[R] Using predict()?

[Daniel Nordlund]

> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf
> Of Uwe Ligges
> Sent: Wednesday, December 12, 2007 5:14 AM
> To: Zembower, Kevin
> Cc: r-help at r-project.org
> Subject: Re: [R] Using predict()?
> 
> 
> 
> Zembower, Kevin wrote:
> > I'm trying to solve a homework problem using R. The problem gives a list
> > of cricket chirps per second and corresponding temperature, and asks to
> > give the equation for the linear model and then predict the temperature
> > to produce 18 chirps per second. So far, I have:
> >
> >> # Homework 11.2.1 and 11.3.3
> >> chirps <- scan()
> > 1: 20
> > 2: 16
> > 3: 19.8
> > 4: 18.4
> > 5: 17.1
> > 6: 15.5
> > 7: 14.7
> > 8: 17.1
> > 9: 15.4
> > 10: 16.2
> > 11: 15
> > 12: 17.2
> > 13: 16
> > 14: 17
> > 15: 14.4
> > 16:
> > Read 15 items
> >> temp <- scan()
> > 1: 88.6
> > 2: 71.6
> > 3: 93.3
> > 4: 84.3
> > 5: 80.6
> > 6: 75.2
> > 7: 69.7
> > 8: 82
> > 9: 69.4
> > 10: 83.3
> > 11: 79.6
> > 12: 82.5
> > 13: 80.6
> > 14: 83.5
> > 15: 76.3
> > 16:
> > Read 15 items
> >> chirps
> >  [1] 20.0 16.0 19.8 18.4 17.1 15.5 14.7 17.1 15.4 16.2 15.0 17.2 16.0
> > 17.0 14.4
> >> temp
> >  [1] 88.6 71.6 93.3 84.3 80.6 75.2 69.7 82.0 69.4 83.3 79.6 82.5 80.6
> > 83.5 76.3
> >> chirps.res <- lm(chirps ~ temp)
> >> summary(chirps.res)
> >
> > Call:
> > lm(formula = chirps ~ temp)
> >
> > Residuals:
> >      Min       1Q   Median       3Q      Max
> > -1.56146 -0.58088  0.02972  0.58807  1.53047
> >
> > Coefficients:
> >             Estimate Std. Error t value Pr(>|t|)
> > (Intercept) -0.31433    3.10963  -0.101 0.921028
> > temp         0.21201    0.03873   5.474 0.000107 ***
> > ---
> > Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
> >
> > Residual standard error: 0.9715 on 13 degrees of freedom
> > Multiple R-Squared: 0.6975,     Adjusted R-squared: 0.6742
> > F-statistic: 29.97 on 1 and 13 DF,  p-value: 0.0001067
> >> # From the linear model summary output above, the equation for the
> > least squares line is:
> >> #    y = -0.3143 + 0.2120*x or chirps = -0.3143 + 0.2120*temp
> >>
> >
> > I can then determine the answer to the prediction, using algebra and R:
> >> pred_temp <- (18+0.3143)/0.2120
> >> pred_temp
> > [1] 86.3882
> >
> > However, I'd like to try to use the predict() function. Since 'chirps'
> > and 'temp' are just vectors of numbers, and not dataframes, these
> > failed:
> > predict(chirps.res, newdata=data.frame(chirp=18))
> > predict(chirps.res, newdata="chirp=18")
> > predict(chirps.res, newdata=18)
> >
> 
> 
> Well, "chirps" (not "chirp", BTW) was your *response* in the lm() call!
> Your new data has to be called "temp", otherwise either the regression
> did not make sense or you are confusing different things.
> 
> Best,
> Uwe Ligges
> 
> > Baltimore, Maryland  21202
> > 410-659-6139
> >

Kevin,

To add to what Uwe Ligges wrote, your model is modeling chirps as a function of temp.  It looks like you are trying to turn that around and use chirps to predict a temp.  predict() won't do that.  Check any introductory regression text of why you probably don't want to do that anyway.  If you want to predict temp from chirps you should run

temp.lm <- lm(temp ~ chirps)
predict(temp.lm, newdata=data.frame(chirps=18))

The slope and intercept will be different from that found in chirps.res (in the absence of a perfect correlation between chirps and temp).

Hope this is helpful,

Dan

Daniel Nordlund
Bothell, WA