1. business and industrial
2. business software

Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
Question 1: Using a little bit of algebra, prove that (4.2) is
equivalent to (4.3). In other words, the logistic function
representation and logit representation for the logistic
regression model are equivalent.
Solution: Equation 4.2 in textbook is
𝑃(𝑥) =
𝑃(𝑥)
1−𝑃(𝑥)
=
=
𝑒 𝛽0+𝛽1
1+ 𝑒 𝛽0+𝛽1
𝛽0 +𝛽1
⁄
(𝑒
)
1+ 𝑒 𝛽0+𝛽1
(𝑒 𝛽0+𝛽1⁄
1−
)
1+ 𝑒 𝛽0+𝛽1
𝛽0 +𝛽1
⁄
(𝑒
)
1+ 𝑒 𝛽0+𝛽1
(1+𝑒 𝛽0+𝛽1 −𝑒 𝛽0 +𝛽1⁄
)
1+ 𝑒 𝛽0+𝛽1
= 𝑒 𝛽0 +𝛽1
Therefore, Hence Proved that Logistic Function
representation and Logit representation are equivalent.
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
Question 2: It was stated in the text that classifying an
observation to the class for which (4.12) is largest is
equivalent to classifying an observation to the class for
which (4.13) is largest. Prove that this is the case. In other
words, under the assumption that the observations in the
kth class are drawn from a N(μk, σ2) distribution, the Bayes’
classifier assigns an observation to the class for which the
discriminant function is maximized.
Solution: In the Linear Discriminant Analysis for p=1,
Take log on the both sides for the equation 𝑃𝑘 (x) in 4.12 and
rearrange the terms,
Log 𝑃𝑘 (x) = Log(
2
(𝑥−𝜇𝑘 )
1
𝜋𝑘
𝑒𝑥𝑝(−
)
2𝜎2
√2𝜋𝜎
2
(𝑥−𝜇𝑙 )
1
𝑘
∑𝑙=1 𝜋𝑙
𝑒𝑥𝑝(−
)
2𝜎2
√2𝜋𝜎
= Log(
2
(𝑥−𝜇𝑘 )
𝜋𝑘 𝑒𝑥𝑝(−
)
2𝜎2
2
(𝑥−𝜇𝑙 )
𝑘
∑𝑙=1 𝜋𝑙 𝑒𝑥𝑝(−
)
2𝜎2
)
)
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
(𝑥−𝜇𝑘 )2
= 𝑙𝑜𝑔 (𝜋𝑘 𝑒𝑥𝑝 (−
)) 2𝜎 2
𝑙𝑜𝑔 (∑𝑘𝑙=1 𝜋𝑙 𝑒𝑥𝑝 (−
(𝑥−𝜇𝑙 )2
2𝜎 2
))
As We know the formula (log(a/b) = log(a) – log(b))
Denominator is going to be the same for all classes as the
variance 𝜎1 2 = ……….. = 𝜎𝑘 2 as assumed.
Which is a variance term shared across all K classes is same.
