Chapter 9 Practical Sheet Solutions

9.1 Practical 1 - Contingency Tables

9.1.1 Contingency Table Construction

We here provide solutions to the practical exercises of Section 8.1.1.4.

These questions involve using the contingency table from the penguin data introduced in Section 8.1.1.3

Use addmargins to add row and column sum totals to the contingency table of penguin data.

penguins_table <- addmargins( penguins_data )
penguins_table

##            Island
## Species     Biscoe Dream Torgersen Sum
##   Adelie        44    56        52 152
##   Chinstrap      0    68         0  68
##   Gentoo       124     0         0 124
##   Sum          168   124        52 344

Use prop.table to obtain a contingency table of proportions.

penguins_prop <- prop.table( penguins_data )
penguins_prop_table <- addmargins( penguins_prop )
penguins_prop_table

##            Island
## Species        Biscoe     Dream Torgersen       Sum
##   Adelie    0.1279070 0.1627907 0.1511628 0.4418605
##   Chinstrap 0.0000000 0.1976744 0.0000000 0.1976744
##   Gentoo    0.3604651 0.0000000 0.0000000 0.3604651
##   Sum       0.4883721 0.3604651 0.1511628 1.0000000

Display the column-conditional probabilities, and use addmargins to add the column sums as an extra row at the bottom of the matrix (note: this should be a row of \(1\)’s).

penguins_prop_1 <- prop.table( penguins_data, 2 )
penguins_prop_1

##            Island
## Species        Biscoe     Dream Torgersen
##   Adelie    0.2619048 0.4516129 1.0000000
##   Chinstrap 0.0000000 0.5483871 0.0000000
##   Gentoo    0.7380952 0.0000000 0.0000000

# Add margin sums only over the rows...
penguins_prop_1_table <- addmargins( penguins_prop_1, margin = 1 )
penguins_prop_1_table

##            Island
## Species        Biscoe     Dream Torgersen
##   Adelie    0.2619048 0.4516129 1.0000000
##   Chinstrap 0.0000000 0.5483871 0.0000000
##   Gentoo    0.7380952 0.0000000 0.0000000
##   Sum       1.0000000 1.0000000 1.0000000

# sum is 1 in each column as expected.

Suppose I want the overall conditional proportions of penguin specie to appear in a final column on the right of the table. How would I achieve this?

# One way is as follows...

# Take the row sums (sum over the columns) on the initial 
# contingency table data.
penguins_prop_2_table <- addmargins( penguins_data, 2 ) 

# Calculate column-conditional proportions from this table.
penguins_prop_2_table <- prop.table( penguins_prop_2_table, 2 ) 

# Add column sums at the bottom of the table.
penguins_prop_2_table <- addmargins( penguins_prop_2_table, 1 ) 

penguins_prop_2_table

##            Island
## Species        Biscoe     Dream Torgersen       Sum
##   Adelie    0.2619048 0.4516129 1.0000000 0.4418605
##   Chinstrap 0.0000000 0.5483871 0.0000000 0.1976744
##   Gentoo    0.7380952 0.0000000 0.0000000 0.3604651
##   Sum       1.0000000 1.0000000 1.0000000 1.0000000

9.1.2 Chi-Square Test of Independence

We here provide solutions to the practical exercise of Section 8.1.2.1.

For the penguin data, apply the \(\chi^2\) test of independence between penguin specie and island of residence, and interpret the results.

# Apply the above tests on the Penguins data.
chisq.test( penguins_data )

## 
##  Pearson's Chi-squared test
## 
## data:  penguins_data
## X-squared = 299.55, df = 4, p-value < 2.2e-16

# Unsurprisingly the p-value is miniscule.

9.1.3 Barplots

We here provide solutions to the practical exercises of Section 8.1.3.1.

Run barplot( DR_prop ). What does the plot show?

barplot( DR_prop )

Figure 9.1: Barplots of the Dose-Result contingency table data.

We see the pair of barplots shown in Figure 9.1. In R, a matrix input to boxplot results in the column categories (in this case, Result) defining the bars while the row categories (Dose) form the stacked levels.

