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1、精选优质文档-倾情为你奉上实验六 重要的参数检验与功效检验【实验类型】验证性【实验学时】2 学时【实验目的】1、掌握假设检验的基本思想；2、掌握重要的参数检验及功效检验的求解方法；3、了解非参数假设检验的基本思想及求解方法。【实验内容】1、参数检验（t 检验、F 检验、二项分布检验和泊松检验等）的计算；2、功效检验的计算；3、非参数检验（符号与秩检验、分布的检验、相关性检验等）的求解。【实验方法或步骤】第一部分、课件例题：#1 例6.2X0.05，不能拒绝原假设，接受H 0 ，即认为平均寿命不大于225h#2 例6.3X-c(78.1, 72.4, 76.2, 74.3, 77.4, 78.4
2、, 76.0, 75.5, 76.7, 77.3)Y-c(79.1, 81.0, 77.3, 79.1, 80.0, 79.1,79.1, 77.3, 80.2, 82.1)t.test(X, Y, al = l, var.equal = T) #H 1 : 1 - 2 0t.test(X, Y, al = l) #使用总体方差不同模型t.test(X, Y, al = l, paired = TRUE) #配对数据检验# 公式形式obtain-data.frame( value = c(78.1, 72.4, 76.2, 74.3, 77.4, 78.4, 76.0, 75.5, 76.7,
3、 77.3, 79.1, 81.0, 77.3, 79.1, 80.0, 79.1, 79.1, 77.3, 80.2, 82.1), group = gl(2, 10) #生成2水平、各有10个元素的因子向量t.test(value group, data = obtain, alternative = less, var.equal = TRUE)由于p值(= 0.000218)0.05，拒绝原假设，即接受H 1 ，再利用 1 - 2 的置信区间，可以说明新操作方法能够提高得率。#3 例6.4X-c(78.1, 72.4, 76.2, 74.3, 77.4, 78.4, 76.0, 75.
4、5, 76.7, 77.3)Y-c(79.1, 81.0, 77.3, 79.1, 80.0, 79.1, 79.1, 77.3, 80.2, 82.1)var.test(X, Y) #方差相等的F检验# 使用公式形式obtain0.05，无法拒绝原假设，所以认为两总体的方差是相同的。从方差比的置信区间0.37, 6.02来看，它包含1，因此，在例6.3中，假设两总体方差相同是合理的#4 例6.5prop.test (400, 10000, p=0.02) #比率(成功的概率)检验#5 例6.6n-c(2000,1500); x-c(652,576)prop.test(x,n) #双样本的比率
5、检验#6 例6.7X - matrix(c(15, 32, 10, 42, 42, 59), nrow=2, byrow=T)colnames(X) - c(30s,24s,20s)rownames(X) - c(Yes,No)XX.yes - XYes,X.total - margin.table(X,2) #计算表格的列和(参数1为行和)prop.test(X.yes, X.total) #三样本的比率检验pairwise.prop.test(X.yes, X.total) #比率的多重比较#7 例6.8binom.test(3, 100, p=0.01, al=g) #二项分布检验(单边
6、)#8 例6.9poisson.test(x=12, T=1.2, r=5, al=g) #Poisson分布检验(单边)#9 例6.10#t 检验X - c(1050, 960, 1120, 1250, 1280); mu0 - 1000t.test(X, mu = mu0, al = g) #原假设H 0 : 1 0 =1000# 计算功效power.t.test(n=length(X), delta=mean(X)-mu0, sd=sd(X), type=one.sample, alternative = one.side)# 计算给定功效下的试验次数power.t.test(power
7、=0.90, delta=mean(X)-mu0, sd=sd(X), type=one.sample, alternative = one.side)#10 例6.12power.prop.test(power=0.9, p1 = 652/2000, p2 = 576/1500)#11 例6.13binom.test( 3, 12, al=l, conf.level = 0.9 )#12 例6.14X 99), length(X)#13 例6.15X 140), n = 19) #n=19是去掉了M=140的项#14 例6.16x - rep(1:4, c(62, 41, 14, 11);
8、y - rep(1:4, c(20, 37, 16, 15)wilcox.test(x, y, conf.int = TRUE)#15 例6.17#计算样本均值与样本标准差，并作为总体参数的估计Z - scan(F:/文档/大学课程/R语言/ch06/exam.data); mu - mean(Z); S - sd(Z)#将整个实轴划分成8个子区间p - seq(from = 0.125, to = 0.875, by = 0.125)q - qnorm(p, mean = mu, sd = S); q#计算落入每个子区间的实际频数Y - table(cut(Z, breaks = c(-In
