From: James Salsman <james_at_bovik.org>

Date: Wed 15 Jun 2005 - 01:08:18 EST

Error in "names<-.default"(`*tmp*`, value = c("I(decade)", "I(decade^2)", :

>>>> main="average yearly inflation-adjusted dollar cost of extreme weather

*>>>> events worldwide")
*

>>>> Call:

>>>> lm(formula = billions ~ poly(decade, 3))

*>>>> 0.2357 -0.9429 1.4143 -0.9429 0.2357
*

*>>>> ---
*

*>>>> Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
*

*>>>> F-statistic: 77.68 on 3 and 1 DF, p-value: 0.08317
*

R-help@stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Wed Jun 15 01:12:31 2005

Date: Wed 15 Jun 2005 - 01:08:18 EST

What I really want now are the 95% confidence intervals that I mistakenly thought that linesHyperb.lm() would provide:

> pm <- lm(billions ~ I(decade) + I(decade^2) + I(decade^3)) > library(sfsmisc) > linesHyperb.lm(pm)

Error in "names<-.default"(`*tmp*`, value = c("I(decade)", "I(decade^2)", :

names attribute [3] must be the same length as the vector [1]
> pm <- lm(billions ~ poly(decade, 3))

> linesHyperb.lm(pm)

Warning message:

'newdata' had 100 rows but variable(s) found have 5 rows

Shouldn't curve(predict(...), add=TRUE) be able to plot confidence interval bands?

Prof Brian Ripley wrote:

> Why do `people' need `to deal with' these, anyway. We have machines to > do that.

Getting a 0.98 adjusted R^2 on the first try, compels me to try to publish the fitted formula.

>>>>> decade <- c(1950, 1960, 1970, 1980, 1990) >>>>> billions <- c(3.5, 5, 7.5, 13, 40) >>>>> # source: http://www.ipcc.ch/present/graphics/2001syr/large/08.17.jpg >>>>> >>>>> pm <- lm(billions ~ poly(decade, 3)) >>>>> >>>>> plot(decade, billions, xlim=c(1950,2050), ylim=c(0,1000),

>>>> main="average yearly inflation-adjusted dollar cost of extreme weather

>>>>> curve(predict(pm, data.frame(decade=x)), add=TRUE) >>>>> # output: http://www.bovik.org/storms.gif >>>>> >>>>> summary(pm) >>>>

>>>> Call:

>>>> lm(formula = billions ~ poly(decade, 3))

>>>>

>>>> Residuals:

>>>> 1 2 3 4 5

>>>>

>>>> Coefficients:

>>>> Estimate Std. Error t value Pr(>|t|)>>>> (Intercept) 13.800 0.882 15.647 0.0406 *>>>> poly(decade, 3)1 25.614 1.972 12.988 0.0489 *>>>> poly(decade, 3)2 14.432 1.972 7.318 0.0865 .>>>> poly(decade, 3)3 6.483 1.972 3.287 0.1880

>>>>

>>>> Residual standard error: 1.972 on 1 degrees of freedom

>>>> Multiple R-Squared: 0.9957, Adjusted R-squared: 0.9829

R-help@stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Wed Jun 15 01:12:31 2005

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