> ####################5.2
> 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,conf.level = 0.95)#单边检验

	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 

> ###########################5.3这是一个经典的两样本比较问题
> 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')#常把我们想要的结果作为备胎h1

	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 

> t.test(X,Y,var.equal=F,alternative='less')#两组样本方差不同

	Welch Two Sample t-test

data:  X and Y
t = -4.2957, df = 17.319, p-value = 0.0002355
alternative hypothesis: true difference in means is less than 0
95 percent confidence interval:
    -Inf -1.9055
sample estimates:
mean of x mean of y 
    76.23     79.43 

> t.test(X,Y,var.equal=TRUE,alternative='less',paired=T)#两组样本数量相同

	Paired t-test

data:  X and Y
t = -4.2018, df = 9, p-value = 0.00115
alternative hypothesis: true difference in means is less than 0
95 percent confidence interval:
      -Inf -1.803943
sample estimates:
mean of the differences 
                   -3.2 

#实战我们将使用MASS包中的UScrime数据集。它包含了1960年美国47个州的刑 罚制度
#对犯罪率影响的信息。我们感兴趣的结果变量为Prob(监禁的概率)、U1(14~24岁年龄 
#段城市男性失业率)和U2(35~39岁年龄段城市男性失业率)。类别型变量So(指示该州是
#否位 于南方的指示变量)将作为分组变量使用

#如果你在美国的南方犯罪，是否更有可能被判监禁?
install.packages("MASS")
library(MASS)
attach(UScrime)
t.test(Prob~So)#默认原假设为监禁概率相同
detach(UScrime)

> ################################5.5方差检验
> 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)
> var.test(X,Y,alternative='two.side')

	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 

> ##############################5.6并非所有的样本总体都服从正态分布的假设！
> 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 

#######5.7
binom.test(1,400,p=0.01,alternative='less')

##################5.8拟合优度检验
X=c(210,312,170,85,223)
chisq.test(X)
length(X)
######################5.12单个总体的检验
X<-c(420, 500, 920, 1380, 1510, 1650, 1760, 2100, 2300, 2350)
ks.test(X, "pexp", 1/1500)
?ks.test

########################两个样本分布的比较
X<-rnorm(100)
Y<-runif(100)

ks.test(X,Y)
t.test(X,Y)
wilcox.test(X,Y)

#################################################5.14列联表的独立性检验
x<-c(60, 3, 32, 11)
dim(x)<-c(2, 2)
x
chisq.test(x, correct = FALSE)#一个逻辑指示在计算2乘2表时的测试统计量时是否应用连续性校正

#实际例子
install.packages("vcd")#关节炎治疗数据
help(package='vcd')
library(vcd)
help("Arthritis")
head(Arthritis)#查看列表的前几项
mytable=xtabs(~Treatment+Improved,data=Arthritis);mytable#创建一个列联表。接受治疗和改善水平之间有关系

chisq.test(mytable)
fisher.test(mytable)
#改善水平和性别的关系
mytable=xtabs(~Improved+Sex,data=Arthritis);mytable
chisq.test(mytable)
fisher.test(mytable)
#多维度列联表的检验;原假设是两变量在第三个变量的每一层中都是条件独立的
mytable=xtabs(~Treatment+Improved+Sex,data=Arthritis)
mantelhaen.test(mytable)

#以上为独立性检验，如果变量独立，那么是否会相关呢？
library(vcd)#二维列联表相关性度量
mytable=xtabs(~Treatment+Improved,data=Arthritis);mytable
assocstats(mytable)

###############################5.22相关检验(相关性的度量)
x<-c(1,2,3,4,5,6); y<-c(6,5,4,3,2,1)
cor.test(x, y, method = "spearman")#还有person相关，kendall相关

#实战
attach(state.x77)#基础包中的数据集，美国50个州在1977年的人口，收入，文盲率，预期寿命
#谋杀率，气温，土地面积等数据，我们想要判断预期寿命和谋杀率的关系
help(state.x77)
#install.packages("XQuartz");library(XQuartz);edit(state.x77);fix(state.x77)
head(state.x77)
states=state.x77[,1:6]#我们打算讨论除了气温和土地面积之外的所有变量的关系
cov(states)
cor(states)
pairs(state.x77)
cor.test(states[,4],states[,5])#预期寿命和谋杀率
detach(state.x77)
#cor.test 一次只能检验一对变量的相关性，而psych包中的corr.test()可以做的更多
library(psych)
corr.test(states,use='complete')#对所有变量做比较,prob 越小越显著

