Test of Independence

Quantal data and trend tests

General test of independence

library(coin)
## Relationship between job satisfaction and income stratified by gender
## Agresti (2002, p. 288, Tab. 7.8)
jobsatisfaction

## Asymptotic generalized Cochran-Mantel-Haenszel test (Agresti, p. 297)
(ct <- cmh_test(jobsatisfaction)) # CMH = 10.2001

## The standardized linear statistic
statistic(ct, type = "standardized")

## The standardized linear statistic for each block
statistic(ct, type = "standardized", partial = TRUE)

## Asymptotic generalized Cochran-Mantel-Haenszel test (Agresti, p. 297)
## Note: 'Job.Satisfaction' as ordinal
cmh_test(jobsatisfaction,
         scores = list("Job.Satisfaction" = c(1, 3, 4, 5))) # L^2 = 9.0342

## Asymptotic linear-by-linear association test (Agresti, p. 297)
## Note: 'Job.Satisfaction' and 'Income' as ordinal
(lt <- lbl_test(jobsatisfaction,
                scores = list("Job.Satisfaction" = c(1, 3, 4, 5),
                              "Income" = c(3, 10, 20, 35))))
statistic(lt)^2 # M^2 = 6.1563

## The standardized linear statistic
statistic(lt, type = "standardized")

## The standardized linear statistic for each block
statistic(lt, type = "standardized", partial = TRUE)

Revisiting Cochran-Armitage Test for Trend in Proportions

## Change in clinical condition and degree of infiltration
## Cochran (1954, Tab. 6)
cochran <- matrix(
    c(11,  7,
      27, 15,
      42, 16,
      53, 13,
      11,  1),
    byrow = TRUE, ncol = 2,
    dimnames = list(
              "Change" = c("Marked", "Moderate", "Slight",
                           "Stationary", "Worse"),
        "Infiltration" = c("0-7", "8-15")
    )
)
cochran <- as.table(cochran)

## Asymptotic Pearson chi-squared test (Cochran, 1954, p. 435)
chisq_test(cochran) # X^2 = 6.88

## Asymptotic Cochran-Armitage test (Cochran, 1954, p. 436)
## Asymptotic Linear-by-Linear Association Test
## Note: 'Change' as ordinal
(ct <- chisq_test(cochran,
                  scores = list("Change" = c(3, 2, 1, 0, -1))))
statistic(ct)^2 # X^2 = 6.66


## Change in size of ulcer crater for two treatment groups
## Armitage (1955, Tab. 2)
armitage <- matrix(
    c( 6, 4, 10, 12,
      11, 8,  8,  5),
    byrow = TRUE, ncol = 4,
    dimnames = list(
        "Treatment" = c("A", "B"),
           "Crater" = c("Larger", "< 2/3 healed",
                        ">= 2/3 healed", "Healed")
    )
)
armitage <- as.table(armitage)

## Approximative (Monte Carlo) Pearson chi-squared test (Armitage, 1955, p. 379)
chisq_test(armitage,
           distribution = approximate(nresample = 10000)) # chi^2 = 5.91

## Approximative (Monte Carlo) Cochran-Armitage test (Armitage, 1955, p. 379)
(ct <- chisq_test(armitage,
                  distribution = approximate(nresample = 10000),
                  scores = list("Crater" = c(-1.5, -0.5, 0.5, 1.5))))
statistic(ct)^2 # chi_0^2 = 5.26

library(exact2x2)
library(coin)

# Example data structure
# Group 1: Control, Group 2: Low dose, Group 3: High dose
responses <- c(2, 5, 8)    # Number with response
totals <- c(10, 12, 15)    # Total in each group
scores <- c(0, 1, 2)       # Dose scores

# Method 1: Using exact2x2 package
exact_ca_test <- function(responses, totals, scores) {
  # Create data matrix
  successes <- responses
  failures <- totals - responses
  
  # Exact Cochran-Armitage test
  result <- cochran.armitage.test(
    x = cbind(successes, failures),
    scores = scores,
    alternative = "two.sided"
  )
  
  return(result)
}

result <- exact_ca_test(responses, totals, scores)
print(result)

library(coin)

# Create data frame
data <- data.frame(
  dose = rep(c(0, 1, 2), times = c(10, 12, 15)),
  response = c(rep(c(0,1), c(8,2)), 
               rep(c(0,1), c(7,5)), 
               rep(c(0,1), c(7,8)))
)
data$dose <- factor(data$dose)
datsum <- data %>%
  group_by(dose) %>%
  summarise(success = sum(response), total = n())
datsum ## need to generate the table for evaluation

# Exact linear-by-linear association test
exact_test <- lbl_test(response ~ dose, 
                       data = datsum, 
                       distribution = "exact",
                        scores = list("dose" = c(1, 2, 3))
                              )
print(exact_test)

     # Example data as a matrix
     dose_data <- matrix(c(10, 9, 10, 7, 0, 1, 0, 3), byrow=TRUE, nrow=2,
                         dimnames=list(resp=0:1, dose=0:3))
     print(dose_data)
        DescTools::CochranArmitageTest(dose_data, alternative = "one.sided")

library(CATTexact)

d <- c(1,2,3,4)
n <- rep(20,4)
r <- c(1,4,3,8)

catt_asy(d, n, r)

catt_exact(d, n, r)


d <- c(d,d)
n <- c(n,n)
r <- c(r,c(0,0,0,0))
catt_exact(d, n, r)