9、f, q, Inf); Y# 作拟合优度检验F - pnorm(q, mean = mu, sd = S); m - length(Y)p - F1; pm - 1 - Fm-1for (i in 2:(m-1) pi-Fi-Fi-1(chi- chisq.test(Y, p = p)#因用样本值代替总体参数, 自由度减少2, 需重新计算p值Pval - 1 - pchisq(chi$statistic, df = m - 3)names(Pval) - P_val; Pval#16 例6.18X - scan(F:/文档/大学课程/R语言/ch06/exam.data)shapiro.tes
10、t(X)#17 例6.19x - matrix(c(60, 3, 32, 11), nc = 2);chisq.test(x)#18 例6.20x - matrix(c(4, 5, 18, 6), nc = 2); fisher.test(x)#19 例6.21X - array(c(19, 0, 132, 9, 11, 6, 52, 97), dim = c(2, 2, 2)mantelhaen.test(X)#20 例6.22rt - read.table(F:/文档/大学课程/R语言/ch06/weight.data); rtwith(rt, cor.test(X,Y)cor.test(
11、 X + Y, data = rt) #公式形式#21 例6.23X - c(86, 77, 68, 91, 70, 71, 85, 87, 63)Y - c(88, 76, 64, 96, 65, 80, 81, 72, 60)cor.test(X, Y, method = kendall)#22 例6.24library(tseries)X-scan(what = ) #注意符号(共40)输入完成后回车一下+ + - + - - - + - + + + - - + - + + - - -+ - - + - - - - + + + + - - + - + + - runs.test(as.f
12、actor(X)第二部分、教材例题：#1 例6.2.1#z.test()函数定义z.test-function(x,n,sigma,alpha,u0=0,alternative=two.sided) options(digits=4) result-list( ) mean-mean(x) z-(mean-u0)/(sigma/sqrt(n) p-pnorm(z,lower.tail=FALSE) result$mean-mean result$z-z result$p.value-p if(alternative=two.sided) p-2*p result$p.value-p else
13、if (alternative = greater|alternative =less ) result$p.value-p else return(your input is wrong) result$conf.int- c( mean-sigma*qnorm(1-alpha/2,mean=0, sd=1, lower.tail = TRUE)/sqrt(n), mean+sigma*qnorm(1-alpha/2,mean=0, sd=1, lower.tail = TRUE)/sqrt(n) resultz.test(0.13, 25, 0.1, 0.05, u0=0.12, alte
14、rnative=less)#2 例6.2.2salt-c(490 , 506, 508, 502, 498, 511, 510, 515 , 512)t.test(salt, mu=500)#3 例6.2.3CaCo3-c(20.9, 20.41, 20.10, 20.00, 20.19, 22.60, 20.99, 20.41, 20, 23, 22)t.test(CaCo3, mu=20.7)#4 例6.2.4#函数chisq.var.test( )可以用来求 2 置信区间chisq.var.test - function (x,var,alpha,alternative=two.side
15、d) options(digits=4) result-list( ) n-length(x) v-var(x) result$var-v chi2-(n-1)*v/var result$chi2-chi2 p-pchisq(chi2,n-1) if(alternative = less|alternative=greater) result$p.value.5) p-1-p p-2*p result$p.value-p else return(your input is wrong) result$conf.int-c( (n-1)*v/qchisq(alpha/2, df=n-1, low