> states=state.x77[,1:6]#我们打算讨论除了气温和土地面积之外的所有变量的关系
> states
               Population Income Illiteracy Life Exp Murder HS Grad
Alabama              3615   3624        2.1    69.05   15.1    41.3
Alaska                365   6315        1.5    69.31   11.3    66.7
Arizona              2212   4530        1.8    70.55    7.8    58.1
Arkansas             2110   3378        1.9    70.66   10.1    39.9
California          21198   5114        1.1    71.71   10.3    62.6
Colorado             2541   4884        0.7    72.06    6.8    63.9
Connecticut          3100   5348        1.1    72.48    3.1    56.0
Delaware              579   4809        0.9    70.06    6.2    54.6
Florida              8277   4815        1.3    70.66   10.7    52.6
Georgia              4931   4091        2.0    68.54   13.9    40.6
Hawaii                868   4963        1.9    73.60    6.2    61.9
Idaho                 813   4119        0.6    71.87    5.3    59.5
Illinois            11197   5107        0.9    70.14   10.3    52.6
Indiana              5313   4458        0.7    70.88    7.1    52.9
Iowa                 2861   4628        0.5    72.56    2.3    59.0
Kansas               2280   4669        0.6    72.58    4.5    59.9
Kentucky             3387   3712        1.6    70.10   10.6    38.5
Louisiana            3806   3545        2.8    68.76   13.2    42.2
Maine                1058   3694        0.7    70.39    2.7    54.7
Maryland             4122   5299        0.9    70.22    8.5    52.3
Massachusetts        5814   4755        1.1    71.83    3.3    58.5
Michigan             9111   4751        0.9    70.63   11.1    52.8
Minnesota            3921   4675        0.6    72.96    2.3    57.6
Mississippi          2341   3098        2.4    68.09   12.5    41.0
Missouri             4767   4254        0.8    70.69    9.3    48.8
Montana               746   4347        0.6    70.56    5.0    59.2
Nebraska             1544   4508        0.6    72.60    2.9    59.3
Nevada                590   5149        0.5    69.03   11.5    65.2
New Hampshire         812   4281        0.7    71.23    3.3    57.6
New Jersey           7333   5237        1.1    70.93    5.2    52.5
New Mexico           1144   3601        2.2    70.32    9.7    55.2
New York            18076   4903        1.4    70.55   10.9    52.7
North Carolina       5441   3875        1.8    69.21   11.1    38.5
North Dakota          637   5087        0.8    72.78    1.4    50.3
Ohio                10735   4561        0.8    70.82    7.4    53.2
Oklahoma             2715   3983        1.1    71.42    6.4    51.6
Oregon               2284   4660        0.6    72.13    4.2    60.0
Pennsylvania        11860   4449        1.0    70.43    6.1    50.2
Rhode Island          931   4558        1.3    71.90    2.4    46.4
South Carolina       2816   3635        2.3    67.96   11.6    37.8
South Dakota          681   4167        0.5    72.08    1.7    53.3
Tennessee            4173   3821        1.7    70.11   11.0    41.8
Texas               12237   4188        2.2    70.90   12.2    47.4
Utah                 1203   4022        0.6    72.90    4.5    67.3
Vermont               472   3907        0.6    71.64    5.5    57.1
Virginia             4981   4701        1.4    70.08    9.5    47.8
Washington           3559   4864        0.6    71.72    4.3    63.5
West Virginia        1799   3617        1.4    69.48    6.7    41.6
Wisconsin            4589   4468        0.7    72.48    3.0    54.5
Wyoming               376   4566        0.6    70.29    6.9    62.9
> cov(states)
              Population      Income   Illiteracy     Life Exp      Murder      HS Grad
Population 19931683.7588 571229.7796  292.8679592 -407.8424612 5663.523714 -3551.509551
Income       571229.7796 377573.3061 -163.7020408  280.6631837 -521.894286  3076.768980
Illiteracy      292.8680   -163.7020    0.3715306   -0.4815122    1.581776    -3.235469
Life Exp       -407.8425    280.6632   -0.4815122    1.8020204   -3.869480     6.312685
Murder         5663.5237   -521.8943    1.5817755   -3.8694804   13.627465   -14.549616
HS Grad       -3551.5096   3076.7690   -3.2354694    6.3126849  -14.549616    65.237894
> cor(states)
            Population     Income Illiteracy    Life Exp     Murder     HS Grad
Population  1.00000000  0.2082276  0.1076224 -0.06805195  0.3436428 -0.09848975
Income      0.20822756  1.0000000 -0.4370752  0.34025534 -0.2300776  0.61993232
Illiteracy  0.10762237 -0.4370752  1.0000000 -0.58847793  0.7029752 -0.65718861
Life Exp   -0.06805195  0.3402553 -0.5884779  1.00000000 -0.7808458  0.58221620
Murder      0.34364275 -0.2300776  0.7029752 -0.78084575  1.0000000 -0.48797102
HS Grad    -0.09848975  0.6199323 -0.6571886  0.58221620 -0.4879710  1.00000000

