It was a professor at the end of the arrangement of work, but also the book "Business data analysis and application" of a homework
table of Contents
Data Description
Data preprocessing
Descriptive statistical analysis
Model analysis (analysis of variance)
Data Description
178 non-degree students of vocational training institutions of the data, the purpose is to understand what students might get better learning results
Data preprocessing
Open data, view data and lock part of the data (this can be used directly after the variable name is specified instead of $ data)
grades=read.table('E:/SWlearning/R/assighment/RegressionAnalysis/Report/ins1.csv',
header=TRUE,sep=',')
head(grades)
attach(grades)
The results show
The variable names into English
names(grades)=c('aveGrades','gender','birth','firmType','eduBG','eduGrd')
The response variable (dependent variable): the dependent variable grade point average (aveGrades).
Arguments: sex (gender), date of birth (birth), enterprise nature (firmType), Highest Level of Education (eduBG), the highest academic graduation time (eduGrd)
Check the appropriate normality of the variables
shapiro.test(aveGrades)
The results show
Shapiro-Wilk normality test
data: aveGrades
W = 0.89736, p-value = 9.286e-10
p-value is very small and therefore reject the null hypothesis, i.e. the data is normally distributed reject the null hypothesis
method continues with BoxCox establish a new variable corresponding to ensure its normality, BoxCox.ar is noted in the function package TSA
library(TSA)
boxcox=BoxCox.ar(aveGrades,lambda = seq(4, 8, 0.1))
View optimal value of lamda
boxcox$mle
Create a new response variable
aveGrades_mod=grades$aveGrades^6.6
The new test response variable normality
shapiro.test(aveGrades_mod)
The results show
Shapiro-Wilk normality test
data: aveGrades_mod
W = 0.99007, p-value = 0.2522
p-value results met our expectations, do not reject the null hypothesis that the response is to accept the new hypothesis of normal distribution
Descriptive statistical analysis
Note that our dependent variable, date of birth (birth) and the highest academic graduation time (eduGrd) is not a discrete variables, we will be ten units of the two classification variables
date of birth (birth) is the largest 1952-6-26 the minimum is 1979-11-10, into the fifties (1) and sixties (2), the seventies (3)
the highest academic graduation time (eduGrd) is the largest 1982-1-1, is the smallest 2004- 3-1, into the eighties (1), the nineties (2), after dated (3)
the first step in
the date of birth (birth) and the highest academic graduation time (eduGrd) become so after the date of variables operations
birthmod=as.Date(grades$birth)
eduGrdmod=as.Date(grades$eduGrd)
The second step
we first classify date of birth
//d1~d4分别是四个时间节点，用来将数据分成五十年代(1），六十年代(2)， 七十年代(3)
d1=as.Date('1950/1/1')
d2=as.Date('1960/1/1')
d3=as.Date('1970/1/1')
d4=as.Date('1980/1/1')
//计算出生日期(birthmod)中的数据个数
s=0
for(i in birthmod){
s=s+1
}
