一些准备
knitr::opts_chunk$set(tidy = TRUE, warning = FALSE, message = FALSE) setwd("C:\\Users\\213yi\\Desktop\\非参数统计\\3-28") library(showtext) #载入库 作业一
区间估计
data=read.table("nerve.txt") fun<-function(data,alpha,fwd){ data<-as.numeric(data) tn<-quantile(data,fwd) B=1000 m=20 assemble<-NULL resolution=as.data.frame(matrix(nrow=3,ncol = 3)) colnames(resolution)<-c("下限","上限","区间长度") rownames(resolution)<-c("正太","枢轴量","分位数") for (i in 1:B){ samp<-sample(data,m,replace = T) assemble<-c(assemble,quantile(samp,fwd)) } #正太置信区间 resolution[1,1]<-tn-qnorm(1-alpha/2)*sd(assemble) resolution[1,2]<-tn+qnorm(1-alpha/2)*sd(assemble) resolution[1,3]<-resolution[1,2]-resolution[1,1] #枢轴量置信区间 resolution[2,1]<-2*tn-quantile(assemble,1-alpha/2) resolution[2,2]<-2*tn-quantile(assemble,alpha/2) resolution[2,3]<-resolution[2,2]-resolution[2,1] #分位数置信区间 resolution[3,1]<-quantile(assemble,alpha/2) resolution[3,2]<-quantile(assemble,1-alpha/2) resolution[3,3]<-resolution[3,2]-resolution[3,1] return(resolution) } #0.75分位点 95%的置信区间 fun(data,0.05,0.75) #0.75分位点 90%的置信区间 fun(data,0.10,0.75) #0.25分位点 95%的置信区间 fun(data,0.05,0.25) #0.25分位点 90%的置信区间 fun(data,0.10,0.25) 检验
#对正态性进行检验 asse<-NULL for (i in 1:1000){ samp<-sample(data,20,replace = T) asse<-c(asse,as.numeric(quantile(samp,0.25))) } shapiro.test(asse)#拒绝了正态性的假设 ks.test(asse,rnorm(,mean = mean(asse),sd=sd(asse)))#拒绝了正态性的假设 #H0:u=u0 norm(asse) line(asse) - 因此，在假设条件满足且置信区间小的前提下，选择枢轴量置信区间
作业二
参数估计
方法：抽出100个点,估计出 μ {\mu} μ= X ˉ {\bar{X}} Xˉ,从 e μ e^{\mu} eμ中抽1000次,每次抽20个点,每次都计算 e X ˉ e^{\bar{X}} eXˉ，存储,枢轴量方法计算出θ
resolution=as.data.frame(matrix(nrow=2,ncol = 3)) colnames(resolution)<-c("下限","上限","区间长度") rownames(resolution)<-c("参数bootstrap","非参数bootstrap") sourced=rnorm(100,mean = 5,sd = 1) xb=mean(sourced) assum<-NULL for (i in 1:1000){ samp<-rnorm(20,mean=xb,sd=1) assum<-c(assum,exp(mean(samp))) } resolution[1,1]=2*exp(xb)-quantile(assum,0.975) resolution[1,2]=2*exp(xb)-quantile(assum,0.025) resolution[1,3]=resolution[1,2]-resolution[1,1] df1<-assum 抽出100个点,作为数据源,抽1000次,每次抽20个点,每次都计算 e X ˉ e^{\bar{X}} eXˉ,存储,枢轴量方法计算出θ
assum<-NULL for (i in 1:1000){ samp<-sample(sourced,20) assum<-c(assum,exp(mean(samp))) } resolution[2,1]=2*exp(xb)-quantile(assum,0.975) resolution[2,2]=2*exp(xb)-quantile(assum,0.025) resolution[2,3]=resolution[2,2]-resolution[2,1] df2<-assum resolution - 非参数Bootstrap方法比参数Bootstrap方法的置信区间长度更短，更为精确
可视化
#分布图 hist(df1,prob=TRUE,xlab="θ",ylab="密度",col="deepskyblue",main="") hist(df2,prob=TRUE,xlab="",ylab="",col="red2",density=30,main="",add=TRUE) legend("topright",legend=c("参数Bootstrap估计","非参数Bootstrap估计"),ncol=1,inset=0.04,col=c("deepskyblue","red2"),density=c(200,50),fill=c("deepskyblue","red2"),cex=0.8) title("Bootstrap估计直方图") 观察得到两者为偏态分布，猜想对数正态，并进行检验
检验
norm(df1,main = "-原始数据-参数") line(df1) norm(log(df1),main = "-log-参数") line(log(df1)) norm(df2,main = "-原始数据-非参数") line(df2) norm(log(df2),main = "-log-非参数") line(log(df2)) - 取对数之后，为正态分布
- 因此，数据为对数正态分布
作业三-正太记分检验
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# 用正态记分检验方法评价人均年消费酒量的中位数 NScoreTest <- function(x, med.val, alternative = 'two-sided') { n <- length(x) x.diff <- x - med.val x.absdiff <- abs(x.diff) x.rank <- rank(x.absdiff) x.sign <- sign(x.diff) # 计算符号正态记分 x.nscore x.nscore <- qnorm((1 + (x.rank) / (n + 1)) / 2, 0, 1) * x.sign df <- data.frame(value = x, absvalue = x.absdiff, rank = x.rank, sign = x.sign, nscore = x.nscore) index <- order(x.absdiff) df <- df[index, ] w <- sum(x.nscore) Tval <- w / sqrt(sum(x.nscore ^ 2)) if (alternative == 'two-sided') { pval = 2 * (1 - pnorm(abs(Tval), 0, 1)) } else if (alternative == 'greater') { pval = 1 - pnorm(Tval, 0, 1) } else { pval = pnorm(Tval, 0, 1) } list(df = df, w = w, Tval = Tval, pval = pval) } alcohol <- c(4.12, 5.81, 7.63, 9.74, 10.39, 11.92, 12.32, 12.89, 13.54, 14.45) NScoreTest(alcohol, med.val = 8, alternative = 'two-sided') - p-value值为0.0,因此在显著性水平0.1的情况下，拒绝原假设，拒绝中位数为8的假设
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