Last term is independent of k, So we exclude that
(𝑥 − 𝜇𝑘 )2
= 𝐿𝑜𝑔 (𝜋𝑘 𝑒𝑥𝑝 (−
))
2𝜎 2
= Log(𝜋𝑘 ) + 𝐿𝑜𝑔𝑒 (𝑒
−(𝑥−𝜇𝑘 )2
2𝜎2
(As log(ab)= log(a) + log(b))
= Log(𝜋𝑘 ) -
(𝑥−𝜇𝑘 )2
2𝜎 2
)
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
(𝑥 2 −2𝑥𝜇𝑘 +𝜇𝑘 2 )
= Log(𝜋𝑘 ) 2𝜎 2
= Log(𝜋𝑘 ) -
(𝑥 2 )
2𝜎 2
+
2𝑥𝜇𝑘
2𝜎 2
-
𝜇𝑘 2
2𝜎 2
The term 𝑥 2 in the above equation is independent of
k, so we can exclude it and we will find k as
𝑙𝑜𝑔(𝜋𝑘 ) +
𝑥𝜇𝑘
𝜎2
-
𝜇𝑘 2
2𝜎 2
Question 3: This problem relates to the QDA model, in
which the observations within each class are drawn from a
normal distribution with a class specific mean vector and a
class specific covariance matrix. We consider the simple case
where p = 1; i.e. there is only one feature. Suppose that we
have K classes, and that if an observation belongs to the kth
class then X comes from a one-dimensional normal
distribution, X ∼ N(μk, 𝜎𝑘 2 ). Recall that the density function
for the one-dimensional normal distribution is given in
(4.11). Prove that in this case, the Bayes’ classifier is not
linear. Argue that it is in fact quadratic. Hint: For this
problem, you should follow the arguments laid out in Section
4.4.2, but without making the assumption that 𝜎1 2 = ……….. =
𝜎𝑘 2
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
Solution:
The solution is same as the above problem but we
cannot skip 𝑥 2 term as 𝜎𝑘 2 varies across classes. So we find
k using the term
Log(𝜋𝑘 ) -
(𝑥 2 )
2𝜎 2
+
2𝑥𝜇𝑘
2𝜎 2
-
𝜇𝑘 2
2𝜎 2
Question 4: When the number of features p is large, there
tends to be a deterioration in the performance of KNN and
other local approaches that perform prediction using only
observations that are near the test observation for which a
prediction must be made. This phenomenon is known as the
curse of dimensionality, and it ties into the fact that curse of
dinon-parametric approaches often perform poorly when p
is large. We mensionality will now investigate this curse.
(a) Suppose that we have a set of observations, each with
measurements on p = 1 feature, X. We assume that X is
uniformly (evenly) distributed on [0, 1]. Associated with each
observation is a response value. Suppose that we wish to
predict a test observation’s response using only observations
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
that are within 10 % of the range of X closest to that test
observation. For instance, in order to predict the response
for a test observation with X = 0.6, we will use observations
in the range [0.55, 0.65]. On average, what fraction of the
available observations will we use to make the prediction?
Solution:
Given there are some set of observations, with p=1.
X is uniformly distributed on [0,1]
There is a response value associated with each observation.
To predict response for a test observation, with X=0.6, use
observations in range [0.55,0.65]
To predict response for a test observation, with X=0.1, use
observations in range [0.05,0.15]
To predict response for a test observation, with X=0.01,
use observations in range [0.005,0.015]
To predict response for a test observation, with X=0.95,
use observations in range [0.90,1]
To predict response for a test observation, with X=0.99,
use observations in range [0.94,1.04]
10% of observations from interval of length 1 fall into
interval of length 0.1
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
50+5(0.1) = 50.5
f(x) = 50+5x, when 0<x<0.1
= 10, when 0.1<= x<= 0.9
= 10-50x, when 0.9<x<=1
We use 9% of observations to make the prediction.
b) Now suppose that we have a set of observations, each
with measurements on p = 2 features, X1 and X2. We
assume that (X1, X2) are uniformly distributed on [0, 1] × [0,
1]. We wish to predict a test observation’s response using
only observations that are within 10 % of the range of X1
and within 10 % of the range of X2 closest to that test
observation. For instance, in order to predict the response
for a test observation with X1 = 0.6 and X2 = 0.35, we will
use observations in the range [0.55, 0.65] for X1 and in the
range [0.3, 0.4] for X2. On average, what fraction of the
available observations will we use to make the prediction?
Solution:
𝑋1 , 𝑋2 are the two observations
Which are uniformly distributed on [0,1] X [0,1]
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
As we take 9% of observations to make prediction in this
case, answer is 9% X 9% = 0.81%
As there are two observations here 𝑋1 , 𝑋2 .
c) Now suppose that we have a set of observations on p =
100 features. Again the observations are uniformly
distributed on each feature, and again each feature ranges
in value from 0 to 1. We wish to predict a test observation’s
response using observations within the 10 % of each
feature’s range that is closest to that test observation. What
fraction of the available observations will we use to make
the prediction?
