3 秩检验

3.1 主要理论

线性秩统计量：定义、基本性质
符号秩统计量：定义、基本性质，Wilcoxon符号秩检验（对称中心检验、成对数据检验）wilcox.test(paired = TRUE)
位置参数的检验：Wilcoxon秩和检验wilcox.test(paired = FALSE)，Mann-Witney检验（统计量与Wilcoxon秩和统计量等价）
尺度参数的检验：Mood检验mood.test、平方秩检验
多个独立样本问题：Kruskal-Waillis检验kruskal.test、Dunn检验DunnTest（DescTools包，当Kruskal-Waillis检验拒绝原假设）、Jonckheere-Terpstra检验JonckheereTerpstraTest（DescTools包）、
区组设计：Friedman检验friedman.test、Page检验、Cochran检验、Duribn检验durbin.test（PMCMRplus包）
相关分析：Spearman秩相关分析、Kendall \(\tau\) 相关检验cor.test、多变量Kendall协同系数检验KendallW（DescTools包）
线性回归的非参数方法cor.test

3.2 函数实现

将上述提到的平方秩检验，Page检验，Cochran检验进行编码实现。

3.2.1 平方秩检验

# 平方秩检验
# 输入：
# x,y:两总体样本向量
# alternative:备择假设方向
square_rank.test <- function(x, y, alternative = c("two.sided", "less", "greater"))
{
  if(!is.vector(x)||!is.vector(y)){
    stop("'x' and 'y' must be vector")
  }
  n1 <- length(x); n2 <- length(y); n <- n1+n2;
  mx <-  mean(x); my <-  mean(y);
  U <- abs(x-mx); V <- abs(y-my); # 绝对离差
  r <- rank(c(U,V),ties.method = "average") # 混合离差秩
  T <- sum(r[1:n1]^2)  # 相应于U的秩平方和
  z <- (T-n2*(n+1)*(2*n+1)/6)/sqrt(n1*n2*(n+1)*(2*n+1)*(8*n+11)/180)
  pl <- pnorm(z)
  pr <- pnorm(z,lower.tail = FALSE)
  Twarning <- (n1<=10|n2<=10)  # 样本数过小的提醒
  s <- ""
  # 如果打结
  if(length(unique(r))!=n){
    mrs <- mean(r^2)
    s2 <- n1*n2*sum(r^4)/(n*(n-1))-n1*n2*mrs^2/(n-1)
    T <- (T-n1*mrs)/sqrt(s2)  # 调整T
    pl <- pnorm(T)
    pr <- pnorm(T,lower.tail = FALSE)
    Twarning <- FALSE
    s <- "(with ties)"
  }
  # p值
  PVAL <- switch(alternative, less = pl, greater = pr, two.sided = 2*min(pl,pr))
  names(T) <- "square rank test statistic"
  names(n1) <- "the number of x"
  names(n2) <- "the number of y"
  names(n) <- "the number of x and y"
  l <- list(statistic = T, 
            parameter = c(n1,n2,n), 
            p.value = PVAL, alternative = alternative, 
            method = paste("Square rank test",s),
            data.name = paste(deparse(substitute(x)),"and",deparse(substitute(y))))
  if(Twarning){
    warning("The sizes of x and y are less than 10. The test may be inaccurate.")
  } 
  structure(l, class = "htest")
}

测试一下：

x <- c(74,75,76,79,82,65,86,58)
y <- c(72,73,69,74,68,75,67,76,66)
square_rank.test(x,y,alternative = "two.sided")
#> Warning in square_rank.test(x, y, alternative = "two.sided"): The sizes of x and
#> y are less than 10. The test may be inaccurate.
#> 
#>  Square rank test
#> 
#> data:  x and y
#> square rank test statistic = 1108, the number of x = 8, the number of y
#> = 9, the number of x and y = 17, p-value = 0.4
#> alternative hypothesis: two.sided

3.2.2 Page检验

备择假设为单调趋势。

# page检验
# 输入：
# x:区组设计矩阵，行处理，列区组
# alternative:备择假设，up为单调上升，down为单调下降
page.test <- function(x, alternative = c("increasing","decreasing"))
{
  if(!is.matrix(x)){
    stop("'x' must be matrix")
  }
  k <- nrow(x)  # 行数，处理
  b <- ncol(x)  # 列数，区组
  r <- apply(x, 2, rank) # 按列求秩
  R <- apply(r,1,sum) # 按行求和
  L <- sum((1:k)*R)  # Page检验统计量
  # 求标准差
  t <- unlist(apply(x,2,table))
  g <- sum(t^3-t) # 打结修正量
  sigma2 <- k*(k+1)*(b*k*(k^2-1)-g)/144
  z <- (L-b*k*(k+1)^2/4)/sqrt(sigma2)
  # p值
  PVAL <- switch(alternative, increasing = pnorm(z,lower.tail = FALSE), 
                 decreasing = pnorm(z))
  
  names(L) <- "Page test statistic"
  l <- list(statistic = L, 
            parameter = list(blocks=b,treatments=k), 
            p.value = PVAL, alternative = alternative, 
            method = "Page test",
            data.name = paste(deparse(substitute(x))))
  structure(l, class = "htest")
}

