R软件与统计建模.docx
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R软件与统计建模.docx
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R软件与统计建模
Ex4.5
>x<-c(54,67,68,78,70,66,67,70,65,69)
>t.test(x)#t.test()做单样本正态分布区间估计
OneSamplet-test
data:
x
t=35.947,df=9,p-value=4.938e-11
alternativehypothesis:
truemeanisnotequalto0
95percentconfidenceinterval:
63.158571.6415
sampleestimates:
meanofx
67.4
平均脉搏点估计为67.4,95%区间估计为63.158571.6415。
>t.test(x,alternative="less",mu=72)#t.test()做单样本正态分布单侧区间估计OneSamplet-test
data:
x
t=-2.4534,df=9,p-value=0.01828
alternativehypothesis:
truemeanislessthan72
95percentconfidenceinterval:
-Inf70.83705
sampleestimates:
meanofx
67.4
p值小于0.05,拒绝原假设,平均脉搏低于常人。
要点:
t.test()函数的用法。
本例为单样本;可做双边和单侧检验。
Ex4.6
>x<-c(140,137,136,140,145,148,140,135,144,141);x
[1]140137136140145148140135144141
>y<-c(135,118,115,140,128,131,130,115,131,125);y
[1]135118115140128131130115131125
>t.test(x,y,var.equal=TRUE)
TwoSamplet-test
data:
xandy
t=4.6287,df=18,p-value=0.0002087
alternativehypothesis:
truedifferenceinmeansisnotequalto0
95percentconfidenceinterval:
7.5362620.06374
sampleestimates:
meanofxmeanofy
140.6126.8
期望差的95%置信区间为7.5362620.06374。
要点:
t.test()可做两正态样本均值差估计。
此例认为两样本方差相等。
ps:
我怎么觉得这题应该用配对t检验?
Ex4.7
>x<-c(0.143,0.142,0.143,0.137)
>y<-c(0.140,0.142,0.136,0.138,0.140)
>t.test(x,y,var.equal=TRUE)
TwoSamplet-test
data:
xandy
t=1.198,df=7,p-value=0.2699
alternativehypothesis:
truedifferenceinmeansisnotequalto095percentconfidenceinterval:
-0.0019963510.006096351
sampleestimates:
meanofxmeanofy
0.141250.13920
期望差的95%的区间估计为-0.0019963510.006096351
Ex4.8
接Ex4.6
>var.test(x,y)
Ftesttocomparetwovariances
data:
xandy
F=0.2353,numdf=9,denomdf=9,p-value=0.04229alternativehypothesis:
trueratioofvariancesisnotequalto195percentconfidenceinterval:
0.058452760.94743902
sampleestimates:
ratioofvariances
0.2353305
要点:
var.test可做两样本方差比的估计。
基于此结果可认为方差不等。
因此,在Ex4.6中,计算期望差时应该采取方差不等的参数。
>t.test(x,y)
WelchTwoSamplet-test
data:
xandy
t=4.6287,df=13.014,p-value=0.0004712
alternativehypothesis:
truedifferenceinmeansisnotequalto095percentconfidenceinterval:
7.35971320.240287
sampleestimates:
meanofxmeanofy
140.6126.8
期望差的95%置信区间为7.35971320.240287。
要点:
t.test(x,y,var.equal=TRUE)做方差相等的两正态样本的均值差估计t.test(x,y)做方差不等的两正态样本的均值差估计
Ex4.9
>x<-c(rep(0,7),rep(1,10),rep(2,12),rep(3,8),rep(4,3),rep(5,2))>n<-length(x)
>tmp<-sd(x)/sqrt(n)*qnorm(1-0.05/2)
>mean(x)
[1]1.904762
>mean(x)-tmp;mean(x)+tmp
[1]1.494041
[1]2.315483
平均呼唤次数为1.9
0.95的置信区间为1.49,2,32
Ex4.10
>x<-c(1067,919,1196,785,1126,936,918,1156,920,948)
>t.test(x,alternative="greater")
OneSamplet-test
data:
x
t=23.9693,df=9,p-value=9.148e-10
alternativehypothesis:
truemeanisgreaterthan0
95percentconfidenceinterval:
920.8443Inf
sampleestimates:
meanofx
997.1
灯泡平均寿命置信度95%的单侧置信下限为920.8443
要点:
t.test()做单侧置信区间估计
Ex5.1
>x<-c(220,188,162,230,145,160,238,188,247,113,126,245,164,231,256,183,190,158,224,175)
>t.test(x,mu=225)
OneSamplet-test
data:
x
t=-3.4783,df=19,p-value=0.002516
alternativehypothesis:
truemeanisnotequalto225
95percentconfidenceinterval:
172.3827211.9173
sampleestimates:
meanofx
192.15
原假设:
油漆工人的血小板计数与正常成年男子无差异。
备择假设:
油漆工人的血小板计数与正常成年男子有差异。
p值小于0.05,拒绝原假设,认为油漆工人的血小板计数与正常成年男子有差异。
上述检验是双边检验。
也可采用单边检验。
备择假设:
油漆工人的血小板计数小于正常成年男子。
>t.test(x,mu=225,alternative="less")
OneSamplet-test
data:
x
t=-3.4783,df=19,p-value=0.001258
alternativehypothesis:
truemeanislessthan225
95percentconfidenceinterval:
-Inf208.4806
sampleestimates:
meanofx
192.15
同样可得出油漆工人的血小板计数小于正常成年男子的结论。
Ex5.2
>pnorm(1000,mean(x),sd(x))
[1]0.5087941
>x
[1]1067919119678511269369181156920948
>pnorm(1000,mean(x),sd(x))
[1]0.5087941
x<=1000的概率为0.509,故x大于1000的概率为0.491.
