# Re: [R] prediction interval for new value

From: Sachin J <sachinj.2006_at_yahoo.com>
Date: Fri 15 Sep 2006 - 19:51:46 GMT

David,

Just confirming, does predict(s.lm,data.frame(x=30000),interval="prediction") gives prediction interval or tolerance interval?

Thanks
Sachin

> 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 <sachinj.2006@yahoo.com> 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:

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David Barron
University of Oxford
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Oxford OX1 1HP

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