Investigate the density argument of the function barplot by both running the commands below, and also looking in the help file.

par(mfrow=c(1,3))
barplot( DR_prop, density = 70 )
barplot( DR_prop, density = 30 )
barplot( DR_prop, density = 0 )

Figure 9.2: Barplots of the Dose-Result contingency table data, investigating the density parameter.

As illustrated in Figure 9.2, we can see that the density argument of boxplot changes the level of shading for the (two) categories defined by the rows (Dose) of the contingency table matrix.

Add a title, and x- and y-axis labels, to the plot above.

We can achieve this with the code below, which yields the plot shown in Figure 9.3.

barplot( DR_prop, density = 30, main = "Comparison of Dose by Response",
         xlab = "treatment outcome", ylab = "proportions" )

Figure 9.3: Barplots of the Dose-Result contingency table data, now with title and axis labels.

Use the help file for barplot to find out how to add a legend to the plot.

We set the argument legend.text to T. The code below thus yields the plot shown in Figure 9.4.

barplot( DR_prop, density = 30, legend.text = T, 
         main = "Comparison of Dose by Response", 
         xlab = "treatment outcome", ylab = "proportions" )

Figure 9.4: Barplots of the Dose-Result contingency table data, now with a legend.

How would we alter the call to barplot in order to view dose proportion levels conditional on result (instead of the overall proportions corresponding to each cell). You may wish to use some of the table manipulation commands from Section 8.1.1.

One possible solution is shown below, which yields the plot shown in Figure 9.5. Note that we can move the legend box using args.legend and setting the usual x and y location arguments for a legend within a list (that is, if we don’t want it covering up part of the plot itself). The exact necessary values of x and y depend on your plotting window.

barplot( prop.table( DR_data, 2 ), density = 30, legend.text = T, 
         main = "Comparison of Dose by Response", 
         xlab = "treatment outcome", ylab = "proportions", 
         args.legend = list( x = 2.5, y = 1.25 )  )

Figure 9.5: Barplots of dose level proportions corresponding to each treatment outcome.

Suppose instead that we wish to display each dose level in a bar, with the proportion of successes and failures illustrated by the shading in each bar. How would we do that?

We can alter the code as shown below, which runs prop.table() on the transposed version of the matrix DR_data, and yields the plot shown in Figure 9.6.

barplot( prop.table( t( DR_data ), 2 ), density = 30, legend.text = T,
         main = "Comparison of Response by Dose", 
         xlab = "dose level", ylab = "proportions", 
         args.legend = list( x = 2.6, y = 1.25 ) )

Figure 9.6: Barplots of treatment outcome proportions corresponding to each dose level.

This is of course the more informative way to display the data, as we are likely more interested in the relative proportion of high dose successes to low dose successes, rather than, say, the proportion of successes which happened to be high dose, for example.

9.1.4 Sieve Diagrams

We here provide solutions to the practical exercises of Section 8.1.3.3.

library(vcd)
sieve( DR_data )

Figure 9.7: Sieve diagram comparing observed frequencies in relation to expected-under-independence frequencies for the Dose-Result data.

What is shown?

We see the plot given in Figure 9.7. A sieve (or parquet) diagram represents for an \(I \times J\) table the expected frequencies under independence as a collection of \(IJ\) rectangles, each containing a set of smaller squares/rectangles. Each of the larger rectangles has height and width proportional to the corresponding row and column marginal frequencies of the contingency table respectively. This way, the area of each rectangle is proportional to the expected-under-independence frequency for the corresponding cell. The number of smaller squares/rectangles in each larger rectangle corresponds to the observed frequency, hence a larger number of smaller squares indicate that the observed frequency was larger.

Now run

sieve( DR_data, shade = T )

Figure 9.8: Sieve diagram comparing observed frequencies in relation to expected-under-independence frequencies for the Dose-Result data.

Does this make the data easier or harder to visualise?

We now see the plot given in Figure 9.8, which is the same plot as Figure 9.7, however now the observed frequency squares are coloured blue (undashed) if the observed frequency is larger than the expected frequency, and red (dashed) if the observed frequency is smaller than the expected frequency.

Finally, run

sieve( DR_data, sievetype = "expected", shade = T )

Figure 9.9: Sieve diagram showing expected-under-independence frequencies for the Dose-Result data.

What is shown now?

