R语言:常用统计检验

来源:互联网 时间:2016-09-25

统计检验是将抽样结果和抽样分布相对照而作出判断的工作。主要分5个步骤:

  1. 建立假设
  2. 求抽样分布
  3. 选择显著性水平和否定域
  4. 计算检验统计量
  5. 判定 —— 百度百科

假设检验(hypothesis test)亦称显著性检验(significant test),是统计推断的另一重要内容,其目的是比较总体参数之间有无差别。假设检验的实质是判断观察到的“差别”是由抽样误差引起还是总体上的不同,目的是评价两种不同处理引起效应不同的证据有多强,这种证据的强度用概率P来度量和表示。除t分布外,针对不同的资料还有其他各种检验统计量及分布,如F分布、X2分布等,应用这些分布对不同类型的数据进行假设检验的步骤相同,其差别仅仅是需要计算的检验统计量不同。

正态总体均值的假设检验

t检验

t.test() => Student's t-Test

require(graphics)

t.test(1:10, y = c(7:20)) # P = .00001855

t.test(1:10, y = c(7:20, 200)) # P = .1245 -- 不在显著

## 经典案例: 学生犯困数据

plot(extra ~ group, data = sleep)

## 传统表达式

with(sleep, t.test(extra[group == 1], extra[group == 2]))

Welch Two Sample t-test

data: extra[group == 1] and extra[group == 2]

t = -1.8608, df = 17.776, p-value = 0.07939

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-3.3654832 0.2054832

sample estimates:

mean of x mean of y

0.75 2.33

## 公式形式

t.test(extra ~ group, data = sleep)

Welch Two Sample t-test

data: extra by group

t = -1.8608, df = 17.776, p-value = 0.07939

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-3.3654832 0.2054832

sample estimates:

mean in group 1 mean in group 2

0.75 2.33

单个总体

  • 某种元件的寿命X(小时)服从正态分布N(mu,sigma^2),其中mu、sigma^2均未知,16只元件的寿命如下;问是否有理由认为元件的平均寿命大于255小时。

X<-c(159, 280, 101, 212, 224, 379, 179, 264,

222, 362, 168, 250, 149, 260, 485, 170)

t.test(X, alternative = "greater", mu = 225)

One Sample t-test

data: X

t = 0.66852, df = 15, p-value = 0.257

alternative hypothesis: true mean is greater than 225

95 percent confidence interval:

198.2321 Inf

sample estimates:

mean of x

241.5

两个总体

  • X为旧炼钢炉出炉率,Y为新炼钢炉出炉率,问新的操作能否提高出炉率?

X<-c(78.1,72.4,76.2,74.3,77.4,78.4,76.0,75.5,76.7,77.3)

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, var.equal=TRUE, alternative = "less")

Two Sample t-test

data: X and Y

t = -4.2957, df = 18, p-value = 0.0002176

alternative hypothesis: true difference in means is less than 0

95 percent confidence interval:

-Inf -1.908255

sample estimates:

mean of x mean of y

76.23 79.43

成对数据t检验

  • 对每个高炉进行配对t检验

X<-c(78.1,72.4,76.2,74.3,77.4,78.4,76.0,75.5,76.7,77.3)

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, alternative = "less")

One Sample t-test

data: X - Y

t = -4.2018, df = 9, p-value = 0.00115

alternative hypothesis: true mean is less than 0

95 percent confidence interval:

-Inf -1.803943

sample estimates:

mean of x

-3.2

正态总体方差的假设检验

var.test() => F Test to Compare Two Variances

x <- rnorm(50, mean = 0, sd = 2)

y <- rnorm(30, mean = 1, sd = 1)

var.test(x, y) # x和y的方差是否相同?

var.test(lm(x ~ 1), lm(y ~ 1)) # 相同.

  • 从小学5年级男生中抽取20名,测量其身高(厘米)如下;问:在0.05显著性水平下,平均值是否等于149,sigma^2是否等于75?