Solution: Given p= 100,
Observations are uniformly distributed, features
ranges from 0 to 1.
We predict test observations response using observations
with in 10% of each feature’s range closest to that test
observation.
9% to the power of 100 = (0.09)100
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
d) Using your answers to parts (a)–(c), argue that a
drawback of KNN when p is large is that there are very few
training observations “near” any given test observation.
Solution:
We use 90% of data to classify that point.
Given a point in 2 dimensions,
We use 90%^2 = 0.81% of the data near that point.
When the dimensions are increased, we exponentially
reduce the number of observations near the point in
consideration.
The drawback of knn is when p is large, there will be few
training observations.
e) Now suppose that we wish to make a prediction for a test
observation by creating a p-dimensional hypercube centered
around the test observation that contains, on average, 10 %
of the training observations. For p = 1, 2, and 100, what is
the length of each side of the hypercube? Comment on your
answer.
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
Note: A hypercube is a generalization of a cube to an
arbitrary number of dimensions. When p = 1, a hypercube is
simply a line segment, when p = 2 it is a square, and when p
= 100 it is a 100-dimensional cube
Solution:
The length of the each side of the hypercube 0.09
Question 5: We now examine the differences between LDA
and QDA.
(a) If the Bayes decision boundary is linear, do we expect
LDA or QDA to perform better on the training set? On the
test set?
Solution: If Bayes decision boundary is linear,
We expect QDA to perform better on training set because of
its higher flexibility yield a closer fit.
LDA to perform better than QDA because QDA would overfit
linearity on the Bayes decision theory on the test set.
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
b) If the Bayes decision boundary is non-linear, do we expect
LDA or QDA to perform better on the training set? On the
test set?
Solution: QDA to perform better both on training, test sets.
c) In general, as the sample size n increases, do we expect
the test prediction accuracy of QDA relative to LDA to
improve, decline, or be unchanged? Why?
Solution: If training set is very large, QDA is more flexible
than LDA and has higher variance.
d) True or False: Even if the Bayes decision boundary for a
given problem is linear, we will probably achieve a superior
test error rate using QDA rather than LDA because QDA is
flexible enough to model a linear decision boundary. Justify
your answer.
Solution: With fewer sample points, variance from using
QDA, lead to overfit, leads to inferior test error rate.
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
Question 6: Suppose we collect data for a group of
students in a statistics class with variables 𝑋1 = hours
studied, 𝑋2 = undergrad GPA, and Y = receive an A. We fit
a logistic regression and produce estimated coefficient,
𝛽̂0 = −6, 𝛽̂1 = 0.05, 𝛽̂2 = 1.
(a) Estimate the probability that a student who studies
for 40 h and has an undergrad GPA of 3.5 gets an A in the
class.
(b) How many hours would the student in part (a) need to
study to have a 50 % chance of getting an A in the class?
Solution: Given 𝑋1 = hours studied
𝑋2 = undergrad GPA
Y = receive an A
𝛽̂0 = −6, 𝛽̂1 = 0.05, 𝛽̂2 = 1
a) Given 𝑋1 = 40 hours, GPA = 3.5
𝑃𝑘 (x) =
=
̂
̂
𝑒 𝛽0+ 𝛽1𝑋1 + 𝑋2
̂
̂
1+ 𝑒 𝛽0+ 𝛽1𝑋1 + 𝑋2
𝑒 −6 + 0.05∗40 + 3.5
1+ 𝑒 −6+ 0.05∗40+ 3.5
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
=
=
𝑒 −0.5
1+ 𝑒 −0.5
06065
1.6065
= 0.3776
Probability is 0.3776
Question 7: Suppose that we wish to predict whether a
given stock will issue a dividend this year (“Yes” or “No”)
based on X, last year’s percent profit.