The plot shown in Figure 9.9, which just shows the expected frequencies under independence for each cell.

9.1.5 Odds Ratios in R

We here provide solutions to the practical exercises of Section 8.1.4.

This section seeks to test your understanding of odds ratios for \(2 \times 2\) contingency tables, as well as your ability to write simple functions in R.

Write a function to compute the odds ratio of the success of event A with probability pA against the sucess of event B with probability pB.

Odds_Ratio <- function(pA, pB){(pA/(1-pA))/(pB/(1-pB))}
Odds_Ratio( 1/3, 1/2 ) # test the function.

## [1] 0.5

Write a function to compute the odds ratio for a \(2 \times 2\) contingency table. Test it on the Dose-response data above.

OR <- function( M ){ ( M[1,1] * M[2,2] ) / ( M[1,2] * M[2,1] ) }
OR( DR_data )

## [1] 1.600601

# M is a 2 by 2 matrix representing contingency table cell counts.

Will there be an issue running your function from part (b) if exactly one of the cell counts of the supplied matrix is equal to zero?

Fine in this case as either 0 is returned if a cell count of 0 appears in the numerator of the odds ratio, or Inf is returned if a cell count of 0 is in the denominator of the odds ratio.

What about if both cells of a particular row or column of the supplied matrix are equal to zero?

We have a problem as there will be a zero in both numerator and denominator of the odds ratio, which is undefined. For example, run…

AB <- matrix( c(3,4,0,0), ncol = 2 ) # Define such a matrix.
OR( AB ) # NaN returned.

## [1] NaN

We consider two possible options for amending the function in this case.

1. First option: ensure that your function terminates and returns a clear error message of what has gone wrong and why when a zero would be found to be in both the numerator and denominator of the odds ratio. Hint: The command stop can be used to halt execution of a function and display an error message.

OR_alt_1 <- function( M ){ 
  
  M_row_sum <- rowSums( M )
  M_col_sum <- colSums( M )
  
  if( prod( M_row_sum ) == 0 | prod( M_col_sum ) == 0 ){
    stop( "At least one row sum or column sum of M is equal 
          to zero, hence the odds ratio is undefined." )
  }
  else{ 
    ( M[1,1] * M[2,2] ) / ( M[1,2] * M[2,1] ) 
  }
}

OR_alt_1( DR_data )  # works fine with DR_data

## [1] 1.600601

OR_alt_1( AB ) # Execution is halted and our error returned.

## Error in OR_alt_1(AB): At least one row sum or column sum of M is equal 
##           to zero, hence the odds ratio is undefined.

1. Second option: in the case that a row or column of zeroes is found, add 0.5 to each cell of the table before calculating the odds ratio in the usual way. Make sure that your function returns a clear warning (as opposed to error) message explaining that an alteration to the supplied table was made before calculating the odds ratio because there was a row or column of zeroes present. Hint: The command warning() can be used to display a warning message (but not halt execution of the function).

OR_alt_2 <- function( M ){ 
  
  M_row_sum <- rowSums( M )
  M_col_sum <- colSums( M )
  
  if( prod( M_row_sum ) == 0 | prod( M_col_sum ) == 0 ){
    M <- M + 0.5 * matrix( 1, nrow = 2, ncol = 2 )
    OR <- ( M[1,1] * M[2,2] ) / ( M[1,2] * M[2,1] )
    warning( "At least one row sum or column sum of the supplied 
             matrix M was equal to zero, hence an amendment of 0.5 was added 
             to the value of each sum prior to calculating the odds ratio." )
    return( OR )
  }
  else{ 
    ( M[1,1] * M[2,2] ) / ( M[1,2] * M[2,1] ) 
  }
}

OR_alt_2( DR_data ) # works fine with DR_data

## [1] 1.600601

OR_alt_2( AB ) # Alternative odds ratio calculated and warning returned.

## Warning in OR_alt_2(AB): At least one row sum or column sum of the supplied 
##              matrix M was equal to zero, hence an amendment of 0.5 was added 
##              to the value of each sum prior to calculating the odds ratio.

## [1] 0.7777778

9.1.6 Mushrooms

We here provide some possible analysis of the mushrooms data, with comments made in R, to get you started. The resulting barplots and sieve diagram are shown in Figures 9.10 and 9.11 respectively.

