In Part 3 we used the lm() command to perform least squares regressions. In Part 4 we will look at more advanced aspects of regression models and see what R has to offer.
One way of checking for non-linearity in your data is to fit a polynomial model and check whether the polynomial model fits the data better than a linear model. However, you may also wish to fit a quadratic or higher model because you have reason to believe that the relationship between the variables is inherently polynomial in nature. Let’s see how to fit a quadratic model in R.
We will use a data set of counts of a variable that is decreasing over time. Cut and paste the following data into your R workspace.
A <- structure(list(Time = c(0, 1, 2, 4, 6, 8, 9, 10, 11, 12, 13,
14, 15, 16, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 29, 30),
Counts = c(126.6, 101.8, 71.6, 101.6, 68.1, 62.9, 45.5, 41.9,
46.3, 34.1, 38.2, 41.7, 24.7, 41.5, 36.6, 19.6,
22.8, 29.6, 23.5, 15.3, 13.4, 26.8, 9.8, 18.8, 25.9, 19.3)), .Names = c("Time", "Counts"),
row.names = c(1L, 2L, 3L, 5L, 7L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 19L, 20L, 21L, 22L, 23L, 25L, 26L, 27L, 28L, 29L, 30L, 31L),
class = "data.frame")
Let’s attach the entire dataset so that we can refer to all variables directly by name.
attach(A)
names(A)
First, let’s set up a linear model, though really we should plot first and only then perform the regression.
linear.model <-lm(Counts ~ Time)
We now obtain detailed information on our regression through the summary() command.
summary(linear.model)
Call:
lm(formula = Counts ~ Time)
Residuals:
Min 1Q Median 3Q Max
-20.084 -9.875 -1.882 8.494 39.445
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 87.1550 6.0186 14.481 2.33e-13 ***
Time -2.8247 0.3318 -8.513 1.03e-08 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 15.16 on 24 degrees of freedom
Multiple R-squared: 0.7512, Adjusted R-squared: 0.7408
F-statistic: 72.47 on 1 and 24 DF, p-value: 1.033e-08
The model explains over 74% of the variance and has highly significant coefficients for the intercept and the independent variable and also a highly significant overall model p-value. However, let’s plot the counts over time and superpose our linear model.
plot(Time, Counts, pch=16, ylab = "Counts ", cex.lab = 1.3, col = "red" )
abline(lm(Counts ~ Time), col = "blue")
Here the syntax cex.lab = 1.3 produced axis labels of a nice size.
The model looks good, but we can see that the plot has curvature that is not explained well by a linear model. Now we fit a model that is quadratic in time. We create a variable called Time2 which is the square of the variable Time.
Time2 <- Time^2
quadratic.model <-lm(Counts ~ Time + Time2)
Note the syntax involved in fitting a linear model with two or more predictors. We include each predictor and put a plus sign between them.
summary(quadratic.model)
Call:
lm(formula = Counts ~ Time + Time2)
Residuals:
Min 1Q Median 3Q Max
-24.2649 -4.9206 -0.9519 5.5860 18.7728
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 110.10749 5.48026 20.092 4.38e-16 ***
Time -7.42253 0.80583 -9.211 3.52e-09 ***
Time2 0.15061 0.02545 5.917 4.95e-06 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 9.754 on 23 degrees of freedom
Multiple R-squared: 0.9014, Adjusted R-squared: 0.8928
F-statistic: 105.1 on 2 and 23 DF, p-value: 2.701e-12
Our quadratic model is essentially a linear model in two variables, one of which is the square of the other. We see that however good the linear model was, a quadratic model performs even better, explaining an additional 15% of the variance. Now let’s plot the quadratic model by setting up a grid of time values running from 0 to 30 seconds in increments of 0.1s.
timevalues <- seq(0, 30, 0.1)
predictedcounts <- predict(quadratic.model,list(Time=timevalues, Time2=timevalues^2))
plot(Time, Counts, pch=16, xlab = "Time (s)", ylab = "Counts", cex.lab = 1.3, col = "blue")
Now we include the quadratic model to the plot using the lines() command.
lines(timevalues, predictedcounts, col = "darkgreen", lwd = 3)
The quadratic model appears to fit the data better than the linear model. We will look again at fitting curved models in our next blog post.
About the Author: David Lillis has taught R to many researchers and statisticians. His company, Sigma Statistics and Research Limited, provides both on-line instruction and face-to-face workshops on R, and coding services in R. David holds a doctorate in applied statistics.
See our full R Tutorial Series and other blog posts regarding R programming.
We were recently fortunate to host a free The Craft of Statistical Analysis Webinar with guest presenter David Lillis. As usual, we had hundreds of attendees and didn’t get through all the questions. So David has graciously agreed to answer questions here.
If you missed the live webinar, you can download the recording here: Ten Data Analysis Tips in R.
Q: Is the M=structure(.list(.., class = “data.frame) the same as M=data.frame(..)? Is there some reason to prefer to use structure(list, … ,) as opposed to M=data.frame?