We examine a large number of companies and discover that
the mean value of X for companies that issued a dividend
was 𝑋̅= 10, while the mean for those that didn’t was 𝑋̅ = 0.
In addition, the variance of X for these two sets of
companies was 𝜎̂ 2 = 36. Finally, 80 % of companies issued
dividends. Assuming that X follows a normal distribution,
predict the probability that a company will issue a dividend
this year given that its percentage profit was X = 4 last year.
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
Hint: Recall that the density function for a normal random
variable is f(x) =
−(𝑥−𝜇)2
𝑒 2𝜎2
√2𝜋𝜎 2
. You will need to use Bayes’
theorem.
Solution:
Mean value of X, who issued a dividend is 𝑋̅ = 10
Mean value of X, who didn’t issue a dividend is 𝑋̅ = 0
Variance of X for 2 sets of companies 𝜎̂ 2 = 36
Assume X follows a normal distribution.
Percentage profit was X = 4 in the last year
Considering probability that X will issue a dividend
𝑃𝑘 (x) =
𝜋𝑘
1
2𝜋𝜎
√
∑𝑘
𝑙=1 𝜋𝑙
𝑃𝑘 (x) =
2
−(𝑥−𝜇𝑘 )
(
)
2𝜎2
𝑒
1
√2𝜋𝜎
From the equation 4.12 in textbook
2
−(𝑥−𝜇𝑘 )
(
)
2𝜎2
𝑒
1
√2𝜋𝜎
−(𝑥−𝜇𝑘 )2
𝑒 2𝜎2 + 𝜋𝑙
𝜋𝑘
1
𝜋𝑘
√2𝜋𝜎
−(𝑥−𝜇𝑙 )2
𝑒 2𝜎2
1
𝑒
√2𝜋𝜎
−(𝑥−𝜇𝑙 )2
2𝜎2
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
Substitute X=4, 𝜇𝑘 = 10, 𝜇𝑙 = 0, 𝜎 2 = 36, 𝜋𝑘 = 0.8, 𝜋𝑙 = 0.2 in the
above equation
=
−(4−10)2
1
0.8∗
∗ 𝑒 36∗2
√2∗3.14∗6
−(4−0)2
−(4−10)2
1
1
0.8∗
∗ 𝑒 36∗2 + 0.2 ∗
∗ 𝑒 36∗2
√2∗3.14∗6
√2∗3.14∗6
−36
=
=
=
=
0.8∗ 𝑒 36∗2
−36
+
−16
0.8∗ 𝑒 36∗2 0.2∗ 𝑒 36∗2
0.8∗ 𝑒 −0.5
0.8∗ 𝑒 −0.5 + 0.8∗ 𝑒 −0.22
0.485
0.485+0.1675
0.485
0.645503
= 0.751
Probability that X will issue a dividend is 0.751
Question 9: This problem has to do with odds.
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
(a) On average, what fraction of people with an odds of 0.37
of defaulting on their credit card payment will in fact
default?
Solution:
𝑝(𝑋)
1−𝑝(𝑋)
= 0.37
p(X) = 0.37(1-p(X))
p(X) = 0.37 – 0.37(p(X))
1.37(p(X)) = 0.37
P(X) = 0.27
On an average, 27% of people are defaulting on their credit
card payment.
(b) Suppose that an individual has a 16 % chance of
defaulting on her credit card payment. What are the odds
that she will default?
Solution:
𝑝(𝑋)
1−𝑝(𝑋)
=
0.16
1−0.16
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
=
0.16
0.84
=
0.19
The odds that she will default is 19%
Question 10: This question should be answered using the
Weekly data set, which is part of the ISLR package. This data
is similar in nature to the Smarket data from this chapter’s
lab, except that it contains 1, 089 weekly returns for 21
years, from the beginning of 1990 to the end of 2010.
10) (a) Produce some numerical and graphical summaries of
the Weekly data. Do there appear to be any patterns?
