《统计学习导论：基于R应用》第2章课后习题参考答案

【第2章课后习题参考答案】
Chapter 2 Exercise
1.
(a) better - a more flexible approach will fit the data closer and with the
large sample size a better fit than an inflexible approach would be obtained

(b) worse - a flexible method would overfit the small number of observations

(d) worse - flexible methods fit to the noise in the error terms and increase variance

2.
(a) regression. inference. quantitative output of CEO salary based on CEO
firm's features.
n - 500 firms in the US
p - profit, number of employees, industry

(b) classification. prediction. predicting new product's success or failure.
n - 20 similar products previously launched
p - price charged, marketing budget, comp. price, ten other variables

(c) regression. prediction. quantitative output of % change
n - 52 weeks of 2012 weekly data
p - % change in US market, % change in British market, % change in German market

3.
(a) See 3a.jpg.

(b)
all 5 lines >= 0

i. (squared) bias - decreases monotonically because increases in flexibility
yield a closer fit

ii. variance - increases monotonically because increases in flexibility yield
overfit

iii. training error - decreases monotonically because increases in flexibility
yield a closer fit

iv. test error - concave up curve because increase in flexibility yields a closer
fit before it overfits

v. Bayes (irreducible) error - defines the lower limit, the test error is bounded
below by the irreducible error due to variance in the error (epsilon) in the output
values (0 <= value). When the training error is lower than the irreducible error,
overfitting has taken place.
The Bayes error rate is defined for classification problems and is determined by
the ratio of data points which lie at the 'wrong' side of the decision boundary,
(0 <= value < 1).

4.
(a) i. stock market price direction, prediction, response: up, down,
input: yesterday's price movement % change, two previous day price movement %
change, etc.

ii. illness classification, inference, response: ill, healthy, input: resting
heart rate, resting breath rate, mile run time

iii. car part replacement, prediction, response: needs to be replace, good,
input: age of part, mileage used for, current amperage

(b) i. CEO salary. inference. predictors: age, industry experience, industry,
years of education. response: salary.

ii. car part replacement. inference. response: life of car part. predictors: age
of part, mileage used for, current amperage.

iii. illness classification, prediction, response: age of death,
input: current age, gender, resting heart rate, resting breath rate, mile run
time.

ii. Netflix movie recommendations. recommend movies based on users who have
watched and rated similar movies.

iii. marketing survey. clustering of demographics for a product(s) to see which
clusters of consumers buy which products.

5.
The advantages for a very flexible approach for regression or classification
are obtaining a better fit for non-linear models, decreasing bias.

The disadvantages for a very flexible approach for regression or classification
are requires estimating a greater number of parameters, follow the noise too
closely (overfit), increasing variance.

A more flexible approach would be preferred to a less flexible approach when we
are interested in prediction and not the interpretability of the results.

A less flexible approach would be preferred to a more flexible approach when we
are interested in inference and the interpretability of the results.

6.
A parametric approach reduces the problem of estimating f down to one of
estimating a set of parameters because it assumes a form for f.

A non-parametric approach does not assume a functional form for f and so
requires a very large number of observations to accurately estimate f.

The advantages of a parametric approach to regression or classification are the
simplifying of modeling f to a few parameters and not as many observations are
required compared to a non-parametric approach.

The disadvantages of a parametric approach to regression or classification
are a potential to inaccurately estimate f if the form of f assumed is wrong or
to overfit the observations if more flexible models are used.

7.
(a) Obs. X1 X2 X3 Distance(0, 0, 0) Y
---------------------------------------------
1 0 3 0 3 Red
2 2 0 0 2 Red
3 0 1 3 sqrt(10) ~ 3.2 Red
4 0 1 2 sqrt(5) ~ 2.2 Green
5 -1 0 1 sqrt(2) ~ 1.4 Green
6 1 1 1 sqrt(3) ~ 1.7 Red

(b) Green. Observation #5 is the closest neighbor for K = 1.

(d) Small. A small K would be flexible for a non-linear decision boundary,
whereas a large K would try to fit a more linear boundary because it takes more
points into consideration.

