R语言基础题及答案——R语言与统计分析第六章课后习题(汤银才)

Posted 2021-10-31 slandarer

R语言与统计分析第六章课后习题(汤银才)

题-1

有一批枪弹, 出厂时, 其初速$v\sim N(950,{\sigma }^2)$（单位：$m/s$）. 经过较长时间储存, 取9发进行测试, 得样本值（单位：$m/s$）如下：$914,920,910,934,953,940,912,924,930$

据经验, 枪弹储存后其初速仍服从正态分布, 且标准差不变, 问是否可认为这批枪弹的初速有显著降低?$(\alpha =0.01)$

speed<-c(914, 920, 910, 934, 953, 940, 912, 924, 930)
t.test(speed, mu=950, conf.level = 0.99)

# p值=0.001<0.01, 不接收原假设
# 初速度有明显变化，且又有926.3 <950
# 可以认为弹药初速度有显著降低

One Sample t-test
.
data: speed
t = -5, df = 8, p-value = 0.001
alternative hypothesis: true mean is not equal to 950
99 percent confidence interval:
910.3 942.3
sample estimates:
mean of x
926.3

题-2

已知维尼纶纤度在正常条件下服从正态分布, 且标准差为0.048. 从某天 生产的产品中抽取5根纤维, 测得其纤度为: $1.32,1.55,1.36,1.40,1.1$, 问这天抽取的维尼纶纤度的总体标准差是否正常?$(\alpha =0.05)$

fiber<-c(1.32,1.55,1.36,1.40,1.1)

# 课本提供的chisq.var.test()函数：
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(alternative == "less"|alternative=="greater"){
    result$p.value<-p
  }else if (alternative=="two.sided") {
    if(p>.5)
    p<-1-p
    p<-2*p
    result$p.value<-p
  }else return("your input is wrong")
    result$conf.int<-c(
    (n-1)*v/qchisq(alpha/2, df=n-1, lower.tail=FALSE),
    (n-1)*v/qchisq(alpha/2, df=n-1, lower.tail=TRUE))
    result
}

chisq.var.test(fiber, 0.048^2, 0.05, alternative="two.sided")
# p值=4.992e-09<0.05, 不接收原假设, 总体标准差不正常

# library(TeachingDemos)
# sigma.test(x,sigmasq=0.048^2,conf.level=0.95,alternative="two.sided")
# 也可以使用TeachingDemos库所带sigma.test函数，计算值为5e-09<0.05

$var
[1] 0.02648
.
$chi2
[1] 45.97
.
$p.value
[1] 4.992e-09
.
$conf.int
[1] 0.009505 0.218654

题-3

下面给出两种型号的计算器充电以后所能使用的时间(单位:小时)的观测值:

型号A	5.5	5.6	6.3	4.6	5.3	5.0	6.2	5.8	5.1	5.2	5.9
型号B	3.8	4.3	4.2	4.9	4.5	5.2	4.8	4.5	3.9	3.7	3.6	2.9

设两样本独立且数据所属的两个总体的密度函数至多差一个平移量. 试问能否认为型号A的计算器平均使用时间比型号B来得长?$(\alpha =0.01)$

# 两个总体的密度函数至多差一个平移量说明方差相同，但未知，使用t检验：
A<-c(5.5,5.6,6.3,4.6,5.3,5.0,6.2,5.8,5.1,5.2,5.9)
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)

t.test(A,B,var.equal = TRUE)
# p值=3e-05<0.05, 故拒绝接收AB时长相等原假设,A所用时间显著比B长 

Two Sample t-test
.
data: A and B
t = 5.3, df = 21, p-value = 3e-05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.7955 1.8212
sample estimates:
mean of x mean of y
5.500 4.192

题-4

测得两批电子器件的样本的电阻$(\Omega )$为:

A批(x)	0.140	0.138	0.143	0.142	0.144	0.137
B批(y)	0.135	0.140	0.142	0.136	0.138	0.130

设这两批器材的电阻值分别服从正态分布$N({\mu }_1,{\sigma }_1^2)$和$N({\mu }_2,{\sigma }_2^2)$, 且两样本独立,

（1）试检验两个总体的方差是否相等?$(\alpha =0.01)$

（2）试检验两个总体的均值是否相等?$(\alpha =0.05)$

x<-c(0.140,0.138,0.143,0.142,0.144,0.137)
y<-c(0.135,0.140,0.142,0.136,0.138,0.130)

var.test(x, y, conf.level = 0.99)
# p值=0.4>0.01,接收H0，认为两个总体的方差相等

# 由于上一问认为方差相同，则这一问设置var.equal = TRUE
t.test(x,y,var.equal = TRUE)
# p值=0.09>0.05,接收H0，认为两个总体的均值相等

F test to compare two variances
.
data: x and y
F = 0.44, num df = 5, denom df = 5, p-value = 0.4
alternative hypothesis: true ratio of variances is not equal to 1
99 percent confidence interval:
0.02964 6.61491
sample estimates:
ratio of variances
0.4428
.
Two Sample t-test
.
data: x and y
t = 1.9, df = 10, p-value = 0.09
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.0007721 0.0084388
sample estimates:
mean of x mean of y
0.1407 0.1368

题-5

有人称某地成年人中大学毕业生比例不低于30%, 为检验之, 随机调查该地15名成年人, 发现有3名大学毕业生, 取$\alpha =0.05$, 问该人的看法是否成立?

binom.test(c(3,12),p=0.3,alternative = "less",conf.level = 0.95)
# p-value=0.3>0.05不拒绝原假设

Exact binomial test
.
data: c(3, 12)
number of successes = 3, number of trials = 15, p-value = 0.3
alternative hypothesis: true probability of success is less than 0.3
95 percent confidence interval:
0.0000 0.4398
sample estimates:
probability of success
0.2