3.2.3 Cochran检验

只取二元数据的完全区组设计的假设检验问题。

# Cochran检验
# 输入：x为bool矩阵，行为处理
cochran.test <- function(x)
{
  if(!is.matrix(x)||!is.logical(x)){
    stop("'x' must be a logical matrix")
  }
  k <- nrow(x)  # 行数，处理
  b <- ncol(x)  # 列数，区组
  Ni <- apply(x, 1, sum)  # 行和
  Lj <- apply(x, 2, sum)  # 列和
  N <- sum(x)  # 总和
  Q <- (k*(k-1)^2*var(Ni))/(k*N-sum(Lj^2))  # 检验统计量
  # p值
  PVAL <- pchisq(Q,k-1,lower.tail = FALSE)
  names(Q) <- "Cochran test statistic"
  l <- list(statistic = Q, 
            parameter = list(df=k-1,blocks=b,treatments=k), 
            p.value = PVAL, 
            method = "Cochran test",
            data.name = deparse(substitute(x)))
  structure(l, class = "htest")
}

3.3 函数测试

对课本上的一些例子使用上述函数进行测试。先载入相关包：

library(DescTools)
library(PMCMRplus)

例5.2.2

Wilcoxon符号秩检验：

x <- c(34.3,35.8,35.4,34.8,35.2,35.1,35.0,35.5)-35
wilcox.test(x,alternative = "greater",mu=0,exact = FALSE)
#> 
#>  Wilcoxon signed rank test with continuity correction
#> 
#> data:  x
#> V = 20, p-value = 0.2
#> alternative hypothesis: true location is greater than 0

例5.2.3

Wilcoxon符号秩检验：

x <- c(42,51,31,61,44,55,48)
y <- c(38,53,36,52,33,49,36)
wilcox.test(x,y,mu = 0,alternative = "greater",paired = TRUE)
#> 
#>  Wilcoxon signed rank test
#> 
#> data:  x and y
#> V = 24, p-value = 0.05
#> alternative hypothesis: true location shift is greater than 0

例5.3.1

Wilcoxon秩和检验：

x <- c(1.20,1.63,2.26,1.87,2.20,1.30)
y <- c(0.94,1.26,1.15)
wilcox.test(x,y,mu = 0,alternative = "greater",paired = FALSE)
#> 
#>  Wilcoxon rank sum test
#> 
#> data:  x and y
#> W = 17, p-value = 0.02
#> alternative hypothesis: true location shift is greater than 0

例5.4.1

Mood检验：

x <- c(4.5,6.5,7.0,10.0,12.0)
y <- c(6.0,7.2,8.0,9.0,9.8)
mood.test(x,y,alternative = "two.sided")
#> 
#>  Mood two-sample test of scale
#> 
#> data:  x and y
#> Z = 2, p-value = 0.1
#> alternative hypothesis: two.sided

例5.4.2

平方秩检验：

x <- c(74,74,76,79,82,65,86,58)
y <- c(72,73,69,74,68,75,67,76,66)
square_rank.test(x,y,alternative = "two.sided")
#> 
#>  Square rank test (with ties)
#> 
#> data:  x and y
#> square rank test statistic = 1, the number of x = 8, the number of y =
#> 9, the number of x and y = 17, p-value = 0.2
#> alternative hypothesis: two.sided

例5.5.1

kruskal检验：

A <- c(73,64,67,62,70)
B <- c(84,80,81,77)
C <- c(82,79,71,75)
x <- list(A,B,C)
kruskal.test(x)
#> 
#>  Kruskal-Wallis rank sum test
#> 
#> data:  x
#> Kruskal-Wallis chi-squared = 8, df = 2, p-value = 0.01

这里课本的\(p\)值好像有误。

Dunn检验进行两两比较：

DunnTest(x)
#> 
#>  Dunn's test of multiple comparisons using rank sums : holm  
#> 
#>     mean.rank.diff   pval    
#> 2-1           7.30 0.0156 *  
#> 3-1           5.05 0.1065    
#> 3-2          -2.25 0.4139    
#> ---
#> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

