# [R] predict problem

Prof Brian Ripley ripley at stats.ox.ac.uk
Mon Apr 9 16:05:25 CEST 2001

```On Mon, 9 Apr 2001, Troels Ring wrote:

> Windows 98
> R : Copyright 2001, The R Development Core Team
> Version 1.2.1 (2001-01-15)
> Dear friends.
> How comes this works and produce a single prediction:
> x <- rnorm(15)
> y <- x + rnorm(15)
> predict(lm(y ~ x))
> new <- data.frame(x = seq(-3, 3, 0.5))
> predict(lm(y ~ x), new, se.fit = TRUE)
> pred.w.plim <- predict(lm(y ~ x), new, interval="confidence")
> new1 <- data.frame(x=3)
> predict(lm(y ~ x), new1, interval="confidence")
>
> while this refuses to take the "new" and predict ?
> lot <- c(30,20,60,80,40,50,60,30,70,60)
> hours <- c(73,50,128,170,87,108,135,69,148,132)
> z1 <- lm(hours~lot)
> new <- data.frame(x=80)
> predict(z1,new,interval="confidence",level=90)

There's no variable called `lot' in your new data frame.

new <- data.frame(lot=80)

And you want a 90% not a 90 confidence interval:

predict(z1,new,interval="confidence",level=0.9)
fit      lwr      upr
[1,] 170 166.9245 173.0755

Now, did you really want a confidence and not prediction interval?

--
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272860 (secr)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

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```