# We create a matrix with the data in.
mushroom_data <- matrix( c(101, 399, 57, 487, 
                           12, 389, 150, 428 ), byrow = TRUE, ncol = 4 )

# Add dimension names as follows.
dimnames( mushroom_data ) <- list( Edibility = c("Edible", "Poisonous"),
                                   Cap_Shape = c("bell", "flat", "knobbed",
                                                 "convex/conical") )

# Have a look.
mushroom_data

##            Cap_Shape
## Edibility   bell flat knobbed convex/conical
##   Edible     101  399      57            487
##   Poisonous   12  389     150            428

# Let's look at the proportion of each shape of mushroom that are 
# edible or poisonous.
mushroom_table <- addmargins( mushroom_data, 2, FUN = mean )
mushroom_table <- prop.table( mushroom_table, 2 )
mushroom_table <- addmargins(mushroom_table, 1 )
mushroom_table

##            Cap_Shape
## Edibility        bell      flat   knobbed convex/conical      mean
##   Edible    0.8938053 0.5063452 0.2753623      0.5322404 0.5160652
##   Poisonous 0.1061947 0.4936548 0.7246377      0.4677596 0.4839348
##   Sum       1.0000000 1.0000000 1.0000000      1.0000000 1.0000000

# Let's perform a chi-square test.
chisq.test( mushroom_data )

## 
##  Pearson's Chi-squared test
## 
## data:  mushroom_data
## X-squared = 113.84, df = 3, p-value < 2.2e-16

# This matches the result found manually in the lectures.

# Barplot
barplot( prop.table( mushroom_data, margin = 2 ), density = 50, 
         main = "Comparison of Edibility by Cap Shape", cex.main = 0.8,
         xlab = "treatment outcome", ylab = "proportions", 
         legend.text = T, args.legend = list( x = 5.1, y = 1.25 ) )

Figure 9.10: Barplots for the mushroom data of edibility proportions corresponding to each cap shape.

# Sieve diagram
sieve( mushroom_data  )

Figure 9.11: Sieve diagram showing expected-under-independence frequencies for the mushroom data.

# Much can be drawn form this diagram.  For example,
# we can see that there are far more edible bell mushrooms in our 
# sample than would be expected under an assumption of independence.

9.2 Practical 2 - Contingency Tables

Practical 2 is designed to be exploratory, hence it is not possible to provide comprehensive solutions. We discuss some analysis of each dataset in accordance with the suggestions provided.

9.2.1 Mushrooms

Inputting the mushrooms data and some exploratory analysis was presented in Section 9.1.6.

We can assign the result of the \(\chi^2\) test to a named object as follows

chisq_Mush <- chisq.test( mushroom_data )
chisq_Mush

## 
##  Pearson's Chi-squared test
## 
## data:  mushroom_data
## X-squared = 113.84, df = 3, p-value < 2.2e-16

This matches the result found manually in the lectures. Very strong evidence to reject the null hypothesis of independence between edibility and cap shape.

9.2.1.1 Residual Analysis

We can use the \(\chi^2\) test result to obtain Pearson residuals…

chisq_Mush$residuals

##            Cap_Shape
## Edibility        bell       flat   knobbed convex/conical
##   Edible     5.589590 -0.3798221 -4.820746      0.6810948
##   Poisonous -5.772167  0.3922285  4.978209     -0.7033419

…and also Adjusted residuals.

chisq_Mush$stdres

##            Cap_Shape
## Edibility        bell       flat   knobbed convex/conical
##   Edible     8.269282 -0.6987975 -7.314099       1.322946
##   Poisonous -8.269282  0.6987975  7.314099      -1.322946

Both residuals suggest that bell-capped mushrooms and knobbed-capped mushrooms contribute most towards the \(X^2\) statistic. For example, there are far more edible bell-capped mushrooms than would be expected if the two variables were independent.

9.2.1.2 GLR Test

We can apply the \(G^2\) GLR test on the mushroom data as follows.

