[R] need 95% confidence interval bands on cubic extrapolation
Marc Schwartz (via MN)
mschwartz at mn.rr.com
Tue Dec 20 23:29:30 CET 2005
On Tue, 2005-12-20 at 13:04 -0800, James Salsman wrote:
> Dear R experts:
>
> I need to get this plot, but also with 95% confidence interval bands:
>
> hour <- c(1, 2, 3, 4, 5, 6)
> millivolts <- c(3.5, 5, 7.5, 13, 40, 58)
>
> plot(hour, millivolts, xlim=c(1,10), ylim=c(0,1000))
>
> pm <- lm(millivolts ~ poly(hour, 3))
>
> curve(predict(pm, data.frame(hour=x)), add=TRUE)
>
> How can the 95% confidence interval band curves be plotted too?
>
> Sincerely,
> James Salsman
>
> P.S. I know I should be using data frames instead of parallel lists.
> This is just a simple example.
There is an example in ?predict.lm.
Given your data, something like the following will work:
hour <- c(1, 2, 3, 4, 5, 6)
millivolts <- c(3.5, 5, 7.5, 13, 40, 58)
pm <- lm(millivolts ~ poly(hour, 3))
# Now create a new dataset with an interval
# of hours that fits your data above
# This is then used in predict.lm() below
# Smaller increments will create smoother lines in the plot
new <- data.frame(hour = seq(1, 6, 0.5))
# Use the new data and generate confidence intervals
# based upon the model
clim <- predict(pm, new, interval = "confidence")
> clim
fit lwr upr
1 4.400794 -17.659582 26.46117
2 2.879712 -12.954245 18.71367
3 2.817460 -14.317443 19.95236
4 4.252232 -12.822969 21.32743
5 7.222222 -8.051125 22.49557
6 11.765625 -2.374270 25.90552
7 17.920635 2.647288 33.19398
8 25.725446 8.650246 42.80065
9 35.218254 18.083351 52.35316
10 46.437252 30.603295 62.27121
11 59.420635 37.360259 81.48101
# Now use matplot to draw the fitted line (black)
# and the CI's (red)
matplot(new$hour, clim,
lty = c(1, 2, 2),
col = c("black", "red", "red"),
type = "l", ylab = "predicted y")
See ?predict.lm and ?matplot for more information.
HTH,
Marc Schwartz
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