A sample R session
18 A sample R session
18.1 A sample session involving regression
For illustrative purposes, a sample
R session may look like the
following. The graphs are not presented to save space and to
encourage the reader to try the problems themself.
Assignment: Explore mtcars.
Here are some things
to do:

Start R.
 First we need to start R. Under Windows this
involves finding the icon and clicking it. Wait a few seconds and
the R application takes over. If you are at a UNIX command
prompt, then typing R will start the R program. If you are using
XEmacs and
ESS then you start XEmacs and then
enter the command Mx R. (That is Alt and x at the same
time, and then R and then enter. Other methods may be applicable to
your case.
 Load the dataset mtcars.
 This dataset is builtin to
R. To access it, we need to tell R we want it. Here is how to do
so, and how to find out the names of the variables. Use
?mtcars to read the documentation on the data set.
> data(mtcars)
> names(mtcars)
[1] "mpg" "cyl" "disp" "hp" "drat" "wt" "qsec" "vs" "am" "gear"
[11] "carb"
 Access the data.
 The data is in a data frame. This makes
it easy to access the data. As a summary, we could access the ``miles
per gallon data'' (mpg) lots of ways. Here are a few: Using
$ as with mtcars$mpg; as a list element as in mtcars[['mpg']]; or as the first column of data using
mtcars[,1]. However, the preferred method is to
attach the dataset to your environment. Then the data is
directly accessible as this illustrates
> mpg # not currently visible
Error: Object "mpg" not found
> attach(mtcars) # attach the mtcars names
> mpg # now it is visible
[1] 21.0 21.0 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 17.8 16.4 17.3 15.2 10.4
[16] 10.4 14.7 32.4 30.4 33.9 21.5 15.5 15.2 13.3 19.2 27.3 26.0 30.4 15.8 19.7
[31] 15.0 21.4
 Categorical Data.
 The value of cylinder is categorical. We can
use table to summarize, and barplot to view the
values. Here is how
> table(cyl)
cyl
4 6 8
11 7 14
> barplot(cyl) # not what you want
> barplot(table(cyl))
If you do so, you will see that for this data 8 cylinder cars are
more common. (This is 1974 car data. Read
more with the help command: help(mtcars) or ?mtcars.)
 Numerical Data.
 The miles per gallon is numeric. What is the
general shape of the distribution? We can view this with a
stem and leaf chart, a histogram, or a boxplot. Here are commands to
do so
> stem(mpg)
The decimal point is at the 
10  44
12  3
14  3702258
16  438
18  17227
20  00445
22  88
24  4
26  03
28 
30  44
32  49
> hist(mpg)
> boxplot(mpg)
From the graphs (in particular the histogram) we can see the miles
per gallon are pretty low. What are the summary statistics including
the mean? (This stem graph is a bit confusing. 33.9, the max value,
reads like 32.9. Using a different scale is better as in stem(mpg,scale=3).)
> mean(mpg)
[1] 20.09062
> mean(mpg,trim=.1) # trim off 10 percent from top, bottom
[1] 19.69615 # for a more resistant measure
> summary(mpg)
Min. 1st Qu. Median Mean 3rd Qu. Max.
10.40 15.43 19.20 20.09 22.80 33.90
So half of the cars get 19.20 miles per gallon or less. What is the
variability or spread? There are several ways to measure this: the standard deviation or the
IQR or mad (the median absolute deviation). Here are all three
> sd(mpg)
[1] 6.026948
> IQR(mpg)
[1] 7.375
> mad(mpg)
[1] 5.41149
They are all different, but measure approximately the same thing 
spread.
 Subsets of the data.

What about the average mpg for cars that have just 4 cylinders? This
can be answered with the mean function as well, but first we
need a subset of the mpg vector corresponding to just the 4
cylinder cars. There is an easy way to do so.
> mpg[cyl == 4]
[1] 22.8 24.4 22.8 32.4 30.4 33.9 21.5 27.3 26.0 30.4 21.4
> mean(mpg[cyl == 4])
[1] 26.66364
Read this like a sentence  ``the miles per gallon of the cars with
cylinders equal to 4''. Just remember ``equals'' is == and
not simply =, and functions use parentheses while accessing data uses
square brackets..
 Bivariate relationships.