Solutions:
>library(ISLR)
>summary(Smarket)
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>pairs(Smarket)
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>cor(Smarket[, -9])
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>attach(Smarket)
>plot(Volume)
(b) Use the full data set to perform a logistic regression with
Direction as the response and the five lag variables plus
Volume as predictors. Use the summary function to print the
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
results. Do any of the predictors appear to be statistically
significant? If so, which ones?
Solution
>glm.fit=glm(Dimension~Lag1+Lag2+Lag3+Lag4+Lag5+
Volume,data=Smarket,family=binomial)
>summary(glm.fit)
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
Lag1 has the lowest p-value. However, p-value is 0.145,
which is relatively large, so there is no real association
between Lag1 and Direction.
10)c) Compute the confusion matrix and overall fraction of
correct predictions. Explain what the confusion matrix is
telling you about the types of mistakes made by logistic
regression.
>glm.probs=predict(glm.fit,type=”response”)
>glm.probs[1:10]
>contrasts(Direction)
>glm.pred=rep(“Down”,1250)
>glm.pred[glm.probs>.5]=”Up”
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>table(glm.pred,Direction)
Percentage of correct predictions on the training data is
=(
=(
145+507
)
145+507+457+141
652
)
1250
=0.5216
= 52.16%
>mean(glm.pred==Direction)
0.5216
The training error rate is 100-52.16 = 47.84%
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
10)d) Now fit the logistic regression model using a training
data period from 1990 to 2008, with Lag2 as the only
predictor. Compute the confusion matrix and the overall
fraction of correct predictions for the held out data (that is,
the data from 2009 and 2010).
Solution:
>train = (Year<2005)
>Smarket.2005 = Smarket[!train,]
>Direction.2005 = Direction[!train]
>glm2.fit =
glm(Direction~Lag2,data=Smarket,family=binomial,subset
=train)
>summary(glm2.fit)
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>probs2=predict(fit.glm2,Smarket.2005,type=”response”)
>glm2.pred=rep(“Down”,length(probs2))
>glm2.pred[probs2>0.5]=”Up”
>table(glm2.pred,Direction.2005)
Percentage of correct predictions on the training data is
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
=(
=(
19+130
)
19 +11+92+130
149
)
252
=0.5088
= 50.88%
>mean(glm2.pred==Direction)
0.5088
The training error rate is 100-50.88 = 49.12%
10)e) Repeat (d) using LDA.
Solution:
>library(MASS)
>fit.lda<lda(Direction~Lag1+Lag2,data=Smarket,subset=train)
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>lda.fit
>plot(lda.fit)
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>lda.pred=predict(lda.fit,Smarket.2005)
>names(lda.pred)
>lda.class=lda.pred$class
>table(lda.class,Direction.2005)
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
Percentage of correct predictions on the test data is
=(
=(
35+106
)
35 +35+76+106
141
)
252
=0.5595
= 55.95%
The test error rate is 100-55.95% = 44.05%
(f) Repeat (d) using QDA.
Solution:
>qda.fit=qda(Direction~Lag1+Lag2,data=Smarket,subset=t
rain)
>qda.fit
>qda.class=predict(qda.fit,Smarket.2005)$class
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>table(qda.class,Direction.2005)
Percentage of correct predictions on the test data is
=(
=(
30+121
)
30 +20+81+121
151
)
252
=0.5992
= 59.92%
The test error rate is 100-59.92% = 40.08%
>mean(qda.class==Direction.2005)
(g) Repeat (d) using KNN with K = 1.
Solution:
>library(class)
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>train.X<-as.matrix(Lag2[train])
>test.X<-as.matrix(Lag2[!train])
train.Direction<-Direction[train]
set.seed(1)
pred.knn<-knn(train.X,test.X,train.Direction,k=1)
table(pred.knn,Direction.2005)
Percentage of correct predictions on the test data is
=(
=(
53+71
)
53 +70+58+71
124
)
252
=0.5
= 50%
The test error rate is 100-50% = 50%
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
10)h) Which of these methods appears to provide the best
results on this data?