例5.5.2

Jonckheere-Terpstra检验：

# 先构建列表
A <- c(260,200,240,170,270,205,190,200,250,200)
B <- c(310,310,190,225,170,210,280,210,280,240)
C <- c(225,260,360,310,270,380,240,295,260,250)
x <- list(A,B,C)
JonckheereTerpstraTest(x,alternative = "increasing")
#> Error in JonckheereTerpstraTest.default(x, alternative = "increasing"): group should be numeric or ordered factor

不知道为什么会报错，查了一下源代码，原因是当参数只给x作为一个列表时，会自动计算g，但是计算出来的g的一个无序的factor，下面的运算都需要g是一个有序的factor，不知道是不是代码的问题。

现手动计算g传入参数：

k <- length(x)
l <- sapply(x,length)
g <-ordered(rep(1:k, l))
x <- unlist(x)
JonckheereTerpstraTest(x,g,alternative = "increasing")
#> Warning in JonckheereTerpstraTest.default(x, g, alternative = "increasing"): Sample size > 100 or data with ties 
#>  p-value based on normal approximation. Specify nperm for permutation p-value
#> 
#>  Jonckheere-Terpstra test
#> 
#> data:  x and g
#> JT = 224, p-value = 0.002
#> alternative hypothesis: increasing

例5.6.1

Friedman检验：

# 先构建矩阵：行为区组，列为处理
# 每个处理的数据
A <- c(14,19,17,17,16,15,18,16)
B <- c(23,25,22,21,24,26,26,22)
C <- c(26,25,29,28,28,27,27,30)
D <- c(30,33,28,27,32,26,36,32)
y <- matrix(c(A,B,C,D),ncol = 4)
y
#>      [,1] [,2] [,3] [,4]
#> [1,]   14   23   26   30
#> [2,]   19   25   25   33
#> [3,]   17   22   29   28
#> [4,]   17   21   28   27
#> [5,]   16   24   28   32
#> [6,]   15   26   27   26
#> [7,]   18   26   27   36
#> [8,]   16   22   30   32

friedman.test(y)
#> 
#>  Friedman rank sum test
#> 
#> data:  y
#> Friedman chi-squared = 21, df = 3, p-value = 1e-04

例5.6.2

Page检验：

x <- c(40,52,80,52,76,100,34,52,51,35,53,65)
x <- matrix(x,3,4)
page.test(x,alternative = "increasing")
#> 
#>  Page test
#> 
#> data:  x
#> Page test statistic = 55, blocks = 4, treatments = 3, p-value = 0.007
#> alternative hypothesis: increasing

例5.6.3

Cochran检验：

x1 <- c(rep(1,8),0,rep(1,5),0)
x2 <- c(1,rep(0,3),1,1,0,1,0,0,0,rep(1,4))
x3 <- c(0,0,0,1,rep(0,7),1,0,0,0)
x <- matrix(as.logical(c(x1,x2,x3)),3,15,byrow = TRUE)
cochran.test(x)
#> 
#>  Cochran test
#> 
#> data:  x
#> Cochran test statistic = 14, df = 2, blocks = 15, treatments = 3,
#> p-value = 9e-04

例5.6.4

Durbin检验：

# 先构建矩阵，无实验数据用NA表示
A <- c(3.5,2.9,3.7,NA)
B <- c(3.7,3.1,NA,4.4)
C <- c(4.1,NA,4.9,5.8)
D <- c(NA,4.5,5.7,5.9)
x <- matrix(c(A,B,C,D),ncol = 4)
durbinTest(x)
#> 
#>  Durbin's rank sum test for a two-way
#>  balanced incomplete block design
#> 
#> data:  x
#> Durbin chi-squared = 8, df = 3, p-value = 0.06

例5.7.1

Spearman秩相关检验：

x <- c(452,318,310,409,405,332,497,321,406,413,334,467)
y <- c(107,147,151,120,123,135,100,143,117,118,141,100)
cor.test(x,y,alternative = "two.sided",method = "spearman",exact = F)
#> 
#>  Spearman's rank correlation rho
#> 
#> data:  x and y
#> S = 563, p-value = 2e-07
#> alternative hypothesis: true rho is not equal to 0
#> sample estimates:
#>   rho 
#> -0.97

例5.7.2

Kendall \(\tau\)相关检验：

x <- c(86,78,65,88,90,90,80,77,76,68,85,70)
y <- c(71,69,62,78,82,75,73,65,66,60,70,61)
cor.test(x,y,alternative = "two.sided",method = "kendall",exact = F)
#> 
#>  Kendall's rank correlation tau
#> 
#> data:  x and y
#> z = 4, p-value = 3e-04
#> alternative hypothesis: true tau is not equal to 0
#> sample estimates:
#>   tau 
#> 0.809