//建立新的数值型变量。因为birthmod是日期型变量，不能直接赋数值型的值如1，2，3
birth_mod=1:s
//开始分类
for(i in 1:s){
fac1=birthmod[i]-d1>0 & birthmod[i]-d2<=0
fac2=birthmod[i]-d2>0 & birthmod[i]-d3<=0
fac3=birthmod[i]-d3>0 & birthmod[i]-d4<=0
if(fac1){birth_mod[i]=1}
if(fac2){birth_mod[i]=2}
if(fac3){birth_mod[i]=3}
}
//给新变量birth_mod三个水平1,2,3
levels(birth_mod)=c(1,2,3)
//将数据类型变成factor，以便之后的统计
birth_mod=as.factor(birth_mod)
Graduate of the highest level of education is the same procedure
d5=as.Date('1990/1/1')
d6=as.Date('2000/1/1')
d7=as.Date('2010/1/1')
s=0
for(i in eduGrdmod){
s=s+1
}
eduGrd_mod=1:s
for(i in 1:s){
fac3=eduGrdmod[i]-d4>0 & eduGrdmod[i]-d5<=0;fac3
fac4=eduGrdmod[i]-d5>0 & eduGrdmod[i]-d6<=0;fac4
fac5=eduGrdmod[i]-d6>0 & eduGrdmod[i]-d7<=0;fac5
if(fac3){eduGrd_mod[i]=1}
if(fac4){eduGrd_mod[i]=2}
if(fac5){eduGrd_mod[i]=3}
}
levels(eduGrd_mod)=c(1,2,3)
eduGrd_mod=as.factor(eduGrd_mod)
第三步
建立新的数据集grades_mod，注意此处的响应变量（aveGrades）没有用之前为了正态性修改的新的响应变量（aveGrades_mod），这里用aveGrades是为了结果好看，且不影响我们进行描述性统计分析
grades_mod=cbind(grades$aveGrades,grades[2],birth_mod,grades[4:5],eduGrd_mod)
summary(grades_mod)结果显示
grades$aveGrades gender birth_mod firmType eduBG eduGrd_mod
Min. :50.00 男:133 1:10 国企:95 本科 :148 1: 48
1st Qu.:77.00 女: 45 2:85 民企:43 大专 : 25 2:104
Median :81.00 3:83 外企:40 硕士 : 2 3: 26
Mean :79.72 硕士或以上: 3
3rd Qu.:84.00
Max. :91.00
第四步
我们还想知道，各个因变量不同水平对应的学员平均成绩
//编写一个输出均值，标准差，最大值，中位数，最小值的函数
stats = function(x){
m = mean(x)
sd= sd(x)
max = max(x)
median = median(x)
min= min(x)
return=c(m,sd,max,median,min)
}
//aggregate是一个重新显示数据的函数，比如在aggdata1中，能显示按性别分类后，男性学员和女性学员对应的平均成绩的均值，标准差，最大值，中位数，最小值，FUN是function函数的意思
aggdata1= aggregate(grades['aveGrades'],
by=list(gender),FUN=stats);aggdata1
aggdata2= aggregate(grades['aveGrades'],
by=list(birth_mod),FUN=stats)
aggdata3= aggregate(grades['aveGrades'],
by=list(firmType),FUN=stats)
aggdata4= aggregate(grades['aveGrades'],
by=list(eduBG),FUN=stats);aggdata
aggdata5= aggregate(grades['aveGrades'],
by=list(eduGrd_mod),FUN=stats)
//按行将数据重叠起来
aggdata=rbind(aggdata1,aggdata2,aggdata3,aggdata4,aggdata5);aggdata
结果显示
模型分析
接下来我们将进行方差分析
第一步
//进行方差分析的函数是aov，~前面是响应变量，注意此时我们得保证响应变量的正态性，所以用的是新的响应变量（aveGrades_mod）而非原始数据，~后面是自变量，在此模型中还包括了所有的交互项
res.ano1=aov(aveGrades_mod~gender+birth_mod+firmType+eduBG+eduGrd_mod+
gender:birth_mod+gender:firmType+gender:eduBG+gender:eduGrd_mod+
birth_mod:firmType+birth_mod:eduBG+birth_mod:eduGrd_mod+
firmType:eduBG+firmType:eduGrd_mod+
eduBG:eduGrd_mod)
//显示方差分析结果
res1=summary(res.ano1);res1
结果显示
第二步
剔除没通过显著性检验的变量， 用剩下的变量再做一次方差分析
res.ano2=aov(aveGrades_mod~gender+birth_mod+eduBG+
gender:firmType+gender:eduGrd_mod+
birth_mod:firmType+
firmType:eduBG)
res2=summary(res.ano2);res2
结果显示
第三步
剔除没通过显著性检验的变量， 用剩下的变量再做一次方差分析
res.ano3=aov(aveGrades_mod~gender+birth_mod+eduBG+
gender:eduGrd_mod+
birth_mod:firmType)
res3=summary(res.ano3);res3
结果显示
性别(gender)，出生日期(birth_mod)，最高学历(eduBG)以及交互作用， 性别:最高学历毕业日期(gender:eduGrd_mod)，出生日期:企业性质(birth_mod:firmType)都通过了在 0.1 水平下的显著性检验
拒绝原假设，即变量的水平不同会显著影响成绩，如性别中，男生和女生的成绩显著不同，而企业性质的不同不影响学员的成绩