 The univariate data on miles per gallon
is interesting, but of course we expect there to be some
relationship with the size of the engine. The engine size is stored
in various ways: with the cylinder size, or the horsepower or even
the displacement. Let's view it two ways. First, cylinder size is a
discrete variable with just a few values, a scatterplot will produce an
interesting graph
> plot(cyl,mpg)
We see a decreasing trend as the number of cylinders increases, and
lots of variation between the different cars. We might be tempted to
fit a regression line. To do so is easy with the command
simple.lm which is a convenient front end to the lm
command. (You need to have loaded the Simple package prior to this.)
> simple.lm(cyl,mpg)
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
37.885 2.876
Which says the slope of the line is 2.876 which in practical terms
means if you step up to the next larger sized engine, your
m.p.g. drops by 5.752 on average.
What are the means for each cylinder size? We did this above for 4
cylinders. If we wanted do this for the 6 and 8 cylinder cars we
could simply replace the 4 in the line above with 6 or 8. If you
wanted a fancy way to do so, you can use tapply which will
apply a function (the mean) to a vector broken down by a
factor:
> tapply(mpg,cyl,mean)
4 6 8
26.66364 19.74286 15.10000
Next, lets investigate the relationship between the numeric variable
horsepower and miles per gallon. The same commands as above will
work, but the scatterplot will look different as horsepower is
essentially a continuous variable.
> simple.lm(hp,mpg)
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
30.09886 0.06823
The fit doesn't look quite as good. We can test this with the
correlation function cor
> cor(hp,mpg)
[1] 0.7761684
> cor(cyl,mpg)
[1] 0.852162
This is the Pearson correlation coefficient, R. Squaring it gives R^{2}.
> cor(hp,mpg)^2
[1] 0.6024373
> cor(cyl,mpg)^2
[1] 0.72618
The usual interpretation is that 72% of the variation is explained by
the linear relationship for the relationship between the number of
cylinders and the miles per gallon.
We can view all three variables together by using different plotting symbols
based on the number of cylinders. The argument pch controls
this as in
> plot(hp,mpg,pch=cyl)
You can add a legend with the legend command to tell the
reader what you did.
> legend(250,30,pch=c(4,6,8),
+ legend=c("4 cylinders","6 cylinders","8 cylinders"))
(Note the + indicates a continuation line.)
 Testing the regression assumptions.

In order to make statistical inferences about the regression line,
we need to ensure that the assumptions behind the statistical model
are appropriate. In this case, we want to check that the residuals
have no trends, and are normallu distributed. We can do
so graphically once we get our hands on the residuals. These are
available through the resid method for the result of an
lm usage.
> lm.res = simple.lm(hp,mpg)
> lm.resids = resid(lm.res) # the residuals as a vector
> plot(lm.resids) # look for change in spread
> hist(lm.resids) # is data bell shaped?
> qqnorm(lm.resids) # is data on straight line?
From the first plot we see that the assumptions are suspicious as
the residuals are basically negative for a while and then they are
mostly positive. This is an indication that the straight line model
is not a good one.
 Clean up.
 There is certainly more to do with this, but not here.
Let's take a break. When leaving an example, you should detach any
data frames. As well, we will quit our R session. Notice you will
be prompted to save your session. If you choose yes, then the next
time you start R in this directory, your session (variables,
functions etc.) will be
restored.
> detach() # clear up namespace
> q() # Notice the parentheses!
The
ttests are the standard test for making statistical
inferences about the center of a dataset.
Assignment: Explore chickwts using ttests.

Start up.
 We start up R as before. If you started in the
same directory, your previous work has been saved. How can you
tell? Try the ls() command to see what is available.
 Attach the data set.
 First we load in the data and attach the
data set
> data(chickwts)
> attach(chickwts)
> names(chickwts)
[1] "weight" "feed"
 EDA
 Let's see what we have. The data is stored in
two columns. The weight and the level of the factor feed
is given for each chick. A boxplot is a good place to look. For
data presented this way, the goal is to make a separate boxplot
for each level of the factor feed. This is done using the
model formula notation of R.
> boxplot(weight ~ feed)
We see that ``casein'' appears to be better than the others. As a
naive check, we
can test its mean against the mean weight.
> our.mu = mean(weight)
> just.casein = weight[feed == 'casein']
> t.test(just.casein,mu = our.mu)
One Sample ttest
data: just.casein
t = 3.348, df = 11, pvalue = 0.006501
alternative hypothesis: true mean is not equal to 261.3099
95 percent confidence interval:
282.6440 364.5226
sample estimates:
mean of x
323.5833
The low pvalue of 0.006501 indicates that the mean weight for the chicks
fed 'casein' is more than the average weight.