G2mush <- G2( mushroom_data )
G2mush$p.value

## [1] 0

G2mush$dev_res

##            Cap_Shape
## Edibility       bell      flat   knobbed convex/conical
##   Edible    10.53326 -3.895337 -8.462185       5.482674
##   Poisonous -6.03326  3.933403 11.005295      -5.394469

G2mush$Gij

##            Cap_Shape
## Edibility        bell      flat   knobbed convex/conical
##   Edible    110.94946 -15.17365 -71.60858       30.05971
##   Poisonous -36.40022  15.47166 121.11651      -29.10030

The \(p\)-value is close to 0, as expected, hence strong evidence to reject \(\mathcal{H}_0\). The deviance residuals are as calculated in lectures. Due to the relation between deviance residuals and the \(G^2\) statistic, classes of individual variables with large marginal sums of deviance residual values contribute most towards the \(G^2\) statistic. In this case, we again notice that the bell cap-shaped mushrooms have a larger contribution toward the \(G^2\) statistic, as evidenced by the large residual values, and also the large marginal sum in residual values over the two classes of edibility for bell-shaped caps.

9.2.1.3 Nominal Odds Ratios

Odds ratios for cells \((1,1)\), \((1,2)\) and \((1,3)\) to reference cell \((I,J)\), as calculated in lectures, are obtained by.

nominal_OR( mushroom_data )

##          [,1]      [,2]     [,3]
## [1,] 7.396988 0.9014427 0.333963

9.2.2 Dose-Result Data

Input of data can be performed as follows:

DoseResult <- matrix( c(47, 25, 12,
                        36, 22, 18,
                        41, 60, 55), byrow = TRUE, ncol = 3 )

dimnames( DoseResult ) <- list( Dose=c("High",
                                       "Medium",
                                       "Low"),
                               Result=c("Success",
                                        "Partial",
                                        "Failure") )

DoseResultTable <- addmargins(DoseResult)

Functions for local and global odds ratios respectively, then applied on the Dose-Result data to produce the results obtained in lectures, are:

local_OR <- function( M ){

  # I and J
  I <- nrow(M)
  J <- ncol(M)

  # Odds ratio matrix.
  OR_local <- matrix( NA, nrow = I-1, ncol = J-1 )
  for( i in 1:(I-1) ){
    for( j in 1:(J-1) ){
      OR_local[i,j] <- OR( M = M[c(i,i+1), c(j,j+1)] )
    }
  }

  return(OR_local)

}

global_OR <- function( M ){

  # I and J
  I <- nrow(M)
  J <- ncol(M)

  # Odds ratio matrix.
  OR_global <- matrix( NA, nrow = I-1, ncol = J-1 )
  for( i in 1:(I-1) ){
    for( j in 1:(J-1) ){
      OR_global[i,j] <- OR( M = matrix( c( sum( M[1:i,1:j] ),
                                           sum( M[(i+1):I,1:j] ),
                                           sum( M[1:i,(j+1):J] ),
                                           sum( M[(i+1):I,(j+1):J] ) ),
                                           nrow = 2 ) )
    }
  }

  return(OR_global)

}

local_OR( DoseResult )

##          [,1]     [,2]
## [1,] 1.148889 1.704545
## [2,] 2.394678 1.120370

global_OR( DoseResult )

##          [,1]     [,2]
## [1,] 2.557038 2.754717
## [2,] 3.023440 2.359736

A plot of multiple fourfold plots can be obtained using the following function, then applied on the Dose-Result data to produce the plots shown in Figure 9.12.

ffold_local <- function ( data ){

  # I and J
  I <- nrow(data)
  J <- ncol(data)

  par( mfrow = c(I-1, J-1) )
  for( i in 1:(I-1) ){
    for( j in 1:(J-1) ){
      sub_data <- data[c(i,i+1),c(j,j+1)]
      fourfoldplot( sub_data )
    }
  }

}

# Apply fourfold local OR plot function.
ffold_local( DoseResult )

Figure 9.12: Four-fold plots of Dose-Result data

To compute the statistic for the linear trend test manually, we could…

score_x <- 1:3
score_y <- 1:3

score_xy <- crossprod( t(score_x), t(score_y) )
n <- sum(DoseResult)

sum_xy <- sum( score_xy * DoseResult )
sum_x <- sum( score_x * rowSums(DoseResult) )
sum_y <- sum( score_y * colSums(DoseResult) )
sum_x2 <- sum( score_x^2 * rowSums(DoseResult) )
sum_y2 <- sum( score_y^2 * colSums(DoseResult) )

rxy <- ( n * sum_xy - sum_x * sum_y ) / ( sqrt( n * sum_x2 - sum_x^2 ) *
                                            sqrt( n * sum_y2 - sum_y^2 ) )

