R软件课后习题第五章.docx
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R软件课后习题第五章.docx
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R软件课后习题第五章
第五章
5.1
####写出求正态总体均值检验的R程序(程序名:
mean.test1.R)
mean.test1<-function(x,mu=0,sigma=-1,side=0){
source("P_value.R")
n<-length(x);xb<-mean(x)
if(sigma>0){
z<-(xb-mu)/(sigma/sqrt(n))
P<-P_value(pnorm,z,side=side)
data.frame(mean=xb,df=n,Z=z,P_value=P)
}
else{
t<-(xb-mu)/(sd(x)/sqrt(n))
P<-P_value(pt,t,paramet=n-1,side=side)
data.frame(mean=xb,df=n-1,T=t,P_value=P)
}
}
####写出求P值的R程序(程序名:
P_value.R)
P_value<-function(cdf,x,paramet=numeric(0),side=0){
n<-length(paramet)
P<-switch(n+1,
cdf(x),
cdf(x,paramet),cdf(x,paramet[1],paramet[2]),
cdf(x,paramet[1],paramet[2],paramet[3])
)
if(side<0)P
elseif(side>0)1-P
else
if(P<1/2)2*P
else2*(1-P)
}
####输入数据,再调用函数mean.test1()
>x<-c(220,188,162,230,145,160,238,188,247,113,126,245,164,231,256,183,190,158,224,175)
>source("mean.test1.R")
>a<-mean.test1(x,mu=225,side=0)
>a
得到:
meandfTP_value
1192.1519-3.4782620.002516436
可知,P值小于0.05,故与正常值存在差异
5.2
####输入数据,再调用函数mean.test1()
>x<-c(1067,919,1196,785,1126,936,918,1156,920,948)
>source("mean.test1.R")
>mean.test1(x,mu=1000,side=1)
得到:
meandfTP_value
1997.19-0.069713220.5270268
所以灯泡寿命为1000小时以上的概率是0.4729732
5.3
####写出两总体均值检验的R程序(程序名:
mean.test2.R)
mean.test2<-function(x,y,
sigma=c(-1,-1),var.equal=FALSE,side=0){
source("P_value.R")
n1<-length(x);n2<-length(y)
xb<-mean(x);yb<-mean(y)
if(all(sigma>0)){
z<-(xb-yb)/sqrt(sigma[1]^2/n1+sigma[2]^2/n2)
P<-P_value(pnorm,z,side=side)
data.frame(mean=xb-yb,df=n1+n2,Z=z,P_value=P)
}
else{
if(var.equal==TRUE){
Sw<-sqrt(((n1-1)*var(x)+(n2-1)*var(y))/(n1+n2-2))
t<-(xb-yb)/(Sw*sqrt(1/n1+1/n2))
nu<-n1+n2-2
}
else{
S1<-var(x);S2<-var(y)
nu<-(S1/n1+S2/n2)^2/(S1^2/n1^2/(n1-1)+S2^2/n2^2/(n2-1))
t<-(xb-yb)/sqrt(S1/n1+S2/n2)
}
P<-P_value(pt,t,paramet=nu,side=side)
data.frame(mean=xb-yb,df=nu,T=t,P_value=P)
}
}
####输入数据,再调用函数mean.test2()
>x<-c(113,120,138,120,100,118,138,123)
>y<-c(138,116,125,136,110,132,130,110)
>source("mean.test2.R")
>mean.test2(x,y,var.equal=TRUE,side=0)
得到:
meandfTP_value
1-3.37514-0.56596720.5803752
P值大于0.05,故接受原假设
5.4
####写出均值已知和均值未知两种情况方差比检验的R程序
(程序名:
var.test2.R)
var.test2<-function(x,y,mu=c(Inf,Inf),side=0){
source("P_value.R")
n1<-length(x);n2<-length(y)
if(all(all(mu Sx2<-sum((x-mu[1])^2)/n1;Sy2<-sum((y-mu[2])^2)/n2 df1=n1;df2=n2 } else{ Sx2<-var(x);Sy2<-var(y);df1=n1-1;df2=n2-1 } r<-Sx2/Sy2 P<-P_value(pf,r,paramet=c(df1,df2),side=side) data.frame(rate=r,df1=df1,df2=df2,F=f,P_value=P) } } ####输入数据 >x<-c(-0.70,-5.60,2.00,2.80,0.70,3.50,4.00,5.80,7.10,-0.50,2.50,-1.60,1.70,3.00,0.40,4.50,4.60,2.50,6.00,-1.40) >a<-shapiro.test(x) >a Shapiro-Wilknormalitytest data: x W=0.9699,p-value=0.7527>0.05 >y<-c(3.70,6.50,5.00,5.20,0.80,0.20,0.60,3.40,6.60,-1.10,6.00,3.80,2.00,1.60,2.00,2.20,1.20,3.10,1.70,-2.00) >b<-shapiro.test(y) >b Shapiro-Wilknormalitytest data: y W=0.971,p-value=0.7754>0.05 由以上可知,两组数据均为正态分布 ####输入数据,再调用函数mean.test2() >x<-c(-0.70,-5.60,2.00,2.80,0.70,3.50,4.00,5.80,7.10,-0.50,2.50,-1.60,1.70,3.00,0.40,4.50,4.60,2.50,6.00,-1.40) >y<-c(3.70,6.50,5.00,5.20,0.80,0.20,0.60,3.40,6.60,-1.10,6.00,3.80,2.00,1.60,2.00,2.20,1.20,3.10,1.70,-2.00) >source("mean.test2.R") >a<-mean.test2(x,y,var.equal=TRUE,side=0);a meandfTP_value 1-0.5638-0.6418720.5248097 >b<-mean.test2(x,y,var.equal=FALSE,side=0);b meandfTP_value 1-0.5636.08632-0.6418720.525013 >c<-t.test(x-y,alternative="two.sided");c OneSamplet-test data: x-y t=-0.6464,df=19,p-value=0.5257 alternativehypothesis: truemeanisnotequalto0 95percentconfidenceinterval: -2.3731461.253146 sampleestimates: meanofx -0.56 以上P值均大于0.05,故均值无差异。 ####输入数据,再调用函数var.test2() >x<-c(-0.70,-5.60,2.00,2.80,0.70,3.50,4.00,5.80,7.10,-0.50,2.50,-1.60,1.70,3.00,0.40,4.50,4.60,2.50,6.00,-1.40) >y<-c(3.70,6.50,5.00,5.20,0.80,0.20,0.60,3.40,6.60,-1.10,6.00,3.80,2.00,1.60,2.00,2.20,1.20,3.10,1.70,-2.00) >source("var.test2.R") >var.test2(x,y) ratedf1df2FP_value 11.59836119191.5983610.3152554 P值大于0.05,故方差无差异。 5.5 ####输入数据,再调用函数var.test2() >a<-shapiro.test(x) >b<-shapiro.test(y) >source("var.test2.R") >c<-var.test2(x,y) >d<-mean.test2(x,y,var.equal=TRUE,side=0) >a;b;c;d Shapiro-Wilknormalitytest data: x W=0.9396,p-value=0.4934 Shapiro-Wilknormalitytest data: y W=0.938,p-value=0.5313 ratedf1df2FP_value 11.9646221191.9646220.3200455 meandfTP_value 1-39.0166720-8.814752.524319e-08 由以上可知,两组数据均为正态分布,且方差相同;且两组数据有差异 5.6 ####调用binom.test()函数 >binom.test(57,400,p=0.147, +alternative=c("two.sided","less","greater"), +conf.level=0.95) Exactbinomialtest data: 57and400 numberofsuccesses=57,numberoftrials=400,p-value=0.8876 alternativehypothesis: trueprobabilityofsuccessisnotequalto0.147 95percentconfidenceinterval: 0.10974770.1806511 sampleestimates: probabilityofsuccess 0.1425 P值大于0.05,故支持结果。 5.7 ####调用binom.test()函数 >binom.test(150,328,p=0.5, +alternative=c("two.sided","less","greater"), +conf.level=0.95) Exactbinomialtest data: 150and328 numberofsuccesses=150,numberoftrials=328,p-value=0.1359 alternativehypothesis: trueprobabilityofsuccessisnotequalto0.5 95percentconfidenceinterval: 0.40247960.5129321 sampleestimates: probabilityofsuccess 0.4573171 P值大于0,故可增大比例。 5.8 ####调用chisq.test()函数 >chisq.test(c(315,101,108,32),p=c(9,3,3,1)/16) Chi-squaredtestforgivenprobabilities data: c(315,101,108,32) X-squared=0.47,df=3,p-value=0.9254 P值大于0.05,故符合自由组合规律。 5.9 5.10 ####输入数据,再调用函数ks.test() >x<-c(2.36,3.14,7.52,3.48,2.76,5.43,6.54,7.41) >y<-c(4.38,4.25,6.53,3.28,7.21,6.55) >ks.test(x,y) Two-sampleKolmogorov-Smirnovtest data: xandy D=0.375,p-value=0.6374 alternativehypothesis: two-sided P值大于0.05,故可认为两样本来自同一总体。 5.11 ####输入数据,再用chisq.test()作检验 >x<-c(358,2492,229,2475) >dim(x)<-c(2,2) >chisq.test(x,correct=FALSE) Pearson'sChi-squaredtest data: x X-squared=24.5865,df=1,p-value=7.105e-07 P值小于0.05,故有关。 5.12 ####输入数据,再用chisq.test()作检验 >x<-c(45,46,28,11,12,20,23,12,10,28,30,35) >dim(x)<-c(4,3) >chisq.test(x,correct=FALSE) Pearson'sChi-squaredtest data: x X-squared=40.401,df=6,p-value=3.799e-07 P值小于0.05,故二者有关,不独立。 