Solution: When we are comparing the test error rates,
qda and lda have minimum error rates, followed by
Logistic regression and knn.
10)i) Experiment with different combinations of predictors,
including possible transformations and interactions, for each
of the methods. Report the variables, method, and
associated confusion matrix that appears to provide the best
results on the held out data. Note that you should also
experiment with values for K in the KNN classifier.
Solutions:
>glm.fit<-glm(Direction~Year,family=binomial)
>summary(glm.fit)
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
#QDA with sqrt(abs(Lag2))
>qda2.fit=qda
>qda2.fit=qda(Direction~Lag1+sqrt(abs(Lag2)),data=Smark
et,subset=train)
>qda2.fit
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>qda2.pred=predict(qda2.fit,Smarket.2005)
>table(qda2.pred$class,Direction.2005)
Percentage of correct predictions on the test data is
=(
=(
34+121
)
34 +20+77+121
155
)
252
=0.5112
= 51.12%
The test error rate is 100-51.12% = 48.88%
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
With k=3, knn classifier,
>pred.knn<-knn(train.X,test.X,train.Direction,k=3)
>table(pred.knn,Direction.2005)
With k=100, knn classifier,
>pred.knn<-knn(train.X,test.X,train.Direction,k=100)
>table(pred.knn,Direction.2005)
Question 11: In this problem, you will develop a model to
predict whether a given car gets high or low gas mileage
based on the Auto data set.
(a) Create a binary variable, mpg01, that contains a 1 if mpg
contains a value above its median, and a 0 if mpg contains a
value below its median. You can compute the median using
the median() function. Note you may find it helpful to use
the data.frame() function to create a single data set
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
containing both mpg01 and the other Auto variables. 172 4.
Classification
Solution:
>attach(Auto)
>mpg01<-rep(0,length(mpg))
>mpg01[mpg>median(mpg)]<-1
>Auto<-data.frame(Auto,mpg01)
(b) Explore the data graphically in order to investigate the
association between mpg01 and the other features. Which
of the other features seem most likely to be useful in
predicting mpg01? Scatterplots and boxplots may be useful
tools to answer this question. Describe your findings.
Solution:
>cor(Auto[, -9])
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>pairs(Auto)
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>boxplot(cylinders~mpg01,data=Auto,main="Cylinders vs
mpg01")
>boxplot(displacement~mpg01,data=Auto,main="Displace
ment vs mpg01")
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>boxplot(horsepower~mpg01,data=Auto,main="Horsepow
er vs mpg01")
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>boxplot(weight~mpg01,data=Auto,main="Weight vs
mpg01")
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>boxplot(acceleration~mpg01,data=Auto,main="Accelerati
on vs mpg01")
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>boxplot(year~mpg01,data=Auto,main="Year vs mpg01")
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
There is a relationship between “mpg01” and “cylinders”,
“weight”, “displacement” and “horsepower”.
(c) Split the data into a training set and a test set.
Solution:
>train<-(year %% 2 ==0)
> Auto.train<-Auto[train,]
> Auto.test<-Auto[!train,]
> mpg01.test<-mpg01[!train]
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
(d) Perform LDA on the training data in order to predict
mpg01 using the variables that seemed most associated
with mpg01 in (b). What is the test error of the model
obtained?
Solution:
>fit.lda<lda(mpg01~cylinders+weight+displacement+horsepower,dat
a=Auto,subset=train)
>fit.lda
>pred.lda<-predict(fit.lda,Auto.test)
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>table(pred.lda$class,mpg01.test)
The test error rate is 12.63%
(e) Perform QDA on the training data in order to predict
mpg01 using the variables that seemed most associated
with mpg01 in (b). What is the test error of the model
obtained?