M2 <- (n-1) * rxy^2

p.value <- 1-pchisq(M2,1)

Or using the function provided we can simply do

linear.trend( table = DoseResult, x = score_x, y = score_y )

## $r
## [1] 0.2709127
## 
## $M2
## [1] 23.11902
## 
## $p.value
## [1] 1.522773e-06

yielding the same result. The test provides evidence to reject the null hypothesis of independence between the two variables.

9.2.3 Titanic

A partial table cross-classifying Class and Sex for Age = Child and Survived = No can be obtained in the usual way of selecting particular elements of an array.

Titanic[,, Age="Child", Survived="No"]

##       Sex
## Class  Male Female
##   1st     0      0
##   2nd     0      0
##   3rd    35     17
##   Crew    0      0

Marginal tables are produced using margin.table() with argument margin denoting the variables of interest (that is, those to still be included in the table). For example, a marginal table of Class and Sex summing over margins 3 and 4 (Age and Survived, thus ignoring them), and a marginal vector for Survived, are obtained as follows

margin.table( Titanic, margin = c(1,2) )

##       Sex
## Class  Male Female
##   1st   180    145
##   2nd   179    106
##   3rd   510    196
##   Crew  862     23

margin.table( Titanic, margin = 4 )

## Survived
##   No  Yes 
## 1490  711

We can use previous odds ratio functions on relevant \(2\times2\) marginal/partial tables. Note that whether we can treat Class as ordinal is debatable, depending on how category Crew might fit into that. If we feel that we can treat Class as ordinal, then global Odds ratios might also be applicable. We calculate nominal, local and global marginal odds ratios for Class and Sex having marginalised over Age and Survived.

nominal_OR( margin.table( Titanic, margin = c(1,2) ) )

##            [,1]
## [1,] 0.03312265
## [2,] 0.04505757
## [3,] 0.06942800

local_OR( margin.table( Titanic, margin = c(1,2) ) )

##           [,1]
## [1,] 0.7351185
## [2,] 0.6489826
## [3,] 0.0694280

global_OR( margin.table( Titanic, margin = c(1,2) ) )

##            [,1]
## [1,] 0.26012139
## [2,] 0.22830253
## [3,] 0.05187198

The nominal marginal odds ratios reflect the fact the proportion of men (or more, precisely, the odds that a randomly selected person of a particular Class is a man) is much lower for all other levels of Class relative to Crew. Both local and global marginal odds ratios provide evidence that there is an association between Class and Sex, particularly for Crew in comparison with any of the other levels. For calculating global odds ratios, we here treat Crew as the fourth category of an ordinal variable Class following on from 3rd. This position in the ordinal ranking (as opposed to being otherwise positioned in the Class ordering) vastly influences the global odds ratios (but less so the local ones, which are applicable for nominal variables anyway).

A Chi square test for independence of Class and Survival, marginalising over Sex and Age, is obtained by

chisq.test( margin.table( Titanic, margin = c(1,4) ) )

## 
##  Pearson's Chi-squared test
## 
## data:  margin.table(Titanic, margin = c(1, 4))
## X-squared = 190.4, df = 3, p-value < 2.2e-16

# An unsurprising overwhelming amount of evidence
# that these two are associated.

A linear trend test - applicability depends if we feel that we can treat Class as an ordinal variable.

linear.trend( table = margin.table( Titanic, margin = c(1,4) ),
              x = 1:4, y = 1:2 )

## $r
## [1] -0.2713954
## 
## $M2
## [1] 162.042
## 
## $p.value
## [1] 0

A sieve and mosaic plot could be obtained by running the following commands.

sieve( Titanic )
mosaic( Titanic )

