Exercise 4.2

library(vcd)
## Warning: package 'vcd' was built under R version 3.5.3
## Loading required package: grid
data("Abortion", package = "vcdExtra")
str(Abortion)
##  'table' num [1:2, 1:2, 1:2] 171 152 138 167 79 148 112 133
##  - attr(*, "dimnames")=List of 3
##   ..$ Sex             : chr [1:2] "Female" "Male"
##   ..$ Status          : chr [1:2] "Lo" "Hi"
##   ..$ Support_Abortion: chr [1:2] "Yes" "No"
structure(Abortion)
## , , Support_Abortion = Yes
## 
##         Status
## Sex       Lo  Hi
##   Female 171 138
##   Male   152 167
## 
## , , Support_Abortion = No
## 
##         Status
## Sex       Lo  Hi
##   Female  79 112
##   Male   148 133
fourfold(Abortion)

b.

fourfold(aperm(Abortion))

c.

oddsratio(Abortion,log = F)
##  odds ratios for Sex and Status by Support_Abortion 
## 
##       Yes        No 
## 1.3614130 0.6338682
oddsratio(aperm(Abortion),log = F)
##  odds ratios for Support_Abortion and Status by Sex 
## 
##    Female      Male 
## 1.7567419 0.8179317
  1. From the fourfolds above, we can tell that when status is low, more females will support abortion. But when status is high, more males will support abortion.

Exersice 4.4 a.

library(vcd)
library(grid)
data("Hospital", package = "vcd")
str(Hospital)
##  'table' num [1:3, 1:3] 43 6 9 16 11 18 3 10 16
##  - attr(*, "dimnames")=List of 2
##   ..$ Visit frequency: chr [1:3] "Regular" "Less than monthly" "Never"
##   ..$ Length of stay : chr [1:3] "2-9" "10-19" "20+"
structure(Hospital)
##                    Length of stay
## Visit frequency     2-9 10-19 20+
##   Regular            43    16   3
##   Less than monthly   6    11  10
##   Never               9    18  16
HS=margin.table(Hospital,1:2)
HS
##                    Length of stay
## Visit frequency     2-9 10-19 20+
##   Regular            43    16   3
##   Less than monthly   6    11  10
##   Never               9    18  16
chisq.test(HS)
## 
##  Pearson's Chi-squared test
## 
## data:  HS
## X-squared = 35.171, df = 4, p-value = 4.284e-07
  1. Answer: based on chi squre value, and p value is very low =4.284e-07<0.0005 level of significance. It indicate that those 2 variables have a strong relation between length of stay and visit frequency
assocstats(HS)
##                     X^2 df   P(> X^2)
## Likelihood Ratio 38.353  4 9.4755e-08
## Pearson          35.171  4 4.2842e-07
## 
## Phi-Coefficient   : NA 
## Contingency Coeff.: 0.459 
## Cramer's V        : 0.365
assoc(HS,shade=TRUE)

Answer: In the association plot, Cell of regular visit-stay 2-9 days, and cell of never visit -stay 20+ days, the observed > expected frequency is colored as blue; cell of regular visit-stay 20+ days, and cell of never visit - stay 2-9 days that contain less than the expected frequency fall below it (and are shaded red). It indicates that strong association between stay period and visit frequency.

library(vcdExtra)
## Warning: package 'vcdExtra' was built under R version 3.5.3
## Loading required package: gnm
## Warning: package 'gnm' was built under R version 3.5.3
CMHtest(HS)
## Cochran-Mantel-Haenszel Statistics for Visit frequency by Length of stay 
## 
##                  AltHypothesis  Chisq Df       Prob
## cor        Nonzero correlation 29.138  1 6.7393e-08
## rmeans  Row mean scores differ 34.391  2 3.4044e-08
## cmeans  Col mean scores differ 29.607  2 3.7233e-07
## general    General association 34.905  4 4.8596e-07

Answer: The p-value is still less than 0.05, and both variables can be considered ordinal, the association between two variables still stay true.

Exercise 4.6 a.

library(vcdExtra)
data("Mammograms")
structure(Mammograms)
##           Reader1
## Reader2    Absent Minimal Moderate Severe
##   Absent       34      10        2      0
##   Minimal       6       8        8      2
##   Moderate      2       5        4     12
##   Severe        0       1        2     14
Kappa(Mammograms)
##             value     ASE      z  Pr(>|z|)
## Unweighted 0.3713 0.06033  6.154 7.560e-10
## Weighted   0.5964 0.04923 12.114 8.901e-34
agreementplot(Mammograms, main = "agreement")

assocstats(Mammograms)
##                     X^2 df   P(> X^2)
## Likelihood Ratio 92.619  9 4.4409e-16
## Pearson          83.516  9 3.2307e-14
## 
## Phi-Coefficient   : NA 
## Contingency Coeff.: 0.657 
## Cramer's V        : 0.503

Answer: The unweighted Kappa value shows a minimal agreement between rater1 and rater2, and the weighted Kappa value suggested stronger.The Likelihood ratio and Pearson Chi Square test also suggested the two raters has different opinions on rating.

Exercise 4.7 a.

data <- matrix(c(24,8,13,8,13,11,10,9,64), nrow=3, ncol = 3, byrow = TRUE)
rownames(data) <- c("Con", "Mixed", "Pro")
colnames(data) <- c("Con", "Mixed", "Pro")
data
##       Con Mixed Pro
## Con    24     8  13
## Mixed   8    13  11
## Pro    10     9  64
Kappa(data)
##             value     ASE     z  Pr(>|z|)
## Unweighted 0.3888 0.05979 6.503 7.870e-11
## Weighted   0.4269 0.06350 6.723 1.781e-11
agreementplot(data, main = "aggreement")