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) best wishes Troels -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
predict problem
6 messages · Troels Ring, Uwe Ligges, Brian Ripley +1 more
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)
What you want is level=0.9 ??? Uwe Ligges -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
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 -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
You need to change "new <- data.frame(x=80)" to read "new <- data.frame(lot=80)" Alan On Mon, 09 Apr 2001 15:43:13 +0200 Troels Ring
<tring at gvdnet.dk> 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) best wishes Troels -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
---------------------- Alan T. Arnholt arnholt at math.appstate.edu -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
On Mon, 09 Apr 2001 15:43:13 +0200 Troels Ring
<tring at gvdnet.dk> wrote:
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)
you will also need to change your level to read .90 versus 90
predict.lm(z1,newdata=new,interval="confidence",level=.90)
fit lwr upr [1,] 170 166.9245 173.0755 Alan ---------------------- Alan T. Arnholt arnholt at math.appstate.edu -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
Faster than the wind several people spotted the errors
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)
new <- data.frame(lot=80) and predict(z1,new,interval="confidence",level=.90)
predict(z1,new,interval="confidence",level=90)
Thank you for your efforts and kindness ! Troels -------------- next part -------------- An HTML attachment was scrubbed... URL: https://stat.ethz.ch/pipermail/r-help/attachments/20010409/73b61a36/attachment.html