9.3 Practical 3 - Contingency Tables and LLMs

9.3.1 Titanic

The command created a sub-table by marginalising over Age and then permuting the dimension order, so that we now have a \(2 \times 2 \times 4\) contingency table array.
Mantel Haenszel Test carried out as follows (continuity correction can be applied similar to previous tests).

mantelhaen.test( TitanicA, correct = FALSE )

## 
##  Mantel-Haenszel chi-squared test without continuity correction
## 
## data:  TitanicA
## Mantel-Haenszel X-squared = 362.67, df = 1, p-value < 2.2e-16
## alternative hypothesis: true common odds ratio is not equal to 1
## 95 percent confidence interval:
##   8.232629 14.185153
## sample estimates:
## common odds ratio 
##          10.80653

An unsurprisingly very small \(p\)-value giving evidence that Sex and Survived are not conditionally independent given Class.

The Fourfold plot given in Figure 9.13 is generated as follows.

par(mfrow = c(1,1))
fourfoldplot( TitanicA )

Figure 9.13: Four-fold plots of Titanic data

Visual evidence that Sex and Survived are not conditionally independent given Class.

The submatrix marginalises out Age and only considers interest to lie in subjects in either of the two categories 3rd or Crew for Class. Note that we do this because parts (e) and (f) consider the question of whether subjects in 3rd class or crew had a greater probability of survival.
We marginalise over Sex as follows and calculate the odds ratio:

TitanicB_margin <- margin.table( TitanicB, c(1,3) )
OR( TitanicB_margin )

## [1] 0.9344041

This suggests that 3rd class passengers had greater odds of survival than did Crew. (Note that the Odds Ratio is defined in terms of the odds of “Surviving: No” in this case).

Now condition on gender and calculate the odds ratios.

OR( TitanicB[,1,] )

## [1] 1.37422

OR( TitanicB[,2,] )

## [1] 7.851852

Both of these are positive, suggesting that both Male and Female Crew had greater odds of survival than did Male or Female 3rd class passengers respectively.

By looking at the appropriate marginal and conditional tables…

prop.table( TitanicB_margin, margin = 1 )

##       Survived
## Class         No       Yes
##   3rd  0.7478754 0.2521246
##   Crew 0.7604520 0.2395480

prop.table( TitanicB[,1,], margin = 1 )

##       Survived
## Class         No       Yes
##   3rd  0.8274510 0.1725490
##   Crew 0.7772622 0.2227378

prop.table( TitanicB[,2,], margin = 1 )

##       Survived
## Class         No       Yes
##   3rd  0.5408163 0.4591837
##   Crew 0.1304348 0.8695652

we can see that indeed the proportion of 3rd class passengers surviving overall was greater than that for Crew, however, when we condition on gender, a higher proportion of Male and Female Crew survived than did Male or Female 3rd Class passengers respectively.

This apparent contradiction occurs because the relationship between survival and class is influenced by the confounding variable Sex, which we ignore in the table marginalising over Sex. The conditional probability tables show that the survival rate was much lower for men compared to women on board the Titanic (perhaps because of the “women and children first” policy) when it came to getting a seat on a lifeboat. If we compare the row-proportional marginal table over survival…

prop.table( margin.table( TitanicB, margin = c(1,2) ), margin = 1 )

##       Sex
## Class       Male    Female
##   3rd  0.7223796 0.2776204
##   Crew 0.9740113 0.0259887

… this shows that there were a greater proportion of women in third class than in the crew – more than a quarter of third class passengers were women whereas less than \(3\%\) of the crew were Female. In this case, the difference in the proportion of women amongst the third class passengers and crew has a substantial impact on the overall survival rate for each group and has the unfortunate effect of masking the third class passengers vs crew effect when gender is ignored in the original table. Hence we have an example of Simpson’s Paradox, where trends within sub-populations can be reversed when the data are aggregated.

9.3.2 Log linear Models: Titanic

Explicitly fit several models (i.e. without using the step function), including the independence model and the fully saturated model. Comment on your findings.

The independence model is fitted as follows:

( Titanic.I.fit <- loglm( ~ Class + Survived + Sex + Age, data=Titanic ) )

## Call:
## loglm(formula = ~Class + Survived + Sex + Age, data = Titanic)
## 
## Statistics:
##                       X^2 df P(> X^2)
## Likelihood Ratio 1243.663 25        0
## Pearson          1637.445 25        0

Unsurprisingly, very strong evidence that Class, Sex and Survived are not all mutually independent.