？？？\(p\)值又跟书本不同。

例5.7.3

Kendall协同系数检验：

# 先构建矩阵，列为评分机构
A <- c(12,9,2,4,10,7,11,6,8,5,3,1)
B <- c(10,1,3,12,8,7,5,9,6,11,4,2)
C <- c(11,8,4,12,2,10,9,7,5,6,3,1)
D <- c(9,1,2,10,12,6,7,4,8,5,11,3)
x <- matrix(c(A,B,C,D),ncol = 4)
KendallW(x,test = TRUE,correct = TRUE)
#> 
#>  Kendall's coefficient of concordance Wt
#> 
#> data:  x
#> Kendall chi-squared = 22, df = 11, subjects = 12, raters = 4, p-value =
#> 0.02
#> alternative hypothesis: Wt is greater 0
#> sample estimates:
#>    Wt 
#> 0.503

3.4 习题5

5.2

解：

对称中心检验问题： \[ \theta=320 \qquad v.s. \qquad \theta>320 \]

# 载入数据
x <- c(310,350,370,375,385,400,415,425,440,295,
          325,295,250,340,295,365,375,360,385)
wilcox.test(x,alternative = "greater",mu = 320,exact = FALSE)
#> 
#>  Wilcoxon signed rank test with continuity correction
#> 
#> data:  x
#> V = 158, p-value = 0.006
#> alternative hypothesis: true location is greater than 320

\(p\)值小于0.05，应当拒绝原假设。

5.3

解：

对称中心检验问题： \[ \theta=10 \qquad v.s. \qquad \theta\ne 10 \]

（1）符号检验：

d <- c(22,9,4,5,1,16,15,26,47,8,31,7)
x <- 10
S1 = sum(d<x) 
S2 = sum(d>x) 
S0 = sum(d==x) 
binom.test(x = S1,n = S1+S2,p = 0.5,alternative = "two.sided")
#> 
#>  Exact binomial test
#> 
#> data:  S1 and S1 + S2
#> number of successes = 6, number of trials = 12, p-value = 1
#> alternative hypothesis: true probability of success is not equal to 0.5
#> 95 percent confidence interval:
#>  0.211 0.789
#> sample estimates:
#> probability of success 
#>                    0.5

\(p\)值巨大，应当接受原假设。

（2）Wilcoxon符号秩检验：

wilcox.test(d,alternative = "two.sided",mu = x,exact = FALSE)
#> 
#>  Wilcoxon signed rank test with continuity correction
#> 
#> data:  d
#> V = 53, p-value = 0.3
#> alternative hypothesis: true location is not equal to 10

\(p\)值大于0.05也是接受原假设，但\(p\)值相对于符号检验要小很多。

5.4

解：

成对数据的检验，采用Wilcoxon符号秩检验。 \[ E(D)=0\qquad v.s. \qquad E(D)\ne 0 \]

x <- c(78,70,67,81,76,72,85,83)
y <- c(62,58,63,77,80,73,82,78)
w <- wilcox.test(x,y,mu = 0,alternative = "two.sided",paired = TRUE)
#> Warning in wilcox.test.default(x, y, mu = 0, alternative = "two.sided", : cannot
#> compute exact p-value with ties
w
#> 
#>  Wilcoxon signed rank test with continuity correction
#> 
#> data:  x and y
#> V = 31, p-value = 0.08
#> alternative hypothesis: true location shift is not equal to 0

有打结：

t <- table(rank(abs(x-y)))
t[t>1]
#> 4 
#> 3

只有1个结，长度为3.本来写好了一个函数用于修正统计量，但是扒了源代码发现关于打结的修正依旧包含在函数中了，只是当样本数少于50或打结时会出现提醒.可以利用命令查看源代码：

stats:::wilcox.test.default

所以在显著性水平0.05下仍然接受原假设，即认为幼儿园生活对孩子的社会知识没有影响。

5.5

解：

成对数据的检验，采用Wilcoxon符号秩检验。 \[ E(D)=0\qquad v.s. \qquad E(D)\ne 0 \]

x <- c(1149,1152,1176,1149,1155,1169,1182,1160,1129,1171)
y <- c(1116,1130,1184,1194,1184,1147,1125,1125,1166,1151)
wilcox.test(x,y,alternative = "two.sided",mu = 0,paired = TRUE)
#> Warning in wilcox.test.default(x, y, alternative = "two.sided", mu = 0, : cannot
#> compute exact p-value with ties
#> 
#>  Wilcoxon signed rank test with continuity correction
#> 
#> data:  x and y
#> V = 32, p-value = 0.7
#> alternative hypothesis: true location shift is not equal to 0

