R语言：常用统计检验
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：
- 建立假设
- 求抽样分布
- 选择显著性水平和否定域
- 计算检验统计量
- 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,sigma2），其中mu、sigma2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”。
反馈与建议
- 作者：ShangFR
- 邮箱：shangfr@foxmail.com
