928510 3.153821 3.100385 3.768770
3.610583
[17] 3.588902 3.180961 3.415033 3.595447 2.826521 3.283875 3.125694
3.558275
[25] 4.625191 3.735673 4.387571 4.884388 3.067845 2.993892 3.068684
4.166771
[33] 4.680322 4.103344 4.340546 3.685950 4.026451 4.161470 3.696610
4.026815
[41] 3.854591 3.399321 3.492201 3.479075 4.417633 3.579751 3.433551
3.248359
[49] 2.874979 3.123579 3.131758 3.447368 3.086075 3.558168 4.537077
> y
 [1] 4.431 4.560 2.920 4.481 3.295 3.679 2.779 2.654 2.837 3.664 3.421
2.693
[13] 2.624 3.057 4.633 3.807 3.886 3.011 3.873 3.717 2.077 3.324 2.064
3.390
[25] 4.170 3.232 3.114 4.701 3.746 2.880 3.723 4.015 4.881 4.483 4.762
3.834
[37] 4.321 3.837 3.624 4.306 3.466 3.492 3.502 3.510 4.236 3.286 3.657
3.218
[49] 2.840 3.011 3.417 3.625 3.524 3.390 4.296
> g <- lm(y ~ x)
> summary(g)

Call:
lm(formula = y ~ x)

Residuals:
    Min      1Q  Median      3Q     Max 
-1.2736 -0.2094  0.0098  0.2654  0.8642 

Coefficients:
              Estimate Std. Error   t value Pr(>|t|)    
(Intercept) -1.274e-15  4.027e-01 -3.16e-15        1    
x            1.000e+00  1.113e-01     8.982 3.13e-12 ***
---
Signif. codes:  0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 

Residual standard error: 0.4179 on 53 degrees of freedom
Multiple R-Squared: 0.6035,     Adjusted R-squared: 0.596 
F-statistic: 80.68 on 1 and 53 DF,  p-value: 3.128e-12 

 

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