Solution:
>fit.qda<qda(mpg01~cylinders+weight+displacement+horsepower,da
ta=Auto,subset=train)
>fit.qda
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
>pred.qda<-predict(fit.qda,Auto.test)
> table(pred.qda$class,mpg01.test)
> mean(pred.qda$class!=mpg01.test)
[1] 0.1318681
The test error rate is 13.18%
(f) Perform logistic regression on the training data in order
to predict mpg01 using the variables that seemed most
associated with mpg01 in (b). What is the test error of the
model obtained?
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
Solution:
> fit.glm<glm(mpg01~cylinders+weight+displacement+horsepower,da
ta=Auto,subset=train)
> summary(fit.glm)
> probs<-predict(fit.glm,Auto.test,type="response")
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
> pred.glm<-rep(0,length(probs))
> pred.glm[probs>0.5]<-1
> table(pred.glm,mpg01.test)
> mean(pred.glm!=mpg01.test)
[1] 0.1263736
The test error rate is 12.63%
(g) Perform KNN on the training data, with several values of
K, in order to predict mpg01. Use only the variables that
seemed most associated with mpg01 in (b). What test errors
do you obtain? Which value of K seems to perform the best
on this data set?
Solution:
>train.X=cbind(cylinders,weight,displacement,horsepower)
[train,]
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
> train.X<cbind(cylinders,weight,displacement,horsepower)[train,]
> test.X<cbind(cylinders,weight,displacement,horsepower)[!train,]
> train.mpg01<-mpg01[train]
> set.seed(1)
> pred.knn<-knn(train.X,test.X,train.mpg01,k=1)
> table(pred.knn,mpg01.test)
> mean(pred.knn!=mpg01.test)
[1] 0.1538462
For k=1, the test error rate is 15.38%
> pred.knn<-knn(train.X,test.X,train.mpg01,k=10)
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
> table(pred.knn,mpg01.test)
> mean(pred.knn!=mpg01.test)
[1] 0.1648352
For k=10, the test error rate is 16.48%
> pred.knn<-knn(train.X,test.X,train.mpg01,k=100)
> table(pred.knn,mpg01.test)
> mean(pred.knn!=mpg01.test)
[1] 0.1428571
For k=100, the test error rate is 14.28%. So for k=100, it
seems to perform the best test.
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
12) This problem involves writing functions.
a) Write a function, Power(), that prints out the result of
raising 2 to the 3rd power. In other words, your function
should compute 23 and print out the results. Hint: Recall
that x^a raises x to the power a. Use the print() function to
output the result.
Solution:
> Power <- function() {
2^3
}
> Power()
[1] 8
(b) Create a new function, Power2(), that allows you to pass
any two numbers, x and a, and prints out the value of x^a.
You can do this by beginning your function with the line >
Power2=function (x,a){ You should be able to call your
function by entering, for instance, > Power2 (3,8) on the
command line. This should output the value of 38, namely, 6,
561.
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
Solution:
> Power2 <- function(x,a) {
x^a
}
> Power2(3,8)
[1] 6561
(c) Using the Power2() function that you just wrote, compute
103, 817, and 1313.
Solution:
d) Now create a new function, Power3(), that actually
returns the result x^a as an R object, rather than simply
printing it to the screen. That is, if you store the value x^a in
an object called result within your function, then you can
simply return() this return() result, using the following line:
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
The line above should be the last line in your function, before
the } symbol.
> Power3 <- function(x,a) {
result<- x^a
return(result)
}
e) Now using the Power3() function, create a plot of f(x) =𝑥 2 .
The x-axis should display a range of integers from 1 to 10,
and the y-axis should display 𝑥 2 . Label the axes
appropriately, and use an appropriate title for the figure.
Consider displaying either the x-axis, the y-axis, or both on
the log-scale. You can do this by using log=‘‘x’’, log=‘‘y’’, or
log=‘‘xy’’ as arguments to the plot() function.