在显著性水平0.05下，接受原假设，即认为这段时间的股票指数的波动程度相同。

5.6

解：

位置参数的检验： \[ \delta=0\qquad v.s\qquad \delta\ne 0 \]

采用Wilcoxon秩和检验：

x <- c(134,146,104,119,124,161,112,83,113,129,97,123)
y <- c(70,118,101,85,107,132,94)
wilcox.test(x,y,alternative = "two.sided",mu = 0,paired = FALSE)  # paired = FALSE
#> 
#>  Wilcoxon rank sum test
#> 
#> data:  x and y
#> W = 63, p-value = 0.08
#> alternative hypothesis: true location shift is not equal to 0

在显著性水平0.05下应当接受原假设，即认为两种饲料对雌鼠的体重增加的影响不显著。

5.7

解：

位置参数的检验： \[ \delta=0\qquad v.s\qquad \delta\ne 0 \]

采用Wilcoxon秩和检验：

x <- c(52,49,54,47,56,55,45,57,55,54)
y <- c(49,48,39,44,40,50,36,41)
wilcox.test(x,y,alternative = "two.sided",mu = 0,paired = FALSE)
#> Warning in wilcox.test.default(x, y, alternative = "two.sided", mu = 0, : cannot
#> compute exact p-value with ties
#> 
#>  Wilcoxon rank sum test with continuity correction
#> 
#> data:  x and y
#> W = 72, p-value = 0.004
#> alternative hypothesis: true location shift is not equal to 0

在显著性水平0.05下应当拒绝原假设，即认为这两个厂产品的寿命不同。

5.8

解：

先载入数据：

x <- c(56,105,63,88,72,112,96,93,65,105,94,87,64,65,68,87)
y <- c(88,94,93,96,99,79,91,94,91,100,99,90,100,110,102,95)

1.成对数据的检验：

wilcox.test(x,y,mu = 0,alternative = "two.sided",paired = TRUE)
#> Warning in wilcox.test.default(x, y, mu = 0, alternative = "two.sided", : cannot
#> compute exact p-value with ties
#> 
#>  Wilcoxon signed rank test with continuity correction
#> 
#> data:  x and y
#> V = 29, p-value = 0.05
#> alternative hypothesis: true location shift is not equal to 0

2.位置参数的检验：

wilcox.test(x,y,mu = 0,alternative = "two.sided",paired = FALSE)
#> Warning in wilcox.test.default(x, y, mu = 0, alternative = "two.sided", : cannot
#> compute exact p-value with ties
#> 
#>  Wilcoxon rank sum test with continuity correction
#> 
#> data:  x and y
#> W = 72, p-value = 0.03
#> alternative hypothesis: true location shift is not equal to 0

3.尺度参数的检验：

mood.test(x,y,alternative = "two.sided")
#> 
#>  Mood two-sample test of scale
#> 
#> data:  x and y
#> Z = 2, p-value = 0.02
#> alternative hypothesis: two.sided

square_rank.test(x,y,alternative = "two.sided")
#> 
#>  Square rank test (with ties)
#> 
#> data:  x and y
#> square rank test statistic = 4, the number of x = 16, the number of y =
#> 16, the number of x and y = 32, p-value = 3e-04
#> alternative hypothesis: two.sided

综上，可以看到在显著性水平0.05下，应当认为两个学科的博士论文页数有显著差异，人均页数显著不同，页数离散程度显著不同。

5.9

解：

先载入数据：

x <- c(83,79,83,74,75,74,86,76,84,73,78,77,80,83,78)
y <- c(75,62,58,89,77,81,27,85,72,85,74,100,43,52,75)

1.成对数据的检验：

wilcox.test(x,y,mu = 0,alternative = "two.sided",paired = TRUE)
#> Warning in wilcox.test.default(x, y, mu = 0, alternative = "two.sided", : cannot
#> compute exact p-value with ties
#> 
#>  Wilcoxon signed rank test with continuity correction
#> 
#> data:  x and y
#> V = 82, p-value = 0.2
#> alternative hypothesis: true location shift is not equal to 0

2.位置参数的检验：

wilcox.test(x,y,mu = 0,alternative = "two.sided",paired = FALSE)
#> Warning in wilcox.test.default(x, y, mu = 0, alternative = "two.sided", : cannot
#> compute exact p-value with ties
#> 
#>  Wilcoxon rank sum test with continuity correction
#> 
#> data:  x and y
#> W = 142, p-value = 0.2
#> alternative hypothesis: true location shift is not equal to 0

