> predict(s.lm,data.frame(x=30000),interval="prediction")
fit lwr upr
[1,] 16073985 -9981352 42129323
> predict(s.lm,data.frame(x=30000),interval="confidence")
fit lwr upr
[1,] 16073985 5978125 26169846
On 15/09/06, Sachin J <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> 1. How do I construct 95% prediction interval for new x values, for
> example - x = 30000?
> 2. How do I construct 95% confidence interval?
>
> my dataframe is as follows :
>
> >dt
>
> structure(list(y = c(26100000,
> 60500000, 16200000, 30700000, 70100000, 57700000, 46700000, 8600000,
> 10000000, 61800000, 30200000, 52200000, 71900000, 55000000, 12700000
> ), x = c(108000, 136000, 35000,
> 77000, 178000, 150000, 126000, 24000, 28000, 214000, 108000,
> 190000, 308000, 252000, 71000)), .Names = c("y",
> "x"), class = "data.frame", row.names = c("1",
> "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13",
> "14", "15"))
>
> my regression eqn is as below:
>
> > s.lm <- lm(y ~ x)
>
> Thanks in advance.
>
>
> ---------------------------------
>
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>
> ______________________________________________
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> PLEASE do read the posting guide
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> and provide commented, minimal, self-contained, reproducible code.
>
--
=================================
David Barron
Said Business School
University of Oxford
Park End Street
Oxford OX1 1HP
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