David,
Thanks for the quick reply.
Just confirming, does predict(s.lm,data.frame(x=30000),interval="prediction")
gives prediction interval or tolerance interval?
Thanks
Sachin
David Barron <[EMAIL PROTECTED]> wrote:
> 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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--
=================================
David Barron
Said Business School
University of Oxford
Park End Street
Oxford OX1 1HP
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