5.13 ####输入数据,并计算Fisher检验 >x<-c(3,6,4,4) >dim(x)<-c(2,2) >fisher.test(x) Fisher'sExactTestforCountData data: x p-value=0.6372 alternativehypothesis: trueoddsratioisnotequalto1 95percentconfidenceinterval: 0.046243825.13272210 sampleestimates: oddsratio 0.521271 P值大于0.05,所以工艺对产品有影响。 5.14 ####输入数据,调用mcnemar.test()函数作McNemar检验 >X<-c(58,1,8,2,42,9,3,7,17) >dim(X)<-c(3,3) >mcnemar.test(X,correct=FALSE) McNemar'sChi-squaredtest data: X McNemar'schi-squared=2.8561,df=3,p-value=0.4144 P值大于0.05,不能认定两结果不同。 5.15 ####采用成对符号检验,输入数据,调用binom.test()作检验 >x<-scan() 1: 13.3213.0614.0211.8613.5813.7713.5114.4214.4415.43 11: Read10items >binom.test(sum(x>14.6),length(x),al="l") Exactbinomialtest data: sum(x>14.6)andlength(x) numberofsuccesses=1,numberoftrials=10,p-value=0.01074 alternativehypothesis: trueprobabilityofsuccessislessthan0.5 95percentconfidenceinterval: 0.00000000.3941633 sampleestimates: probabilityofsuccess 0.1 P—值为0.01074<0.05,拒绝原假设,所以长度在中位数之上。 ####输入数据,调用wilcox.test()函数 >x<-scan() 1: 13.3213.0614.0211.8613.5813.7713.5114.4214.4415.43 11: Read10items >wilcox.test(x,mu=14.6,alternative="less",exact=FALSE,correct=FALSE,conf.int=TRUE) Wilcoxonsignedranktest data: x V=4.5,p-value=0.00949 alternativehypothesis: truelocationislessthan14.6 95percentconfidenceinterval: -Inf14.24504 sampleestimates: (pseudo)median 13.74995 P—值为0.00949<0.05,拒绝原假设,所以长度在中位数之上。 5.16 ####采用成对符号检验,输入数据,调用binom.test()作检验 >x<-scan() 1: 40.833.037.548.042.540.042.036.011.322.036.027.314.232.152.038.017.320.021.046.1 21: Read20items >y<-scan() 1: 37.041.023.417.031.540.031.036.05.711.521.06.126.521.344.528.022.620.011.022.3 21: Read20items >binom.test(sum(x Exactbinomialtest data: sum(x numberofsuccesses=3,numberoftrials=20,p-value=0.002577 alternativehypothesis: trueprobabilityofsuccessisnotequalto0.5 95percentconfidenceinterval: 0.032070940.37892683 sampleestimates: probabilityofsuccess 0.15 P—值为0.002577<0.05,接受原假设,2种方法有差异。 ####输入数据,调用wilcox.test()函数 >x<-c(40.8,33.0,37.5,48.0,42.5,40.0,42.0,36.0,11.3,22.0,36.0,27.3,14.2,32.1,52.0,38.0,17.3,20.0,21.0,46.1) >y<-c(37.0,41.0,23.4,17.0,31.5,40.0,31.0,36.0,5.7,11.5,21.0,6.1,26.5,21.3,44.5,28.0,22.6,20.0,11.0,22.3) >wilcox.test(x,y,alternative="greater",paired=TRUE) Wilcoxonsignedranktestwithcontinuitycorrection data: xandy V=134,p-value=0.003477 alternativehypothesis: truelocationshiftisgreaterthan0 P—值为0.003477<0.05,接受原假设,2种方法有差异。 >wilcox.test(x-y,alternative="greater") 具有相同的结果 >binom.test(sum(x>y),length(x),alternative="greater") Exactbinomialtest data: sum(x>y)andlength(x) numberofsuccesses=14,numberoftrials=20,p-value=0.05766 alternativehypothesis: trueprobabilityofsuccessisgreaterthan0.5 95percentconfidenceinterval: 0.49218161.0000000 sampleestimates: probabilityofsuccess 0.7 结果还为有差异 >x<-c(40.8,33.0,37.5,48.0,42.5,40.0,42.0,36.0,11.3,22.0,36.0,27.3,14.2,32.1,52.0,38.0,17.3,20.0,21.0,46.1) >y<-c(37.0,41.0,23.4,17.0,31.5,40.0,31.0,36.0,5.7,11.5,21.0,6.1,26.5,21.3,44.5,28.0,22.6,20.0,11.0,22.3) >wilcox.test(x,y,alternative="less",exact=FALSE,correct=FALSE) Wilcoxonranksumtest data: xandy W=272.5,p-value=0
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