> -----Original Message----- > From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf > Of Uwe Ligges > Sent: Wednesday, December 12, 2007 5:14 AM > To: Zembower, Kevin > Cc: r-help@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 ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