# 8. (a)
college = read.csv("../data/College.csv")
# 8. (b)
fix(college)
rownames(college) = college[,1]
college = college[,-1]
fix(college)
# 8. (c)
# i.
summary(college)
# ii.
pairs(college[,1:10])
# iii.
plot(college$Private, college$Outstate)
# iv.
Elite = rep("No", nrow(college))
Elite[college$Top10perc>50] = "Yes"
Elite = as.factor(Elite)
college = data.frame(college, Elite)
summary(college$Elite)
plot(college$Elite, college$Outstate)
# v.
par(mfrow=c(2,2))
hist(college$Apps)
hist(college$perc.alumni, col=2)
hist(college$S.F.Ratio, col=3, breaks=10)
hist(college$Expend, breaks=100)
# vi.
par(mfrow=c(1,1))
plot(college$Outstate, college$Grad.Rate)
# High tuition correlates to high graduation rate.
plot(college$Accept / college$Apps, college$S.F.Ratio)
# Colleges with low acceptance rate tend to have low S:F ratio.
plot(college$Top10perc, college$Grad.Rate)
# Colleges with the most students from top 10% perc don't necessarily have
# the highest graduation rate. Also, rate > 100 is erroneous!

# 9.
Auto = read.csv("../data/Auto.csv", header=T, na.strings="?")
Auto = na.omit(Auto)
dim(Auto)
summary(Auto)

#9.(a)
# quantitative: mpg, cylinders, displacement, horsepower, weight,
# acceleration, year
# qualitative: name, origin

#9.(b)
# apply the range function to the first seven columns of Auto
sapply(Auto[, 1:7], range)
# mpg cylinders displacement horsepower weight acceleration year
# [1,] 9.0 3 68 46 1613 8.0 70
# [2,] 46.6 8 455 230 5140 24.8 82

# 9.(c)
sapply(Auto[, 1:7], mean)
# mpg cylinders displacement horsepower weight acceleration
# 23.445918 5.471939 194.411990 104.469388 2977.584184 15.541327
# year
# 75.979592

sapply(Auto[, 1:7], sd)
# mpg cylinders displacement horsepower weight acceleration
# 7.805007 1.705783 104.644004 38.491160 849.402560 2.758864
# year
# 3.683737

#9.(d)
newAuto = Auto[-(10:85),]
dim(newAuto) == dim(Auto) - c(76,0)
newAuto[9,] == Auto[9,]
newAuto[10,] == Auto[86,]

sapply(newAuto[, 1:7], range)
# mpg cylinders displacement horsepower weight acceleration year
# [1,] 11.0 3 68 46 1649 8.5 70
# [2,] 46.6 8 455 230 4997 24.8 82
sapply(newAuto[, 1:7], mean)
# mpg cylinders displacement horsepower weight acceleration
# 24.404430 5.373418 187.240506 100.721519 2935.971519 15.726899
# year
# 77.145570
sapply(newAuto[, 1:7], sd)
# mpg cylinders displacement horsepower weight acceleration
# 7.867283 1.654179 99.678367 35.708853 811.300208 2.693721
# year
# 3.106217

# 9.(e)
pairs(Auto)
plot(Auto$mpg, Auto$weight)
# Heavier weight correlates with lower mpg.
plot(Auto$mpg, Auto$cylinders)
# More cylinders, less mpg.
plot(Auto$mpg, Auto$year)
# Cars become more efficient over time.

#9.(f)
pairs(Auto)
# See descriptions of plots in (e).
# All of the predictors show some correlation with mpg. The name predictor has
# too little observations per name though, so using this as a predictor is
# likely to result in overfitting the data and will not generalize well.

# 10.
# 10.(a)
library(MASS)
?Boston
dim(Boston)
# 506 rows, 14 columns
# 14 features, 506 housing values in Boston suburbs

#10.(b)
pairs(Boston)
# X correlates with: a, b, c
# crim: age, dis, rad, tax, ptratio
# zn: indus, nox, age, lstat
# indus: age, dis
# nox: age, dis
# dis: lstat
# lstat: medv

#10.(c)
plot(Boston$age, Boston$crim)
# Older homes, more crime
plot(Boston$dis, Boston$crim)
# Closer to work-area, more crime
plot(Boston$rad, Boston$crim)
# Higher index of accessibility to radial highways, more crime
plot(Boston$tax, Boston$crim)
# Higher tax rate, more crime
plot(Boston$ptratio, Boston$crim)
# Higher pupil:teacher ratio, more crime

# 10.(d)
par(mfrow=c(1,3))
hist(Boston$crim[Boston$crim>1], breaks=25)
# most cities have low crime rates, but there is a long tail: 18 suburbs appear
# to have a crime rate > 20, reaching to above 80
hist(Boston$tax, breaks=25)
# there is a large divide between suburbs with low tax rates and a peak at 660-680
hist(Boston$ptratio, breaks=25)
# a skew towards high ratios, but no particularly high ratios