Solution:
> x <- 1:10
> plot(x, Power3(x,2), log= "xy", xlab="Log of x", ylab=
"Log of x^2", main= "Log of x^2 vs Log of x")
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
(f) Create a function, PlotPower(), that allows you to create
a plot of x against x^a for a fixed a and for a range of values
of x. For instance, if you call > PlotPower (1:10,3) then a plot
should be created with an x-axis taking on values 1, 2,..., 10,
and a y-axis taking on values 13 , 23 ,..., 103
> PlotPower <- function(x,a) {
plot(x, Power3(x,a))
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
}
> PlotPower(1:10,3)
13. Using the Boston data set, fit classification models in
order to predict whether a given suburb has a crime rate
above or below the median. Explore logistic regression, LDA,
and KNN models using various subsets of the predictors.
Describe your findings.
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
Solution:
> library(MASS)
> attach(Boston)
> crim01 <- rep(0, length(crim))
> crim01[crim > median(crim)] <- 1
> Boston <- data.frame(Boston, crim01)
> train <- 1:(length(crim)/2)
> test <- (length(crim) / 2 + 1):length(crim)
> Boston.train <- Boston[train, ]
> Boston.test <- Boston[test, ]
> crim01.test <- crim01[test]
> fit.glm <- glm(crim01~. - crim01 - crim, data = Boston,
family = binomial, subset=train)
Warning message:
glm.fit: fitted probabilities numerically 0 or 1 occurred
> probs <- predict(fit.glm, Boston.test, type = "response")
> pred.glm <- rep(0, length(probs))
> pred.glm[probs > 0.5] <- 1
> table(pred.glm, crim01.test)
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
crim01.test
pred.glm 0 1
0 68 24
1 22 139
> mean(pred.glm!=crim01.test)
[1] 0.1818182
The test error rate for logistic regression is 18.18%
> fit.glm <- glm(crim01~. -crim01 - crim - chas - nox, data =
Boston, family = binomial, subset=train)
Warning message:
glm.fit: fitted probabilities numerically 0 or 1 occurred
> probs <- predict(fit.glm, Boston.test, type = "response")
> pred.glm <- rep(0, length(probs))
> pred.glm[probs > 0.5] <- 1
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
> table(pred.glm, crim01.test)
crim01.test
pred.glm 0 1
0 78 28
1 12 135
> mean(pred.glm!=crim01.test)
[1] 0.1581028
The test error rate for this logistic regression is 15.81%
> fit.lda <- lda(crim01~. - crim01 - crim, data = Boston,
subset = train)
> pred.lda <- predict(fit.lda, Boston.test)
> table(pred.lda$class, crim01.test)
> mean(pred.lda$class !=crim01.test)
[1] 0.1343874
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
The test error rate for lda is 13.43%
> fit.lda <- lda(crim01~. - crim01 - crim - chas - nox, data =
Boston, subset = train)
> pred.lda <- predict(fit.lda, Boston.test)
> table(pred.lda$class, crim01.test)
> mean(pred.lda$class !=crim01.test)
[1] 0.1501976
The test error rate for this lda is 15.01%
> train.X=cbind(zn, indus, chas, nox, rm, age, dis, rad, tax,
ptratio, black, lstat, medv)[train,]
> train.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax,
ptratio, black, lstat, medv)[train,]
> test.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax,
ptratio, black, lstat, medv)[test,]
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
> train.crim01 <- crim01[train]
> set.seed(1)
> pred.knn <- knn(train.X, test.X, train.crim01, k=1)
> mean(pred.knn !=crim01.test)
[1] 0.458498
For this KNN (k=1), the test error rate is 45.84%
> pred.knn <- knn(train.X, test.X, train.crim01, k=10)
> table(pred.knn, crim01.test)
crim01.test
pred.knn 0 1
0 83 23
1 7 140
Chapter 4
Solutions for Classification
Text book: An Introduction to Statistical Learning with Applications in R
For this KNN (k=10), the test error rate is 11.85%
> pred.knn <- knn(train.X, test.X, train.crim01, k=100)
> table(pred.knn, crim01.test)
> mean(pred.knn !=crim01.test)
[1] 0.4901186
For this KNN (k=100), the test error rate is 49.01%