16、er.tail=F), (n-1)*v/qchisq(alpha/2, df=n-1, lower.tail=T) resulttime-c(42, 65, 75, 78, 59, 71, 57, 68, 54, 55)chisq.var.test(time, 80, 0.05, alternative=less)#5 例6.3.1x-c(20.5, 19.8, 19.7, 20.4, 20.1, 20.0, 19.0, 19.9)y-c(20.7, 19.8, 19.5, 20.8, 20.4, 19.6, 20.2)t.test(x, y, var.equal=TRUE)#6 例6.3.2
17、x-c(20.5, 19.8, 19.7, 20.4, 20.1, 20.0, 19.0, 19.9)y-c(20.7, 19.8, 19.5, 20.8, 20.4, 19.6, 20.2)var.test(x, y)#7 例6.4.1x-c(20.5, 18.8, 19.8, 20.9, 21.5, 19.5, 21.0, 21.2)y-c(17.7, 20.3, 20.0, 18.8, 19.0, 20.1, 20.0, 19.1)t.test(x, y, paired=TRUE)library(DAAG)data.x-c(20.5, 18.8, 19.8, 20.9, 21.5, 19
18、.5, 21.0, 21.2)data.y-c(17.7, 20.3, 20.0, 18.8, 19.0, 20.1, 20.0, 19.1)z-data.frame(data.x, data.y)onesamp(z, x=data.y, y=data.x)#8 例6.5.1binom.test(c(7, 5), p=0.4)prop.test(7, 12, p=0.4, correct=TRUE)#9 例6.5.2prop.test(35, 120, p=0.25, conf.level=0.975, correct=TRUE)#10 例6.6.1success-c(23, 25)total
19、-c(102, 135)prop.test(success, total)第三部分、课后习题：6.1#z.test()函数定义z.test-function(x,n,sigma,alpha,u0=0,alternative=two.sided) options(digits=4) result-list( ) mean-mean(x) z-(mean-u0)/(sigma/sqrt(n) p-pnorm(z,lower.tail=FALSE) result$mean-mean result$z-z result$p.value-p if(alternative=two.sided) p-2*p
20、 result$p.value-p else if (alternative = greater|alternative =less ) result$p.value-p else return(your input is wrong) result$conf.int- c( mean-sigma*qnorm(1-alpha/2,mean=0, sd=1, lower.tail = TRUE)/sqrt(n), mean+sigma*qnorm(1-alpha/2,mean=0, sd=1, lower.tail = TRUE)/sqrt(n) resultx-c(914,920,910,93
21、4,953,940,912,924,930)t.test(x, mu=950)结果如下：结论：因为p值=0.001 = 0.01, 故认为这批枪弹的初速有显著降低6.2#函数chisq.var.test( )可以用来求 2 置信区间chisq.var.test - function (x,var,alpha,alternative=two.sided) options(digits=4) result-list( ) n-length(x) v-var(x) result$var-v chi2-(n-1)*v/var result$chi2-chi2 p-pchisq(chi2,n-1) if
22、(alternative = less|alternative=greater) result$p.value.5) p-1-p p-2*p result$p.value-p else return(your input is wrong) result$conf.int-c( (n-1)*v/qchisq(alpha/2, df=n-1, lower.tail=F), (n-1)*v/qchisq(alpha/2, df=n-1, lower.tail=T) resultx = 0.05, 所以接受原假设，认为抽取的维尼纶纤度的总体标准差正常6.3a-c(5.5,5.6,6.3,4.6,5.
23、3,5.0,6.2,5.8,5.1,5.2,5.9)b-c(3.8,4.3,4.2,4.9,4.5,5.2,4.8,4.5,3.9,3.7,3.6,2.9)var.test(a,b)t-(mean(a)-mean(b)/(sqrt(10*var(a)+11*var(b)/21)*sqrt(1/11+1/12) #求t检验统计量t p = 0.05, 故接收原假设, 认为型号A的计算器平均使用时间比型号B长。6.4a1-c(0.140,0.138,0.143,0.142,0.144,0.137)b1 = 0.01, 故接收原假设, 认为两个总体的方差相等.结论: 因为p值=0.04549 = 0.05, 故接收原假设, 认为成年人中大学毕业生比例不低于30%.专心-专注-专业
