数学建模第五次作业

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1、1.区间估计解：R程序：X<-c(1067,919,1196,785,1126,936,918,1156,920,948);n<-length(X);xb<-mean(X);S2<-var(X);S<-sqrt(S2);T=(xb-1000)/(S/sqrt(n));结果：meandfab1997.19902.99651091.203p-value=0.5270268由此可知：置信区间为902.996到1096.203，该批灯泡能使用1000小时以上的概率为52.70268%。2.假设检测I解：假设H0<=225，H1>225此问题是单边检验输入

2、数据，调用函数t.test()R程序：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,alternative="greater",mu=225)结果：data:Xt=-3.4783,df=19,p-value=0.9987alternativehypothesis:truemeanisgreaterthan22595percentconfidenceinterval:175.8194Infsampleesti

3、mates:meanofx192.15由此可知：P值为0.9987，不能拒绝原假设，接受H0，即认为油漆工人的血小板计数与正常成年男子有差异，油漆作业对人体血小板计数有影响。3.假设检测II解：（1）两总体方差相同模型选择t检验法R程序：X<-c(113,120,138,120,100,118,138,123)Y<-c(138,116,125,136,110,132,130,110)t.test(X,Y,alternative="less",var.equal=TRUE)t.test(X,Y,alternative="less")结果：TwoSa

4、mplet-testdata:XandYt=-0.566,df=14,p-value=0.2902alternativehypothesis:truedifferenceinmeansislessthan095percentconfidenceinterval:-Inf7.12812sampleestimates:meanofxmeanofy121.250124.625（2）两总体方差不同模型R程序：X<-c(113,120,138,120,100,118,138,123)Y<-c(138,116,125,136,110,132,130,110)

5、var.test(X,Y)结果：Ftesttocomparetwovariancesdata:XandYF=1.2278,numdf=7,denomdf=7,p-value=0.7935alternativehypothesis:trueratioofvariancesisnotequalto195percentconfidenceinterval:0.24581036.1327511sampleestimates:ratioofvariances1.227800（3）成对数据模型R程序：X<-c(113,120,138,120,100,118,

6、138,123)Y<-c(138,116,125,136,110,132,130,110)var.test(X,Y,alternative="less",exact=F)结果：Ftesttocomparetwovariancesdata:XandYF=1.2278,numdf=7,denomdf=7,p-value=0.6033alternativehypothesis:trueratioofvariancesislessthan195percentconfidenceinterval:0.0000004.649733sampleestimate

7、s:ratioofvariances1.227800由此可知：（1）p=0.29(2)p=0.79(3)p=0.60，置信区间（1）>（2）>（3）。1.假设检测III解：由题可知所检验的问题为H0:p=p0=0.147,H1:p≠p0。调用binom.test()函数：binom.test(57,400,p=0.147)结果：data:57and400numberofsuccesses=57,numberoftrials=400,p-value=0.8876alternativehypothesis:trueprobabilityofsuc

8、cessisnotequalto0.14795percentconfidenceinterval:0.10974770.1806511s