X<-scan()

136 144 143 157 137 159 135 158 147 165

158 142 159 150 156 152 140 149 148 155

var.test(X,Y)

F test to compare two variances

data: X and Y

F = 34.945, num df = 19, denom df = 9, p-value = 6.721e-06

alternative hypothesis: true ratio of variances is not equal to 1

95 percent confidence interval:

9.487287 100.643093

sample estimates:

ratio of variances

34.94489

  • 对炼钢炉的数据进行分析

X<-c(78.1,72.4,76.2,74.3,77.4,78.4,76.0,75.5,76.7,77.3)

Y<-c(79.1,81.0,77.3,79.1,80.0,79.1,79.1,77.3,80.2,82.1)

var.test(X,Y)

F test to compare two variances

data: X and Y

F = 1.4945, num df = 9, denom df = 9, p-value = 0.559

alternative hypothesis: true ratio of variances is not equal to 1

95 percent confidence interval:

0.3712079 6.0167710

sample estimates:

ratio of variances

1.494481

二项分布的总体检验

  • 有一批蔬菜种子的平均发芽率为P=0.85,现在随机抽取500粒,用种衣剂进行浸种处理,结果有445粒发芽,问种衣剂有无效果。

binom.test(445,500,p=0.85)

Exact binomial test

data: 445 and 500

number of successes = 445, number of trials = 500, p-value = 0.01207

alternative hypothesis: true probability of success is not equal to 0.85

95 percent confidence interval:

0.8592342 0.9160509

sample estimates:

probability of success

0.89

  • 按照以往经验,新生儿染色体异常率一般为1%,某医院观察了当地400名新生儿,有一例染色体异常,问该地区新生儿染色体是否低于一般水平?

binom.test(1,400,p=0.01,alternative="less")

Exact binomial test

data: 1 and 400

number of successes = 1, number of trials = 400, p-value = 0.09048

alternative hypothesis: true probability of success is less than 0.01

95 percent confidence interval:

0.0000000 0.0118043

sample estimates:

probability of success

0.0025

非参数检验

数据是否正态分布的Neyman-Pearson 拟合优度检验-chisq

  • 5种品牌啤酒爱好者的人数如下

    A 210

    B 312

    C 170

    D 85

    E 223

    问不同品牌啤酒爱好者人数之间有没有差异?

X<-c(210, 312, 170, 85, 223)

chisq.test(X)

Chi-squared test for given probabilities

data: X

X-squared = 136.49, df = 4, p-value < 2.2e-16

  • 检验学生成绩是否符合正态分布

X<-scan()

25 45 50 54 55 61 64 68 72 75 75

78 79 81 83 84 84 84 85 86 86 86

87 89 89 89 90 91 91 92 100

A<-table(cut(X, br=c(0,69,79,89,100)))

#cut 将变量区域划分为若干区间

#table 计算因子合并后的个数

p<-pnorm(c(70,80,90,100), mean(X), sd(X))

p<-c(p[1], p[2]-p[1], p[3]-p[2], 1-p[3])

chisq.test(A,p=p)

Chi-squared test for given probabilities

data: A

X-squared = 8.334, df = 3, p-value = 0.03959

#均值之间有无显著区别

大麦的杂交后代芒性状的比例 无芒:长芒: 短芒=9:3:4,而实际观测值为335:125:160 ,检验观测值是否符合理论假设?

chisq.test(c(335, 125, 160), p=c(9,3,4)/16)

Chi-squared test for given probabilities

data: c(335, 125, 160)

X-squared = 1.362, df = 2, p-value = 0.5061

  • 现有42个数据,分别表示某一时间段内电话总机借到呼叫的次数,

    接到呼叫的次数 0 1 2 3 4 5 6

    出现的频率 7 10 12 8 3 2 0

    问:某个时间段内接到的呼叫次数是否符合Possion分布?

x<-0:6

y<-c(7,10,12,8,3,2,0)

mean<-mean(rep(x,y))

q<-ppois(x,mean)

n<-length(y)

p[1]<-q[1]

p[n]<-1-q[n-1]

for(i in 2:(n-1))

p[i]<-1-q[i-1]

chisq.test(y, p= rep(1/length(y), length(y)) )

Chi-squared test for given probabilities

data: y

X-squared = 19.667, df = 6, p-value = 0.003174

Z<-c(7, 10, 12, 8)

n<-length(Z); p<-p[1:n-1]; p[n]<-1-q[n-1]

chisq.test(Z, p= rep(1/length(Z), length(Z)))

Chi-squared test for given probabilities

data: Z

X-squared = 1.5946, df = 3, p-value = 0.6606

P值越小越有理由拒绝无效假设,认为总体之间有差别的统计学证据越充分。需要注意:不拒绝H0不等于支持H0成立,仅表示现有样本信息不足以拒绝H0。

传统上,通常将P>0.05称为“不显著”,0.0l<P≤0.05称为“显著”,P≤0.0l称为“非常显著”。

注:本文参考来自张金龙科学网博客。

反馈与建议

相关阅读:
Top