> cor.test(states[,4],states[,5])#预期寿命和谋杀率

	Pearson's product-moment correlation

data:  states[, 4] and states[, 5]
t = -8.6596, df = 48, p-value = 2.26e-11
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
 -0.8700837 -0.6420442
sample estimates:
       cor 
-0.7808458 #所以是负相关

> corr.test(states,use='complete')#对所有变量做比较,prob 越小越显著
Call:corr.test(x = states, use = "complete")
Correlation matrix 
           Population Income Illiteracy Life Exp Murder HS Grad
Population       1.00   0.21       0.11    -0.07   0.34   -0.10
Income           0.21   1.00      -0.44     0.34  -0.23    0.62
Illiteracy       0.11  -0.44       1.00    -0.59   0.70   -0.66
Life Exp        -0.07   0.34      -0.59     1.00  -0.78    0.58
Murder           0.34  -0.23       0.70    -0.78   1.00   -0.49
HS Grad         -0.10   0.62      -0.66     0.58  -0.49    1.00
Sample Size 
[1] 50
Probability values (Entries above the diagonal are adjusted for multiple tests.) 
           Population Income Illiteracy Life Exp Murder HS Grad
Population       0.00   0.59       1.00      1.0   0.10       1
Income           0.15   0.00       0.01      0.1   0.54       0
Illiteracy       0.46   0.00       0.00      0.0   0.00       0
Life Exp         0.64   0.02       0.00      0.0   0.00       0
Murder           0.01   0.11       0.00      0.0   0.00       0
HS Grad          0.50   0.00       0.00      0.0   0.00       0

 To see confidence intervals of the correlations, print with the short=FALSE option

#以上大都是在讲两组比较，若是多组比较则需要方差分析的方法。这里仅介绍单因素方差分析
#多元方差分析参见《R语言实战》
#单因素方差分析中，你感兴趣的是比较分类因子定义的两个或多个组别中的因变量均值。以
#multcomp包中的cholesterol数据集为例(取自Westfall，Tobia，Rom，Hochberg，1999)，
#50 个患者均接受降低胆固醇药物治疗(trt)五种疗法中的一种疗法。其中三种治疗条件使用药
#物相同，分别是20 mg一天一次(1time)、10 mg一天两次(2times)和5 mg一天四次(4times)。
#剩下的两种方式(drugD和drugE)代表候选药物。哪种药物疗法降低胆固醇(响应变量)最多呢?
install.packages('multcomp')
library(multcomp)
attach(cholesterol)
head(cholesterol)
table(trt)
fit=aov(response~trt);summary(fit)#方差分析只能看出它们之间的不同，不能说明具体细节

#绘图表示细节
boxplot(response~trt)
install.packages('gplots')
library(gplots)
plotmeans(response~trt,xlab='Treatment',ylab='Response',main='difference between 5 treatment')
detach(cholesterol)