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《多元统计及R语言建模》第2章王斌会2020.2.1rm(list=ls())#清理内存options(digits=4)#输出结果位数par(mar=c(4,4,2,1))#设置图片输出位置library(openxlsx)library(knitr)2.1　　对下面的相关系数矩阵，试用R语言求其逆矩阵、特征根和特征向量。要求写出R语言计算函数。R=1.000.800.260.670.340.801.000.330.590.340.260.331.000.370.210.670.590.371.000.350.340.340.210.351.00R=matrix(c(1.00,0.80,0.26,0.67,0.34,0.80,1.00,0.33,0.59,0.34,0.26,0.33,1.00,0.37,0.21,0.67,0.59,0.37,1.00,0.35,0.34,0.34,0.21,0.35,1.00),nrow=5,ncol=5);R#生成矩阵R[,1][,2][,3][,4][,5][1,]1.000.800.260.670.34[2,]0.801.000.330.590.34[3,]0.260.331.000.370.21[4,]0.670.590.371.000.35[5,]0.340.340.210.351.00R.=solve(R);R.[,1][,2][,3][,4][,5][1,]3.3881-2.12220.23706-1.0685-0.10623[2,]-2.12222.9421-0.33593-0.1331-0.16164[3,]0.2371-0.33591.20699-0.3764-0.08812[4,]-1.0685-0.1331-0.376372.0091-0.21562[5,]-0.1062-0.1616-0.08812-0.21561.18505R.e=eigen(R,symmetric=T);R.eeigen()decomposition$values[1]2.79230.82630.77910.42060.1818$vectors[,1][,2][,3][,4][,5][1,]-0.52550.34022-0.16650.159380.74494[2,]-0.51870.23435-0.17780.50823-0.62142[3,]-0.3131-0.90308-0.22870.149430.10844[4,]-0.49660.03869-0.1186-0.83116-0.21673[5,]-0.3318-0.110840.93500.056160.013552.2　　某厂对50个计件工人某月份工资进行登记，获得以下原始资料（单位：元）。试按组距为300编制频数，计算频数、频率和累计频率，并绘制直方图。14651760198522702980137517351940222026701405175519652240282012951645188021102550135517101910219026001265162518652095252012251605184520402430117518352030237010001535181020102290112515751815203023202.2.1　　写出R语言程序。d2.2=read.xlsx('mvexer5.xlsx',sheet='E2.2');tail(d2.2)#读取数据并显示后六行X452290461125471575481815492030502320　　先对数据进行简单的统计分析：min(d2.2)[1]1000max(d2.2)[1]2980　　可以得到某月份最低工资为1000元,最高工资为2980元。　　画出组距为300的直方图：H1=hist(d2.2$X,breaks=seq(900,3000,300))#以组距为300画直方图　　编制频数表，列出频数、频率和累计频率：names(H1)[1]"breaks""counts""density""mids""xname""equidist"data.frame('组中值'=H1$mids,'频数'=H1$counts,'频率'=H1$density*300,'累计频率'=cumsum(H1$density*300))组中值频数频率累计频率1105030.060.062135070.140.2031650100.200.4041950150.300.705225080.160.866255050.100.967285020.041.002.2.3　　用R语言作正态概率图并分析之：qqnorm(d2.2$X);qqline(d2.2$X)　　由正态概率图可以看出数据点基本是一条直线，因此数据可以近似看作正态分布。2.3　　以下是一份关学生是否抽烟与每天学习时间长短关系的调查数据，试用R语言对其进行基本统计分析。d2.3=read.xlsx('mvexer5.xlsx',sheet='E2.3');d2.3#读取数据编号是否抽烟每天学习时间11是少于5小时22否5-10小时33否5-10小时44是超过10小时55否超过10小时66是少于5小时77是5-10小时88是少于5小时99否超过10小时1010是5-10小时　　对数据进行简单的统计分析：dim(d2.3);names(d2.3)[1]103[1]"编号""是否抽烟""每天学习时间"str(d2.3)'data.frame':10obs.of3variables:$编号:num12345678910$是否抽烟:chr"是""否""否""是"...$每天学习时间:chr"少于5小时""5-10小时""5-10小时""超过10小时"...summary(d2.3)编号是否抽烟每天学习时间Min.:1.00Length:10Length:101stQu.:3.25Class:characterClass:characterMedian:5.50Mode:characterMode:characterMean:5.503rdQu.:7.75Max.:10.00　　可以看到抽取学生中4位不抽烟，6位抽烟，每天学习时间小于5小时的有3位，5-10小时的有4位，超过10小时的有3位，抽烟情况和学习时间分别的分布比较均匀。2.4　　　　　　试编制进行计量数据频数表分析的R语言函数。ch=function(X){H=hist(X)n=H$breaks[2]-H$breaks[1]c=data.frame('组中值'=H$mids,'频数'=H$counts,'频率'=H$density*n,'累计频率'=cumsum(H$density*n))c}ch(d2.2$X)组中值频数频率累计频率11250100.200.2021750200.400.6032250140.280.884275060.121.00