3.尺度参数的检验：

mood.test(x,y,alternative = "two.sided")
#> 
#>  Mood two-sample test of scale
#> 
#> data:  x and y
#> Z = -3, p-value = 0.007
#> alternative hypothesis: two.sided

square_rank.test(x,y,alternative = "two.sided")
#> 
#>  Square rank test (with ties)
#> 
#> data:  x and y
#> square rank test statistic = -3, the number of x = 15, the number of y
#> = 15, the number of x and y = 30, p-value = 5e-04
#> alternative hypothesis: two.sided

综上，可以看到在显著性水平0.05下，应当认为两个学校的学生成绩没有显著差异，平均成绩没有显著不同，离散程度有显著不同。

5.10

解：

可以考虑尺度参数的双边检验，方差比较低说明加工精度比较高。

x <- c(18.0,17.1,16.4,16.9,16.9,16.7,16.7,17.2,17.5,16.9)
y <- c(17.0,16.9,17.0,16.9,17.2,17.1,16.8,17.1,17.1,16.2)

Mood检验：

mood.test(x,y,alternative = "two.sided")
#> 
#>  Mood two-sample test of scale
#> 
#> data:  x and y
#> Z = 2, p-value = 0.1
#> alternative hypothesis: two.sided

平方秩检验：

square_rank.test(x,y,alternative = "two.sided")
#> 
#>  Square rank test (with ties)
#> 
#> data:  x and y
#> square rank test statistic = 1, the number of x = 10, the number of y =
#> 10, the number of x and y = 20, p-value = 0.2
#> alternative hypothesis: two.sided

\(p\)值都大于0.05，应当接受原假设，即认为他们的水平（加工精度）一致。

5.11

解：

尺度参数的双边检验。

x <- c(8.8,8.2,5.6,4.9,8.9,4.2,3.6,7.1,5.5,8.6,6.3,3.9)
y <- c(13.0,14.5,22.8,20.7,19.6,18.4,21.3,24.2,19.6,11.7)

Mood检验：

mood.test(x,y,alternative = "two.sided")
#> 
#>  Mood two-sample test of scale
#> 
#> data:  x and y
#> Z = -0.5, p-value = 0.6
#> alternative hypothesis: two.sided

平方秩检验：

square_rank.test(x,y,alternative = "two.sided")
#> 
#>  Square rank test (with ties)
#> 
#> data:  x and y
#> square rank test statistic = -2, the number of x = 12, the number of y
#> = 10, the number of x and y = 22, p-value = 0.03
#> alternative hypothesis: two.sided

在显著性水平0.05下，mood检验表示应当接受原假设，而平方秩检验表示应当拒绝原假设。

但是mood检验需要要求两总体位置参数相等，利用位置参数的检验：

wilcox.test(x,y,alternative = "two.sided",mu = 0,paired = FALSE)
#> Warning in wilcox.test.default(x, y, alternative = "two.sided", mu = 0, : cannot
#> compute exact p-value with ties
#> 
#>  Wilcoxon rank sum test with continuity correction
#> 
#> data:  x and y
#> W = 0, p-value = 9e-05
#> alternative hypothesis: true location shift is not equal to 0

\(p\)值很小，应当拒绝原假设，即认为两总体的位置参数不等。为了使用mood检验，我们需要估计两总体位置参数，进而平移使他们相等。

# 平移两组样本
x1 <- x-mean(x)
y1 <- y-mean(y)
# 再进行mood检验
mood.test(x1,y1,alternative = "two.sided")
#> 
#>  Mood two-sample test of scale
#> 
#> data:  x1 and y1
#> Z = -2, p-value = 0.04
#> alternative hypothesis: two.sided

\(p\)值小于0.05，应当拒绝原假设。综上，应当认为两组数据的方差存在差异。

5.12

解：

尺度参数的双边检验。

x <- c(8.2,10.7,7.5,14.6,6.3,9.2,11.9,5.6,12.8,5.2,4.9,13.5)
y <- c(4.7,6.3,5.2,6.8,5.6,4.2,6.0,7.4,8.1,6.5)

先进行位置参数的检验：

wilcox.test(x,y,alternative = "two.sided",mu = 0,paired = FALSE)
#> Warning in wilcox.test.default(x, y, alternative = "two.sided", mu = 0, : cannot
#> compute exact p-value with ties
#> 
#>  Wilcoxon rank sum test with continuity correction
#> 
#> data:  x and y
#> W = 92, p-value = 0.03
#> alternative hypothesis: true location shift is not equal to 0