# 10.(e)
dim(subset(Boston, chas == 1))
# 35 suburbs

# 10.(f)
median(Boston$ptratio)
# 19.05

# 10.(g)
> t(subset(Boston, medv == min(Boston$medv)))
# 399 406
# crim 38.3518 67.9208 above 3rd quartile
# zn 0.0000 0.0000 at min
# indus 18.1000 18.1000 at 3rd quartile
# chas 0.0000 0.0000 not bounded by river
# nox 0.6930 0.6930 above 3rd quartile
# rm 5.4530 5.6830 below 1st quartile
# age 100.0000 100.0000 at max
# dis 1.4896 1.4254 below 1st quartile
# rad 24.0000 24.0000 at max
# tax 666.0000 666.0000 at 3rd quartile
# ptratio 20.2000 20.2000 at 3rd quartile
# black 396.9000 384.9700 at max; above 1st quartile
# lstat 30.5900 22.9800 above 3rd quartile
# medv 5.0000 5.0000 at min
summary(Boston)
# Not the best place to live, but certainly not the worst.

# 10.(h)
dim(subset(Boston, rm > 7))
# 64
dim(subset(Boston, rm > 8))
# 13
summary(subset(Boston, rm > 8))
summary(Boston)
# relatively lower crime (comparing range), lower lstat (comparing range)

【资源来源】
代码、数据：
https://github.com/asadoughi/stat-learning/archive/refs/heads/master.zip
https://www.statlearning.com/resources-first-edition

课后答案：
https://blog.princehonest.com/stat-learning/

网上视频：
https://www.dataschool.io/15-hours-of-expert-machine-learning-videos/

《统计学习导论：基于R应用》第2章课后习题参考答案相关推荐

《统计学习导论-基于R应用》第二章：统计学习（代码）
A = matrix(seq(1,16),4,4) A 1 5 9 13 2 6 10 14 3 7 11 15 4 8 12 16 A[1,] 1 5 9 13 A[-c(1,3),] 2 6 10 ...
统计学习导论 - 基于R的应用学习笔记1
统计学习导论 - 基于R的应用学习笔记 Chapter 1 导论统计学习是什么: 关于估计ƒ的一系列方法 Y=f(x)+ξ f:X 提供给 Y 的系统信息,是 x 的函数 ξ:随机误差项:均值为0 ...
最优化理论c语言代码,《统计学习导论基于R应用》PDF代码导图+《最优化理论与算法第2版》PDF习题指导...
要想深入理解机器学习,或者对人工智能的某个领域有所研究,都必须掌握统计学.最优化.矩阵及其应用等知识. 推荐<统计学习导论:基于R应用>,适合运用统计学习前沿技术分析数据的人士.读起来不费 ...
【医学信息学】《统计学习导论-基于r应用》的学习总结
以下仅为笔者在学习<统计学习导论-基于r应用>过程中的理解总结,如有错误,敬请指正统计学习导论概述什么是统计学习? 对一系列观测值(自变量/预测变量/X,因变量/响应变量/Y)之间的关 ...
《统计学习导论-基于R应用》
机器学习更底层的东西在<统计学习理论的本质>里, <统计学习理论的本质>更现代化.更通俗的理解在这里 <统计学习导论-基于R应用> Gareth James Dan ...
《统计学习导论：基于R应用》第3章课后习题参考答案
[第3章课后习题参考答案] Chapter 3 Exercise 1. In Table 3.4, the null hypothesis for "TV" is that in ...
mysql 三阶多项式拟合,《统计学习导论-基于R应用》第三章：线性回归（代码）...
库library 库:一组不含在基础R配置内的函数和数据集 library(MASS) # 加载库 library(ISLR)# 安装库 install.packages("ISLR&quo ...
统计学习导论-基于R应用学习笔记
目录误差假设检验 F-检验分类classification 线性判别分析(LDA) Threshold 分类阀值 resampling 重采样留一法交叉验证(LOOCV) The Bootst ...
统计学习导论基于R应用——作业 3
7. 解: (a)欧几里德距离: |x| = sqrt( x[1]^2 + x[2]^2 + - + x[n]^2 ) 所以计算结果是: 1.3. 2.2. 3.sqrt(10). 4.sqrt(5) ...

《统计学习导论：基于R应用》第2章课后习题参考答案

《统计学习导论：基于R应用》第2章课后习题参考答案相关推荐

最新文章

热门文章