要点:
pnorm计算正态分布的分布函数。
在R软件中,计算值均为下分位点。
Ex5.3
>A<-c(113,120,138,120,100,118,138,123)
>B<-c(138,116,125,136,110,132,130,110)
>t.test(A,B,paired=TRUE)
Pairedt-test
data:
AandB
t=-0.6513,df=7,p-value=0.5357
alternativehypothesis:
truedifferenceinmeansisnotequalto0
95percentconfidenceinterval:
-15.628898.87889
sampleestimates:
meanofthedifferences
-3.375
p值大于0.05,接受原假设,两种方法治疗无差异。
Ex5.4
(1)
正态性W检验:
>x<-c(-0.7,-5.6,2,2.8,0.7,3.5,4,5.8,7.1,-0.5,2.5,-1.6,1.7,3,0.4,4.5,4.6,2.5,6,-1.4)
>y<-c(3.7,6.5,5,5.2,0.8,0.2,0.6,3.4,6.6,-1.1,6,3.8,2,1.6,2,2.2,1.2,3.1,1.7,-2)
>shapiro.test(x)
Shapiro-Wilknormalitytest
data:
x
W=0.9699,p-value=0.7527
>shapiro.test(y)
Shapiro-Wilknormalitytest
data:
y
W=0.971,p-value=0.7754
ks检验:
>ks.test(x,"pnorm",mean(x),sd(x))
One-sampleKolmogorov-Smirnovtest
data:
x
D=0.1065,p-value=0.977
alternativehypothesis:
two-sided
Warningmessage:
Inks.test(x,"pnorm",mean(x),sd(x)):
cannotcomputecorrectp-valueswithties
>ks.test(y,"pnorm",mean(y),sd(y))
One-sampleKolmogorov-Smirnovtest
data:
y
D=0.1197,p-value=0.9368
alternativehypothesis:
two-sided
Warningmessage:
Inks.test(y,"pnorm",mean(y),sd(y)):
cannotcomputecorrectp-valueswithties
pearson拟合优度检验,以x为例。
>sort(x)
[1]-5.6-1.6-1.4-0.7-0.50.40.71.72.02.52.52.83.03.54.0[16]4.54.65.86.07.1
>x1<-table(cut(x,br=c(-6,-3,0,3,6,9)))
>p<-pnorm(c(-3,0,3,6,9),mean(x),sd(x))
>p
[1]0.048947120.249900090.620022880.900758560.98828138
>p<-c(p[1],p[2]-p[1],p[3]-p[2],p[4]-p[3],1-p[4]);p
[1]0.048947120.200952980.370122780.280735680.09924144
>chisq.test(x1,p=p)
Chi-squaredtestforgivenprobabilities
data:
x1
X-squared=0.5639,df=4,p-value=0.967
Warningmessage:
Inchisq.test(x1,p=p):
Chi-squaredapproximationmaybeincorrect
p值为0.967,接受原假设,x符合正态分布。
(2)
方差相同模型t检验:
>t.test(x,y,var.equal=TRUE)
TwoSamplet-test
data:
xandy
t=-0.6419,df=38,p-value=0.5248
alternativehypothesis:
truedifferenceinmeansisnotequalto095percentconfidenceinterval:
-2.3261791.206179
sampleestimates:
meanofxmeanofy
2.0652.625
方差不同模型t检验:
>t.test(x,y)
WelchTwoSamplet-test
data:
xandy
t=-0.6419,df=36.086,p-value=0.525
alternativehypothesis:
truedifferenceinmeansisnotequalto095percentconfidenceinterval:
-2.329261.20926
sampleestimates:
meanofxmeanofy
2.0652.625
配对t检验:
>t.test(x,y,paired=TRUE)
Pairedt-test
data:
xandy
t=-0.6464,df=19,p-value=0.5257
alternativehypothesis:
truedifferenceinmeansisnotequalto095percentconfidenceinterval:
-2.3731461.253146
sampleestimates:
meanofthedifferences
-0.56
三种检验的结果都显示两组数据均值无差异。
(3)
方差检验:
>var.test(x,y)
Ftesttocomparetwovariances
data:
