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  1. PO
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    <p>As pointed out above, this is a bit tricky for a multi-dimensional model.</p> <p>Get some data (you neglected to include a reproducible example: see <a href="http://tinyurl.com/reproducible-000" rel="nofollow noreferrer">http://tinyurl.com/reproducible-000</a> ...)</p> <pre><code>library(foreign) dat &lt;- read.arff(url("http://www.cs.umb.edu/~rickb/files/UCI/autos.arff")) </code></pre> <p>Split into training and test data sets:</p> <pre><code>train &lt;- dat[1:150,] test &lt;- dat[151:nrow(dat),] </code></pre> <p>The variable names are a bit awkward for R (the dashes are interpreted as minus operators, so we have to use back-quotes to protect the names):</p> <pre><code>fit &lt;- lm(`city-mpg` ~ `peak-rpm`+horsepower+`wheel-base`,data=train) temp_test &lt;- predict(fit,test) </code></pre> <p>Plot the predictions vs peak RPM:</p> <pre><code>par(las=1,bty="l") ## cosmetic plot(test[["peak-rpm"]],temp_test,xlab="peak rpm",ylab="predicted") </code></pre> <p>In order to add the line, we have to adjust the intercept according to some baseline values of the other parameters: we'll use the mean (another alternative is to center all the predictor variables before fitting the model):</p> <pre><code>cf &lt;- coef(fit) abline(a=cf["(Intercept)"]+ mean(test$horsepower)*cf["horsepower"]+ mean(test$`wheel-base`)*cf["`wheel-base`"], b=coef(fit)["`peak-rpm`"]) </code></pre> <p>Another way to do this is to use <code>predict()</code>:</p> <pre><code>newdat &lt;- with(test, data.frame(horsepower=mean(horsepower), "wheel-base"=mean(`wheel-base`), "peak-rpm"=seq(min(`peak-rpm`), max(`peak-rpm`), length=41), check.names=FALSE)) newdat["city-mpg"] &lt;- predict(fit,newdat) with(newdat,lines(`peak-rpm`,`city-mpg`,col=4)) </code></pre> <p>(41 points is silly for a straight line -- we could have used just 2 -- but will work well if you want to plot something curved, like confidence intervals or a nonlinear fit.)</p> <p>Alternatively you could just fit the marginal model, but the actual fitted line is somewhat different (it will only be the same if all the predictors are orthogonal to each other):</p> <pre><code>fit2 &lt;- lm(`city-mpg` ~ `peak-rpm`,data=train) abline(fit2,col="red") </code></pre> <p><img src="https://i.stack.imgur.com/Ty90y.png" alt="enter image description here"></p>
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    1. POAbline not working with Linear regression Model
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    1. USBen Bolker
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      1. VTUpMod
    1. COThanks alot for your response. But what does predicted value tells here ? Does it tells cityMpg value ??? And in the table '-' is not there, the words are combined, so I do not face any issue regaring the space. Can u tell what does plot(test[["peak-rpm"]],temp_test,xlab="peak rpm",ylab="predicted") represents, a graph of peak rpm and predicted value(what will be dis value)
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    2. CO(1) yes, the predicted value is the same as the response variable of the fitted model. (2) the plot puts peak RPM on the x axis and the predicted value of city MPG from the model on the y axis.
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    3. COOh, ok. Means if i want to predict cityMpg value from 3 attributes, then I cant do that. I can only predict its vaalue from one other attribute ?
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