[R] interpretation of lm

[dechao wang]

Hello,

I have not receive the response for this. Sorry for
putting many questions together. I am newer using R.
Some questions may not be good, however, answer to any
question will be very appreciated. 

By the way can you tell me is it possible to do
polynomail regression and response Surface Regression?

thanks,
 --- dechao wang <dechwang at yahoo.co.uk> wrote: > Dear
statistitians / R users,
> 
> I was told to analyse the effects of the four
> factors
> using lm or update for aov. The following is the
> result from lm. As you can see that in the last few
> lines enclosed coefficients. 
> 
> 1)Does this mean I can write a formula, like 
> 
> yield =0.26 + 0.03*H + 1.48*T + 0.04*L + 0.004*C
> 
> 2) in the two levels design, is there any difference
> between I use (-1, 1) to represent the lower and
> higher levels and I use (5000, 10000) as the real
> levels?
> 
> 3) how can I extract the coefficients after
> analysis,
> I mean can I write the coefficients as a vector?
> 
> 4) how can I call R function from C code, for
> example,
> can I call lm(...) from C functiomns?
> 
> 5) after experiment designs, we can analyse which
> factors have big effects, can we construct in R a
> response surface so that we can optimise the levels
> of
> the factors by maxmising the response surface, can
> you
> advise me how to do that in R?
> 
> 6) if the number of factors is over 1000, is it
> possible to analyse the effect of each factor using
> the same way as above?
> 
> Many thanks
> Dechao
> > > 
> > > Dear R users,
> > > 
> > > I did 16 experiments, with 16 responses (yield),
> > and 4
> > > factors(hidden, theres, lrate, cycl) with each
> > having
> > > 2 levels as shown below. I want to do analysis
> of
> > > variance to see which factor affect the
> response.
> > The
> > > coses are as follows:
> > > 
> > >
> hidden<-c(2,2,10,10,2,2,10,10,2,2,10,10,2,2,10,10)
> > > theres<-c(0.0001, 0.1, 0.0001, 0.1, 0.0001, 0.1,
> > > 0.0001, 0.1, 0.0001, 0.1, 0.0001, 0.1, 0.0001,
> > 0.1,
> > > 0.0001, 0.1)
> > >
> >
>
lrate<-c(0.2,0.2,0.2,0.2,0.2,0.2,0.2,0.2,0.9,0.9,0.9,0.9,0.9,0.9,0.9,0.9)
> > >
> >
>
cycl<-c(5000,5000,5000,5000,10000,10000,10000,10000,5000,5000,5000,5000,10000,10000,10000,10000)
> > > yield<-c(0.26, 1.77, 0.29, 1.75, 0.26, 1.77,
> 0.31,
> > > 1.75, 0.29, 1.78, 0.35, 1.83, 0.29, 1.79, 0.35,
> > 1.83)
> > > 
> > >     npk <- data.frame(H=factor(hidden),
> > > T=factor(theres),
> > >                        L=factor(lrate),
> > > C=factor(cycl), yield=yield)
> > > 
> > >      ( npk.aov <- aov(yield ~ H*T*L*C, npk) )
> > >      summary(npk.aov)
> > > 
> > > when I run it, I got the following results: but
> > there
> > > is no any F values and p-values. Can you please
> > advise
> > > me what I missed? Many thanks in advance, dechao
> > 
> > 
> > Your model is much to complex for only 16
> > observations.
> > 
> > The following e.g. works:
> > 
> >  tmp <- lm(yield ~ H+T+L+C, npk)
> > > summary(tmp)
> > 
> > Call:
> > lm(formula = yield ~ H + T + L + C, data = npk)
> > 
> > Residuals:
> >       Min        1Q    Median        3Q       Max 
> > -0.029375 -0.009687 -0.001250  0.011563  0.025625 
> > 
> > Coefficients:
> >             Estimate Std. Error t value Pr(>|t|)  
>  
> > (Intercept) 0.260625   0.010618  24.545 5.88e-11
> ***
> > H10         0.031250   0.009497   3.290 0.007200
> ** 
> > T0.1        1.483750   0.009497 156.228  < 2e-16
> ***
> > L0.9        0.043750   0.009497   4.607 0.000757
> ***
> > C10000      0.003750   0.009497   0.395 0.700503  
>  
> > ---
> > Signif. codes:  0 `***' 0.001 `**' 0.01 `*' 0.05
> `.'
> > 0.1 ` ' 1 
> > 
> 
> 
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