Multiple Linear Regression
14 Multiple Linear Regression
Linear regression was used to model the effect one variable, an
explanatory variable
, has on another, the response variable.
In particular, if one variable changed by some amount, you assumed the
other changed by a multiple of that same amount. That multiple being
the slope of the regression line. Multiple
linear regression
does the same, only there are multiple
explanatory variables or regressors.
There are many situations where intuitively this is the correct
model. For example, the price of a new home depends on many factors  the
number of bedrooms, the number of bathrooms, the location of the
house, etc. When a house is built, it costs a certain amount for the
builder to build an extra room and so the cost of house reflects
this. In fact, in some new developments, there is a pricelist for
extra features such as $900 for an upgraded fireplace.
Now, if you are buying an older house it isn't so clear what the price
should be. However, people do develop rules of thumb to help figure
out the value. For example, these may be add $30,000 for an extra bedroom and $15,000 for an
extra bathroom, or
subtract $10,000 for the busy street. These are intuitive uses of a
linear model to explain the cost of a house based on several
variables. Similarly, you might develop such insights when buying a
used car, or a new computer. Linear regression is also used to predict
performance. If you were accepted to college, the college may have
used some formula to assess your application based on highschool GPA,
standardized test scores such as SAT, difficulty of highschool
curriculum, strength of your letters of recommendation, etc. These
factors all may predict potential performance. Despite there being no
obvious reason for a linear fit, the tools are easy to use and so may
be used in this manner.
14.1 The model
The standard regression model modeled the response variable
y_{i}
based on
x_{i} as
y_{i} = b_{0} + b_{1} x_{i} + e_{i}
where the
e are i.i.d. N(0,
s^{2}). The task at hand was
to estimate the parameters
b_{0},
b_{1},
s^{2} using the data
(
x_{i},
y_{i}). In multiple regression, there are potentially many
variables and each one needs one (or more) coefficients. Again we use
b but more subscripts. The basic model is
y_{i} = b_{0} + b_{1} x_{i1} + b_{2} x_{i2} + ··· + b_{p} x_{ip} + e_{i}
where the
e are as before. Notice, the subscript on the
x's involves the
ith sample and the number of the variable 1, 2,
..., or
p.
A few comments before continuing
The method of least squares is typically used to find the coefficients
b_{j},
j = 0,1,...,
p. As with simple regression, if we have estimated the
b's with
b's then the estimator for
y_{i} is


_{i} = b_{0} + b_{1} x_{i1} + ··· + b_{n} x_{ip}

and the residual
amount is
e_{i} =
y_{i} 
y^{^}_{i}. Again the
method of least squares finds the values for the
b's which minimize the
squared residuals,
å(
y_{i} 
y^{^}_{i})
^{2}.
If the model is appropriate for the data, statistical inference can be made as
before. The estimators,
b_{i} (also known as
b_{i}), have known standard
errors, and the distribution of (
b_{i} 
b_{i})/
SE(
b_{i}) will be the
tdistribution with
n(
p+1) degrees of freedom. Note, there are
p+1
terms estimated these being the values of
b_{0},
b_{1},...,
b_{p}.
Example: Multiple linear regression with known answer
Let's investigate the model and its implementation in
R with data
which we generate ourselves so we know the answer. We explicitly
define the two regressors, and then the response as a linear
function of the regressors with some normal noise added. Notice,
linear models are still solved with the
lm function, but we
need to recall a bit more about the
model formula syntax.
> x = 1:10
> y = sample(1:100,10)
> z = x+y # notice no error term  sigma = 0
> lm(z ~ x+y) # we use lm() as before
... # edit out Call:...
Coefficients:
(Intercept) x y
4.2e15 1.0e+00 1.0e+00
# model finds b_0 = 0, b_1 = 1, b_2 = 1 as expected
> z = x+y + rnorm(10,0,2) # now sigma = 2
> lm(z ~ x+y)
...
Coefficients:
(Intercept) x y
0.4694 0.9765 0.9891
# found b_0 = .4694, b_1 = 0.9765, b_2 = 0.9891
> z = x+y + rnorm(10,0,10) # more noise  sigma = 10
> lm(z ~ x+y)
...
Coefficients:
(Intercept) x y
10.5365 1.2127 0.7909
Notice that as we added more noise the guesses got worse and
worse as expected. Recall that the difference
between
b_{i} and
b_{i} is controlled by the standard error of
b_{i} and the standard deviation of
b_{i} (which the standard error
estimates) is related to
s^{2} the variance of the
e_{i}.
In short, the more noise the worse the confidence, the more data the
better the confidence.
The model formula syntax is pretty easy to use in this case. To add
another explanatory variable you just ``add'' it to the right side of
the formula. That is to add
y we use
z ~ x + y
instead of simply
z ~ x as in simple regression. If you
know for sure that there is no intercept term (
b_{0} = 0) as it is
above, you can explicitly remove this by adding 1 to the formula
> lm(z ~ x+y 1) # no intercept beta_0
...
Coefficients:
x y
2.2999 0.8442
Actually, the
lm command only returns the coefficients (and
the formula call) by default. The two methods
summary and
anova can yield more information. The output of
summary
is similar as for simple regression.
> summary(lm(z ~ x+y ))
Call:
lm(formula = z ~ x + y)
Residuals:
Min 1Q Median 3Q Max
16.793 4.536 1.236 7.756 14.845
Coefficients:
Estimate Std. Error t value Pr(>t)
(Intercept) 10.5365 8.6627 1.216 0.263287
x 1.2127 1.4546 0.834 0.431971
y 0.7909 0.1316 6.009 0.000537 ***

Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
Residual standard error: 11.96 on 7 degrees of freedom
Multiple RSquared: 0.8775, Adjusted Rsquared: 0.8425
Fstatistic: 25.08 on 2 and 7 DF, pvalue: 0.000643
First,
summary returns the method that was used with
lm, next is a fivenumber summary of the residuals. As before,
the residuals are available with the
residuals command. More
importantly, the regression coefficients are presented in a table
which includes their estimates (under
Estimate), their
standard error (under
Std. Error), the
tvalue for a
hypothesis test that
b_{i} = 0 under
t value and the
corresponding
pvalue for a twosided test. Small
pvalues are
flagged as shown with the 3 asterisks,
***, in the
y
row. Other tests of hypotheses are easily done knowing the first two
columns and the degrees of freedom. The standard error for the residuals is given along with its
degrees of freedom. This allows one to investigate the residuals
which are again available with the
residuals method. The
multiple and adjusted
R^{2} is given.
R^{2} is interpreted as the
``fraction of the variance explained by the model.'' Finally the
F
statistic is given. The
pvalue for this is from a hypotheses test
that
b_{1} = 0 =
b_{2} = ··· =
b_{p}. That is, the
regression is not appropriate. The theory for this comes
from that of the analysis of variance (ANOVA).
Example: Sale prices of homes
The
homeprice
dataset contains information about homes that
sold in a town of New Jersey in the year 2001. We wish to develop
some rules of thumb in order to help us figure out what are
appropriate prices for homes. First, we need to explore the data a
little bit. We will use the
lattice graphics package for the
multivariate analysis. First we define the useful
panel.lm
function for our graphs.
> library(lattice);data(homeprice);attach(homeprice)
> panel.lm = function(x,y) {
+ panel.xyplot(x,y)
+ panel.abline(lm(y~x))}
> xyplot(sale ~ rooms  neighborhood,panel= panel.lm,data=homeprice)
## too few points in some of the neighborhoods, let's combine
> nbd = as.numeric(cut(neighborhood,c(0,2,3,5),labels=c(1,2,3)))
> table(nbd) # check that we partitioned well
nbd
1 2 3
10 12 7
> xyplot(sale ~ rooms  nbd, panel= panel.lm,layout=c(3,1))
Figure 53: lattice graphic with sale price by number of rooms and
neighborhood
The last graph is plotted in
figure
53.
We compressed the
neighborhood variable as the data was to thinly
spread out. We kept it numeric using
as.numeric as
cut returns a
factor. This is not necessary for
R
to do the regression, but to fit
the above model without modification we need to use a numeric
variable and not a categorical one. The figure shows the
regression lines for the 3 levels of the neighborhood. The multiple
linear regression model assumes that the regression line should have
the same slope for all the levels.
Next, let's find the coefficients for the model. If you are still
unconvinced that the linear relationships are appropriate, you might
try some more plots. The
pairs(homeprice) command gives a
good start.
We'll begin with the regression on bedrooms and neighborhood.
> summary(lm(sale ~ bedrooms + nbd))
...
Coefficients:
Estimate Std. Error t value Pr(>t)
(Intercept) 58.90 48.54 1.213 0.2359
bedrooms 35.84 14.94 2.400 0.0239 *
nbd 115.32 15.57 7.405 7.3e08 ***
This would say an extra bedroom is worth $35 thousand, and a better
neighborhood $115 thousand. However, what does that negative
intercept mean? If there are 0 bedrooms (a small house!) then the
house is worth
> 58.9 + 115.32*(1:3) # nbd is 1, 2 or 3
[1] 56.42 171.74 287.06
This is about correct, but looks funny.
Next, we know that home buyers covet bathrooms. Hence, they should
add value to a house. How much?
> summary(lm(sale ~ bedrooms + nbd + full))
...
Coefficients:
Estimate Std. Error t value Pr(>t)
(Intercept) 67.89 47.58 1.427 0.1660
bedrooms 31.74 14.77 2.149 0.0415 *
nbd 101.00 17.69 5.709 6.04e06 ***
full 28.51 18.19 1.567 0.1297
...
That is $28 thousand dollars per full bathroom. This seems a little high, as the
construction cost of a new bathroom is less than this. Could it
possibly be $15 thousand?
To test this we will do a formal hypothesis test  a onesided test
to see if this
b is 15 against the alternative it is greater
than 15.
> SE = 18.19
> t = (28.51  15)/SE
> t
[1] 0.7427158
> pt(t,df=25,lower.tail=F)
[1] 0.232288
We accept the null hypothesis in this case. The standard error is
quite large.
Before rushing off to buy or sell a home, try to do some of the
problems on this dataset.
Example: Quadratic regression
In 1609 Galileo proved that the trajectory of a body falling with a
horizontal component is a parabola.
^{15} In the course of gaining insight
into this fact, he set up an experiment which measured two variables,
a height and a distance, yielding the following data
height (punti) 
100 
200 
300 
450 
600 
800 
1000 
dist (punti) 
253 
337 
395 
451 
495 
534 
574 
In plotting the data, Galileo apparently saw the parabola and with
this insight proved it mathematically.
Our modern eyes, now expect parabolas. Let's see if linear
regression can help us find the coefficients.
> dist = c(253, 337,395,451,495,534,574)
> height = c(100,200,300,450,600,800,1000)
> lm.2 = lm(dist ~ height + I(height^2))
> lm.3 = lm(dist ~ height + I(height^2) + I(height^3))
> lm.2
...
(Intercept) height I(height^2)
200.211950 0.706182 0.000341
> lm.3
...
(Intercept) height I(height^2) I(height^3)
1.555e+02 1.119e+00 1.254e03 5.550e07
Notice we need to use the construct
I(height2), The
I
function allows us to use the usual notation for powers. (The
^ is used differently in the model notation.)
Looking at a plot of the data with the quadratic curve and the cubic
curve is illustrative.
> quad.fit = 200.211950 + .706182 * pts 0.000341 * pts^2
> cube.fit = 155.5 + 1.119 * pts  .001234 * pts^2 + .000000555 * pts^3
> plot(height,dist)
> lines(pts,quad.fit,lty=1,col="blue")
> lines(pts,cube.fit,lty=2,col="red")
> legend(locator(1),c("quadratic fit","cubic fit"),lty=1:2,col=c("blue","red"))
All this gives us figure
54.
Figure 54: Galileo's data with quadratic and cubic least squares fit.
Both curves seem to fit well. Which to choose? A hypothesis test of
b_{3} = 0 will help decide between the two choices. Recall this is
done for us with the
summary command
> summary(lm.3)
...
Coefficients:
Estimate Std. Error t value Pr(>t)
(Intercept) 1.555e+02 8.182e+00 19.003 0.000318 ***
height 1.119e+00 6.454e02 17.332 0.000419 ***
I(height^2) 1.254e03 1.360e04 9.220 0.002699 **
I(height^3) 5.550e07 8.184e08 6.782 0.006552 **

Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
...
Notice the
pvalue is quite small (0.006552) and so is flagged
automatically by
R. This says the null hypothesis (
b_{3}=0)
should be rejected and the alternative (
b_{3} ¹ 0) is accepted.
We are tempted to attribute this cubic presence to resistance which
is ignored in the mathematical solution which finds the quadratic
relationship.
Some Extra Insight: Easier plotting
To plot the quadratic and cubic lines above was a bit of typing. You
might expect the computer to do this stuff for you. Here is an
alternative which can be generalized, but requires much more sophistication (and just as much
typing in this case)
> pts = seq(min(height),max(height),length=100)
> makecube = sapply(pts,function(x) coef(lm.3) %*% x^(0:3))
> makesquare = sapply(pts,function(x) coef(lm.2) %*% x^(0:2))
> lines(pts,makecube,lty=1)
> lines(pts,makesquare,lty=2)
The key is using the function which takes the coefficients returned
by
coef and ``multiplies'' (
%*%) by the
appropriate powers of
x, namely 1,
x,
x^{2} and
x^{3}. Then this
function is applied to each value of
pts using
sapply
which finds the value of the function for each value of
pts.
14.2 Problems

14.1
 For the
homeprice
dataset, what does a half
bathroom do for the sale price?
 14.2
 For the
homeprice
dataset, how do the
coefficients change if you force the intercept, b_{0} to be 0?
(Use a 1 in the model formula notation.) Does it make any
sense for this model to have no intercept term?
 14.3
 For the
homeprice
dataset, what is the effect of
neighborhood on the difference between sale price and list price? Do
nicer neighborhoods mean it is more likely to have a house go over the
asking price?
 14.4
 For the
homeprice
dataset, is there a
relationship between houses which sell for more than
predicted (a positive residual) and houses which sell for more than
asking? (If so, then perhaps the real estate agents aren't pricing
the home correctly.)
 14.5
 For the
babies
dataset, do a multiple regression
of birthweight with regressors the mothers age, weight and
height. What is the value of R^{2}? What are the coefficients? Do
any variables appear to be 0?
Copyright © John Verzani, 20012. All rights reserved.