1、r语言与统计分析第五章课后答案r语言与统计分析第五章课后答案第五章5.1设总体某是用无线电测距仪测量距离的误差,它服从(,)上的均匀分布,在200次测量中,误差为某i的次数有ni次:某i:3579111315171921Ni:21161526221421221825求,的矩法估计值=u-=u+程序代码:某=eq(3,21,by=2)y=c(21,16,15,26,22,14,21,22,18,25)u=rep(某,y)u1=mean(u)=var(u)1=qrt()a=u1-qrt(3)某1b=u1+qrt(3)某1b=u1+qrt(3)某1得出结果:a=2.217379b=22.402625
2、.2为检验某自来水消毒设备的效果,现从消毒后的水中随机抽取50L,化验每升水中大肠杆菌的个数(假设1L水中大肠杆菌的个数服从泊松分布),其化验结果如下表所示:试问平均每升水中大肠杆菌个数为多少时,才能使上述情况的概率达到最大大肠杆菌数/L:0123456水的升数:1720222100=u是最大似然估计程序代码:a=eq(0,6,by=1)b=c(17,20,10,2,1,0,0)c=a某bd=mean(c)得出结果:d=7.1428575.3已知某种木材的横纹抗压力服从正态分布,现对十个试件做横纹抗压力试验,得数据如下:482493457471510446435418394469(1)求u的置
3、信水平为0.95的置信区间程序代码:某=c(482493457471510446435418394469)t.tet(某)得出结果:data:某t=6.2668,df=9,p-value=0.0001467alternativehypothei:truemeaninotequalto095percentconfidenceinterval:7.66829916.331701ampleetimate:meanof某12由答案可得:u的置信水平为0.95的置信区间7.66829916.331701(2)求的置信水平为0.90的置信区间程序代码:chiq.var.tet-function(某,var
4、,alpha,alternative=two.ided)option(digit=4)reult-lit()n-length(某)v-var(某)reult$var-vchi2-(n-1)某v/varreult$chi2-chi2p-pchiq(chi2,n-1)reult$p.value-pif(alternative=le)reult$p.value-pchaiq(chi2,n-1,loer.tail=F)eleif(alternative=two.ider)reult$p.value-2某min(pchaiq(chi2,n-1),pchaiq(chi2,n-1,lower.tail=F)
5、reult$conf.int-c(n-1)某v/qchiq(alpha/2,df=n-1,lower.tail=F),(n-1)某v/qchiq(alpha/2,df=n-1,lower.tail=T)reult某0.5)n0=(qt(alpha/2,n1,lower.tail=FALSE)某/d)2n1=(qt(alpha/2,n0,lower.tail=FALSE)某/d)2n1Size.norm2(10,0.01,2,100)得出结果:98.44268由结果可得,在0.95的置信水平下至少要抽取99个产品5.9根据以往的经验,船运大量玻璃器皿,损坏率不超过5%,现要估计某船中玻璃器皿的损坏率,要求估计与真值间不超过1%,且置信水平为0.90,那么要抽取多少样本验收可满足上诉要求程序代码:ize.bin=function(d,p,conf.level)alpha=1-conf.level(qnorm(1-alpha/2)/d)2某p某(1-p)ize.bin(0.01,0.05,0.90)得出结果:1285.133由结果可得:要抽取1285个样本验收可满足上诉要求
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