应当拒绝原假设，即认为他们的位置参数不相等。平移数据后再进行mood检验：

# 平移两组样本
x1 <- x-mean(x)
y1 <- y-mean(y)
# 再进行mood检验
mood.test(x1,y1,alternative = "two.sided")
#> 
#>  Mood two-sample test of scale
#> 
#> data:  x1 and y1
#> Z = 3, p-value = 0.004
#> alternative hypothesis: two.sided

平方秩检验：

square_rank.test(x,y,alternative = "two.sided")
#> Warning in square_rank.test(x, y, alternative = "two.sided"): The sizes of x and
#> y are less than 10. The test may be inaccurate.
#> 
#>  Square rank test
#> 
#> data:  x and y
#> square rank test statistic = 3110, the number of x = 12, the number of
#> y = 10, the number of x and y = 22, p-value = 1e-04
#> alternative hypothesis: two.sided

两个检验都表明应当拒绝原假设，即认为他们的尿酸浓度变异不相同。

5.13

解：

关于位置参数的多样本双边检验问题。

A <- c(80,203,236,252,284,368,457,393)
B <- c(133,180,100,160)
C <- c(156,295,320,448,465,481,279)
D <- c(194,214,272,330,386,475)
x <- list(A,B,C,D)

Kruskal-Waillis检验：

kruskal.test(x)
#> 
#>  Kruskal-Wallis rank sum test
#> 
#> data:  x
#> Kruskal-Wallis chi-squared = 8, df = 3, p-value = 0.04

在显著性水平0.05下应当拒绝原假设，即认为这四种药物的治疗效果不同。

5.14

解：

关于位置参数的多样本双边检验问题。

A <- c(830,910,940,890,890,960,910,920,900)
B <- c(910,900,810,830,840,830,880,910,890,840)
C <- c(1010,1000,910,930,960,950,940)
D <- c(780,820,810,770,790,810,800,810)
x <- list(A,B,C,D)
kruskal.test(x)
#> 
#>  Kruskal-Wallis rank sum test
#> 
#> data:  x
#> Kruskal-Wallis chi-squared = 26, df = 3, p-value = 1e-05

\(p\)值很小，应当拒绝原假设，即认为每种培育方法的水稻产量不相同。为了比较任意两种方法的水稻产量之间的差异，需要用的Dunn检验：

DunnTest(x)
#> 
#>  Dunn's test of multiple comparisons using rank sums : holm  
#> 
#>     mean.rank.diff    pval    
#> 2-1          -6.53  0.2438    
#> 3-1           7.74  0.2438    
#> 4-1         -17.02  0.0021 ** 
#> 3-2          14.27  0.0141 *  
#> 4-2         -10.49  0.0778 .  
#> 4-3         -24.76 8.6e-06 ***
#> ---
#> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

在显著性水平0.05下，方法4跟方法1、方法3跟方法2、方法4跟方法3的水稻产量有显著差异。

5.15

关于位置参数的多样本趋势性检验问题。

采用Jonckheere-Terpstra检验：

A <- c(125,136,116,101,105,109)
B <- c(122,114,132,120,119,127)
C <- c(128,142,128,134,135,132,140,129)
x <- list(A,B,C)
k <- length(x)
l <- sapply(x,length)
g <-ordered(rep(1:k, l))
x <- unlist(x)
JonckheereTerpstraTest(x,g)
#> Warning in JonckheereTerpstraTest.default(x, g): Sample size > 100 or data with ties 
#>  p-value based on normal approximation. Specify nperm for permutation p-value
#> 
#>  Jonckheere-Terpstra test
#> 
#> data:  x and g
#> JT = 112, p-value = 0.002
#> alternative hypothesis: two.sided

\(p\)值小于显著性水平0.05，应当拒绝原假设，即认为结论可靠。

5.16

解：

关于位置参数的多样本趋势性检验问题。

采用Jonckheere-Terpstra检验：

A <- c(40,35,38,43,44,41)
B <- c(38,40,47,44,40,42)
C <- c(48,40,45,43,46,48,44)
x <- list(A,B,C)
k <- length(x)
l <- sapply(x,length)
g <-ordered(rep(1:k, l))
x <- unlist(x)
JonckheereTerpstraTest(x,g)
#> Warning in JonckheereTerpstraTest.default(x, g): Sample size > 100 or data with ties 
#>  p-value based on normal approximation. Specify nperm for permutation p-value
#> 
#>  Jonckheere-Terpstra test
#> 
#> data:  x and g
#> JT = 91, p-value = 0.02
#> alternative hypothesis: two.sided