A: They are not the same. The structure( .. .) syntax is a short-hand way of storing a data set. If you have a data set called M, then the command dput(M) provides a shorthand way of storing the dataset. You can then reconstitute it later as follows: M <- structure( . . . .). Try it for yourselves on a rectangular dataset. For example, start off with (more…)
In Part 2 of this series, we created two variables and used the lm() command to perform a least squares regression on them, treating one of them as the dependent variable and the other as the independent variable. Here they are again.
height = c(176, 154, 138, 196, 132, 176, 181, 169, 150, 175)
bodymass = c(82, 49, 53, 112, 47, 69, 77, 71, 62, 78)
Today we learn how to obtain useful diagnostic information about a regression model and then how to draw residuals on a plot. As before, we perform the regression.
lm(height ~ bodymass)
Call:
lm(formula = height ~ bodymass)
Coefficients:
(Intercept) bodymass
98.0054 0.9528
Now let’s find out more about the regression. First, let’s store the regression model as an object called mod and then use the summary() command to learn about the regression.
mod <- lm(height ~ bodymass)
summary(mod)
Here is what R gives you.
Call:
lm(formula = height ~ bodymass)
Residuals:
Min 1Q Median 3Q Max
-10.786 -8.307 1.272 7.818 12.253
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 98.0054 11.7053 8.373 3.14e-05 ***
bodymass 0.9528 0.1618 5.889 0.000366 ***
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 9.358 on 8 degrees of freedom
Multiple R-squared: 0.8126, Adjusted R-squared: 0.7891
F-statistic: 34.68 on 1 and 8 DF, p-value: 0.0003662
R has given you a great deal of diagnostic information about the regression. The most useful of this information are the coefficients themselves, the Adjusted R-squared, the F-statistic and the p-value for the model.
Now let’s use R’s predict() command to create a vector of fitted values.
regmodel <- predict(lm(height ~ bodymass))
regmodel
Here are the fitted values:
1 2 3 4 5 6 7 8 9 10
176.1334 144.6916 148.5027 204.7167 142.7861 163.7472 171.3695 165.6528 157.0778 172.3222
Now let’s plot the data and regression line again.
plot(bodymass, height, pch = 16, cex = 1.3, col = "blue", main = "HEIGHT PLOTTED AGAINST BODY MASS", xlab = "BODY MASS (kg)", ylab = "HEIGHT (cm)")
abline(lm(height ~ bodymass))
We can plot the residuals using R’s for loop and a subscript k that runs from 1 to the number of data points. We know that there are 10 data points, but if we do not know the number of data we can find it using the length() command on either the height or body mass variable.
npoints <- length(height)
npoints
[1] 10
Now let’s implement the loop and draw the residuals (the differences between the observed data and the corresponding fitted values) using the lines() command. Note the syntax we use to draw in the residuals.
for (k in 1: npoints) lines(c(bodymass[k], bodymass[k]), c(height[k], regmodel[k]))
Here is our plot, including the residuals.
In part 4 we will look at more advanced aspects of regression models and see what R has to offer.
About the Author: David Lillis has taught R to many researchers and statisticians. His company, Sigma Statistics and Research Limited, provides both on-line instruction and face-to-face workshops on R, and coding services in R. David holds a doctorate in applied statistics.
See our full R Tutorial Series and other blog posts regarding R programming.
In Part 1 we installed R and used it to create a variable and summarize it using a few simple commands. Today let’s re-create that variable and also create a second variable, and see what we can do with them.
As before, we take height to be a variable that describes the heights (in cm) of ten people. Type the following code to the R command line to create this variable.
height = c(176, 154, 138, 196, 132, 176, 181, 169, 150, 175)
Now let’s take bodymass to be a variable that describes the weight (in kg) of the same ten people. Copy and paste the following code to the R command line to create the bodymass variable.
bodymass = c(82, 49, 53, 112, 47, 69, 77, 71, 62, 78)
Both variables are now stored in the R workspace. To view them, enter:
height bodymass
We can now create a simple plot of the two variables as follows:
plot(bodymass, height)
However, this is a rather simple plot and we can embellish it a little. Type the following code into the R workspace:
plot(bodymass, height, pch = 16, cex = 1.3, col = "red", main = "MY FIRST PLOT USING R", xlab = "Body Mass (kg)", ylab = "HEIGHT (cm)")
[Note: R is very picky about the quotation marks you use. If the font that is displaying this post shows the beginning and ending quotation marks as facing in different directions, it won’t work in R. They both have to look the same–just straight lines. You may have to retype them within R rather than cutting and pasting.]
In the above code, the syntax pch = 16 creates solid dots, while cex = 1.3 creates dots that are 1.3 times bigger than the default (where cex = 1). More about these commands later.
Now let’s perform a linear regression on the two variables by adding the following text at the command line:
lm(height~bodymass)
We see that the intercept is 98.0054 and the slope is 0.9528. By the way – lm stands for “linear model”.
Finally, we can add a best fit line to our plot by adding the following text at the command line:
abline(98.0054, 0.9528)
None of this was so difficult!
In Part 3 we will look again at regression and create more sophisticated plots.
About the Author: David Lillis has taught R to many researchers and statisticians. His company, Sigma Statistics and Research Limited, provides both on-line instruction and face-to-face workshops on R, and coding services in R. David holds a doctorate in applied statistics.
See our full R Tutorial Series and other blog posts regarding R programming.