xandy
F=1.5984,numdf=19,denomdf=19,p-value=0.3153
alternativehypothesis:
trueratioofvariancesisnotequalto195percentconfidenceinterval:
0.63265054.0381795
sampleestimates:
ratioofvariances
1.598361
接受原假设,两组数据方差相同。
Ex5.5
>a<-c(126,125,136,128,123,138,142,116,110,108,115,140)>b<-c(162,172,177,170,175,152,157,159,160,162)
正态性检验,采用ks检验:
>ks.test(a,"pnorm",mean(a),sd(a))
One-sampleKolmogorov-Smirnovtest
data:
a
D=0.1464,p-value=0.9266
alternativehypothesis:
two-sided
>ks.test(b,"pnorm",mean(b),sd(b))
One-sampleKolmogorov-Smirnovtest
data:
b
D=0.2222,p-value=0.707
alternativehypothesis:
two-sided
Warningmessage:
Inks.test(b,"pnorm",mean(b),sd(b)):
cannotcomputecorrectp-valueswithties
a和b都服从正态分布。
方差齐性检验:
>var.test(a,b)
Ftesttocomparetwovariances
data:
aandb
F=1.9646,numdf=11,denomdf=9,p-value=0.3200alternativehypothesis:
trueratioofvariancesisnotequalto195percentconfidenceinterval:
0.50219437.0488630
sampleestimates:
ratioofvariances
1.964622
可认为a和b的方差相同。
选用方差相同模型t检验:
>t.test(a,b,var.equal=TRUE)
TwoSamplet-test
data:
aandb
t=-8.8148,df=20,p-value=2.524e-08
alternativehypothesis:
truedifferenceinmeansisnotequalto0
95percentconfidenceinterval:
-48.24975-29.78358
sampleestimates:
meanofxmeanofy
125.5833164.6000
可认为两者有差别。
Ex5.6
二项分布总体的假设检验:
>binom.test(57,400,p=0.147)
Exactbinomialtest
data:
57and400
numberofsuccesses=57,numberoftrials=400,p-value=0.8876alternativehypothesis:
trueprobabilityofsuccessisnotequalto0.14795percentconfidenceinterval:
0.10974770.1806511
sampleestimates:
probabilityofsuccess
0.1425
P值>0.05,故接受原假设,表示调查结果支持该市老年人口的看法
Ex5.7
二项分布总体的假设检验:
>binom.test(178,328,p=0.5,alternative="greater")
Exactbinomialtest
data:
178and328
numberofsuccesses=178,numberoftrials=328,p-value=0.06794alternativehypothesis:
trueprobabilityofsuccessisgreaterthan0.595percentconfidenceinterval:
0.49576161.0000000
sampleestimates:
probabilityofsuccess
0.5426829
不能认为这种处理能增加母鸡的比例。
Ex5.8
利用pearson卡方检验是否符合特定分布:
>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
接受原假设,符合自由组合定律。
Ex5.9
利用pearson卡方检验是否符合泊松分布:
>n<-length(z)
>y<-c(92,68,28,11,1,0)
>x<-0:
5
>q<-ppois(x,mean(rep(x,y)));n<-length(y)
>p[1]<-q[1];p[n]=1-q[n-1]
>chisq.test(y,p=p)
Chi-squaredtestforgivenprobabilities
data:
y
X-squared=2.1596,df=5,p-value=0.8267
Warningmessage:
Inchisq.test(y,p=p):
Chi-squaredapproximationmaybeincorrect重新分组,合并频数小于5的组:
>z<-c(92,68,28,12)
>n<-length(z);p<-p[1:
n-1];p[n]<-1-q[n-1]
>chisq.test(z,p=p)
Chi-squaredtestforgivenprobabilities
data:
z
X
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