\(p\)值小于显著性水平0.05，应当拒绝原假设，即认为研究者的经验可靠。

5.17

解：

完全区组设计的Friedman检验：

A <- c(73,75,67,61,69,79)
B <- c(83,81,99,82,85,87)
C <- c(73,60,73,77,68,74)
D <- c(58,64,64,71,77,74)
E <- c(77,75,73,59,85,82)
y <- matrix(c(A,B,C,D,E),ncol = 5)
friedman.test(y)
#> 
#>  Friedman rank sum test
#> 
#> data:  y
#> Friedman chi-squared = 15, df = 4, p-value = 0.006

\(p\)值小于显著性水平0.05，应当拒绝原假设，即认为面积大小有差异。

5.18

解：

二元区组设计的Cochran检验：

A <- c(rep(1,9),0)
B <- c(1,1,0,1,0,1,0,0,1,1)
C <- c(0,0,1,1,0,0,0,0,0,1)
x <- as.logical(c(A,B,C))
x <- matrix(x,nrow = 3,byrow = TRUE)
cochran.test(x)
#> 
#>  Cochran test
#> 
#> data:  x
#> Cochran test statistic = 6, df = 2, blocks = 10, treatments = 3,
#> p-value = 0.05

\(p\) 值略小于显著性水平0.05，应当拒绝原假设，即认为顾客对这三种糕点的爱好并不相同。

5.19

解：

完全区组设计的Page检验，单调上升：

x1 <- c(36,51,71,63,82,128)
x2 <- c(62,91,40,51,33,81)
x3 <- c(53,81,67,75,116,38)
x4 <- c(105,63,49,65,107,33)
x5 <- c(36,46,62,63,42,104)
x6 <- c(118,65,126,96,122,112)
x7 <- c(42,108,123,32,69,102)
x8 <- c(51,63,55,86,41,121)
x9 <- c(114,51,30,109,97,86)
x <- matrix(c(x1,x2,x3,x4,x5,x6,x7,x8,x9),ncol=9)
page.test(x,alternative = "increasing")
#> 
#>  Page test
#> 
#> data:  x
#> Page test statistic = 685, blocks = 9, treatments = 6, p-value = 0.2
#> alternative hypothesis: increasing

\(p\)值大于显著性水平0.05，应当接受原假设，即认为论断不正确。

或者利用DescTools的PageTest函数：

PageTest(t(x))
#> 
#>  Page test for ordered alternatives
#> 
#> data:  t(x)
#> L = 685, p-value = 0.2

\(p\)值相差不大。

5.20

解：

平衡的不完全区组设计的Durbin检验：

A <- c(73,74,NA,71)
B <- c(NA,75,67,72)
C <- c(74,75,68,NA)
D <- c(75,NA,72,75)
x <- matrix(c(A,B,C,D),ncol = 4)
durbinTest(x)
#> 
#>  Durbin's rank sum test for a two-way
#>  balanced incomplete block design
#> 
#> data:  x
#> Durbin chi-squared = 7, df = 3, p-value = 0.06

\(p\)值大于显著性水平0.05，应当接受原假设，即认为四种饲料的品质无差别。

5.21

解：

5.22

解：

5.23

解：

Kendall协同系数检验：

A <- c(9,2,4,10,7,6,8,5,3,1)
B <- c(10,1,3,8,7,5,9,6,4,2)
C <- c(8,4,2,10,9,7,5,6,3,1)
D <- c(9,1,2,10,6,7,4,8,5,3)
x <- matrix(c(A,B,C,D),ncol = 4)
KendallW(x,test = TRUE,correct = TRUE)
#> 
#>  Kendall's coefficient of concordance Wt
#> 
#> data:  x
#> Kendall chi-squared = 31, df = 9, subjects = 10, raters = 4, p-value =
#> 3e-04
#> alternative hypothesis: Wt is greater 0
#> sample estimates:
#>    Wt 
#> 0.853

\(p\)值很小，应当拒绝原假设，即认为这些排序产生较为一致的效果。

5.24

解：

线性相关分析：

x <- c(58.8,61.4,71.3,74.4,76.7,70.7,57.5,46.4,39.1,48.5,70.0,70.1)
y <- c(8.4,9.27,8.73,6.36,8.50,7.82,9.14,8.24,9.57,9.58,8.11,6.83)
cor.test(x,y,alternative = "two.sided",method = "pearson")
#> 
#>  Pearson's product-moment correlation
#> 
#> data:  x and y
#> t = -3, df = 10, p-value = 0.03
#> alternative hypothesis: true correlation is not equal to 0
#> 95 percent confidence interval:
#>  -0.8822 -0.0789
#> sample estimates:
#>    cor 
#> -0.625

\(p\)值小于显著性水平0.05，应当拒绝原假设，即认为两者存在线性关系。