The 'sunflower' feed also looks higher. Is it similar to the
'casein' feed? A twosample ttest will tell us
> t.test(weight[feed == 'casein'],weight[feed == 'sunflower'])
Welch Two Sample ttest
data: weight[feed == "casein"] and weight[feed == "sunflower"]
t = 0.2285, df = 20.502, pvalue = 0.8215
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
53.94204 43.27538
sample estimates:
mean of x mean of y
323.5833 328.9167
Notice the pvalue now is 0.8215 which indicates that the null
hypothesis should be accepted. That is, there is no indication
that the mean weights are not the same.
The feeds 'linseed' and 'soybean' appear to be the same, and have
the same spread. We can test for the equivalence of the mean, and
in addition use the pooled estimate for the standard
deviation. This is done as follows using var.equal=TRUE
> t.test(weight[feed == 'linseed'],weight[feed == 'soybean'],
+ var.equal=TRUE)
Two Sample ttest
data: weight[feed == "linseed"] and weight[feed == "soybean"]
t = 1.3208, df = 24, pvalue = 0.1990
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
70.92996 15.57282
sample estimates:
mean of x mean of y
218.7500 246.4286
The null hypothesis of equal means is accepted as the pvalue is
not too small.
 Close up.
 Again, lets close things down just for practice.
> detach() # clear up namespace
> q() # Notice the parentheses!
18.3 A simulation example
The
tstatistic for a data sets
X_{1},
X_{2},...,
X_{n} is
where
X^{} is the sample mean and
s is the sample standard deviation. If the
X_{i}'s are normal, the distribution is the
tdistribution. What if
the
X_{i} are not normal?
Assignment: How robust is the
distribution of t to changes in the distribution of X_{i}?
This is the job of simulation. We use the computer to generate random
data and investigate the results. Suppose our
R session is already
running and just want to get to work.
First, let's define a useful function to create the
t statistic for
a data set.
> make.t = function(x,mu) (mean(x)mu)/( sqrt(var(x)/length(x)))
Now, when we want to make the
tstatistic we simply use this function.
> mu = 1;x=rnorm(100,mu,1)
> make.t(x,mu)
[1] 1.552077
That shows the
t statistic for one random sample of size 100 from
the normal distribution with mean 1 and standard deviation 1.
We need to use a different distribution, and create lots of random
samples so that we can investigate the distribution of the
t
statistic. Here is how to create samples when the
X_{i} are
exponential
> mu = 10;x=rexp(100,1/mu);make.t(x,mu)
[1] 1.737937
Now, we need to create lots of such samples and store them
somewhere. We use a for loop, but first we define a variable to store
our data
> results = c() # initialize the results vector
> for (i in 1:200) results[i] = make.t(rexp(100,1/mu),mu)
That's it. Our numbers are now stored in the variable
results.
We could have spread this out over a few lines, but instead we
combined a few functions together. Now we can view the distribution
using graphical tools such as histograms, boxplots and probability
plots.
> hist(res) # histogram looks bell shaped
> boxplot(res) # symmetric, not longtailed
> qqnorm(res) # looks normal
When
n=100, the data looks approximately normal. Which is good as the
tdistribution does too.
What about if
n is small? Then the
tdistribution has
n1
degrees of freedom and is longtailed. Will this change things? Let's
see with
n=8.
> for (i in 1:200) res[i] = make.t(rexp(8,1/mu),mu)
> hist(res) # histogram is not bell shaped
> boxplot(res) # asymmetric, longtailed
> qqnorm(res) # not close to $t$ or normal.
We see a marked departure from a symmetric distribution. We conclude
that the
t is not this robust.
So, if the underlying distribution is skewed and
n is small, the
t
distribution is not robust, what about if the underlying distribution
is symmetric, but longtailed? Let's think of a random distribution
that is like this? Hmmm, how about the
t distribution with a few
degrees of freedom? Let's see. We will look at the shape of the
tstatistic with
n=8 (7 degrees of freedom) when the underlying
distribution is the
tdistribution with 5 degrees of freedom.
> for (i in 1:200) res[i] = make.t(rt(8,5),0)
> hist(res) # histogram is bell shaped
> boxplot(res) # symmetric, longtailed
> qqnorm(res) # not close to normal.
> qqplot(res,rt(200,7) # close to t with 7 degrees of freedom
We see a symmetric, longtailed distribution which is not normal, but
is close to the
tdistribution with 7 degrees of freedom. We
conclude that the
tstatistic is robust to this amount of change in
the tails of the underlying population distribution.
Copyright © John Verzani, 20012. All rights reserved.