to start with, you should now what kinds of independent variables in your
data=Rail.
it looks like you depend variable is "travel".
is "Rail" the only IV in your model? it seems has been classifed to L, Q, C.

from the output, it seems the final model should be:
E(travel)=66.5+54*Rail.L-4.7 Rail.Q-2.6* Rail.C-0.57 *Rail^4+12*Rail^5
but, in order to get this output, your function should be
lm(travel~Rail.L+ Rail.Q+ Rail.C+ Rail^4+Rail^5, data=Rail)  ???
On Wed, Feb 6, 2013 at 7:00 AM, jing tang <gimmytang@gmail.com> wrote:

> Hi,
> I am reading the book "Mixed Effects Models in S and S-Plus" and come
> across an example with the Rail data.
> I tried to use lm(travel~Rail,data=Rail) and got the following result:
>
> Call:
> lm(formula = travel ~ Rail, data = Rail)
>
> Residuals:
>     Min      1Q  Median      3Q     Max
> -6.6667 -1.0000  0.1667  1.0000  6.3333
>
> Coefficients:
>             Estimate Std. Error t value Pr(>|t|)
> (Intercept)  66.5000     0.9477  70.169  < 2e-16 ***
> Rail.L       54.3032     2.3214  23.392 2.22e-11 ***
> Rail.Q       -4.6917     2.3214  -2.021 0.066161 .
> Rail.C       -2.6584     2.3214  -1.145 0.274458
> Rail^4       -0.5669     2.3214  -0.244 0.811181
> Rail^5       11.1919     2.3214   4.821 0.000418 ***
> ---
> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>
>
> I wonder how the model is represented. Is it correct to write: Travel =
> beta1 + beta2*Rail + error? Beta1 should be Intercept which is clear to me.
> But how to link beta2 with the coefficients Rail.L, Rail.Q, Rail.C, Rail^4
> and Rail^5 shown in the result?
>
> Thanks for your comments,
>
> Best,
> Jing
>
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>
>
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