Hi:

On Wed, Jul 21, 2010 at 8:46 AM, vijaysheegi <vijay.sheegi@gmail.com> wrote:

>
> Hi R-community,
> I have the code as follows,i Fitted model as follows
> lbeer<-log(beer_monthly)
> t<-seq(1956,1995.2,length=length(beer_monthly)) #beer_monthly contains 400+
> entries
>

This is unnecessary:

> t2=t^2
> beer_fit_parabola=lm(lbeer~t+t2)
>

Try

beer_fit_parabola <- lm(lbeer ~ poly(t, 2))

instead. poly() is a safer function to use for polynomial regression. See
?poly for the reasons why.

>
> Below is not working for me.
> Please help me in preparing the new data set for the below prediction
>
>
> tpred <- data.frame(t = seq(1995, 1998, length = 20))
predict(beer_fit_parabola, newdata = tpred)

If that doesn't work, try putting lbeer and t in a data frame and use it as
the data = argument of lm() [same model]. Then try predict() again with the
same code as above. This shouldn't be necessary, though.

Here's a quick made up example:

x <- 1:10
y <- 3 + 2 * x - x^2 + rnorm(10)
plot(x, y)
mm <- lm(y ~ poly(x, 2))
predict(mm, newdata = data.frame(x = c(11, 12)))
         1          2
 -96.10648 -117.04534

If you look at summary(mm), you'll see why polynomial models are not very
useful for prediction, particularly when extrapolating outside the range of
the observed data. (Compare the parameter estimates to the function that
produced y as a crude check.) A better choice might be a natural or
restricted cubic spline to reduce the standard error of prediction.

HTH,
Dennis

predict(beer_fit_parabola,newdata=data.frame(t=seq(1995,1998,length=20),t2=seq(1995,1998,length=20))
>
> #it is listing all 400+ entries ,but not 20 ahead prediction.
>
> Thanks In advance for your help
> --
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