The fully saturated model is fitted as follows:

( Titanic.sat.fit <- loglm( ~ Class*Survived*Sex*Age, data=Titanic ) )

## Call:
## loglm(formula = ~Class * Survived * Sex * Age, data = Titanic)
## 
## Statistics:
##                  X^2 df P(> X^2)
## Likelihood Ratio   0  0        1
## Pearson          NaN  0        1

Note the NaN appears under Pearson because the chi-squared statistic is ill-defined if there are expected values equal to zero.

Pick two distinct nested models, and calculate the \(G^2\) likelihood ratio test statistic testing the hypothesis \(\mathcal{H}_0: \, \mathcal{M}_0\) against alternative \(\mathcal{H}_1: \, \mathcal{M}_1\).

Let’s consider \(\mathcal{H}_0: \, \mathcal{M}_0\) to be the all two-way interactions model and the \(\mathcal{H}_1: \, \mathcal{M}_1\) to be the all three-way interactions model. Then we have that:

( Three.fit <- loglm( ~ Class*Survived*Sex
                      + Class*Survived*Age
                      + Class*Sex*Age
                      + Survived*Sex*Age, data = Titanic ) )

## Call:
## loglm(formula = ~Class * Survived * Sex + Class * Survived * 
##     Age + Class * Sex * Age + Survived * Sex * Age, data = Titanic)
## 
## Statistics:
##                           X^2 df  P(> X^2)
## Likelihood Ratio 0.0001904416  3 0.9999993
## Pearson                   NaN  3       NaN

( Two.fit <- loglm( ~ Class*Survived + Survived*Sex
                    + Survived*Age + Class*Sex
                    + Sex*Age + Class*Age, data = Titanic ) )

## Call:
## loglm(formula = ~Class * Survived + Survived * Sex + Survived * 
##     Age + Class * Sex + Sex * Age + Class * Age, data = Titanic)
## 
## Statistics:
##                       X^2 df P(> X^2)
## Likelihood Ratio 116.5881 13        0
## Pearson               NaN 13      NaN

( Titanic.DG2 <- Two.fit$deviance - Three.fit$deviance )

## [1] 116.5879

# The p-value for testing H_0: M_0 against alternative H_1: M_1 is then given by
( p.value <- 1 - pchisq(Titanic.DG2, 10) )

## [1] 0

# Evidence of three-way interactions between the variables.

Use backwards selection to fit a log-linear model using AIC. Which model is chosen? What does this tell you?

step( Titanic.sat.fit, direction="backward", test = "Chisq" )

## Start:  AIC=64
## ~Class * Survived * Sex * Age
## 
##                          Df AIC        LRT Pr(>Chi)
## - Class:Survived:Sex:Age  3  58 0.00019044        1
## <none>                       64                    
## 
## Step:  AIC=58
## ~Class + Survived + Sex + Age + Class:Survived + Class:Sex + 
##     Survived:Sex + Class:Age + Survived:Age + Sex:Age + Class:Survived:Sex + 
##     Class:Survived:Age + Class:Sex:Age + Survived:Sex:Age
## 
##                      Df     AIC    LRT  Pr(>Chi)    
## - Survived:Sex:Age    1  57.685  1.685    0.1942    
## <none>                   58.000                     
## - Class:Sex:Age       3  61.783  9.783    0.0205 *  
## - Class:Survived:Age  3  89.263 37.262 4.049e-08 ***
## - Class:Survived:Sex  3 117.013 65.013 4.984e-14 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Step:  AIC=57.69
## ~Class + Survived + Sex + Age + Class:Survived + Class:Sex + 
##     Survived:Sex + Class:Age + Survived:Age + Sex:Age + Class:Survived:Sex + 
##     Class:Survived:Age + Class:Sex:Age
## 
##                      Df     AIC    LRT  Pr(>Chi)    
## <none>                   57.685                     
## - Class:Sex:Age       3  71.953 20.268 0.0001494 ***
## - Class:Survived:Age  3  95.899 44.214 1.359e-09 ***
## - Class:Survived:Sex  3 126.904 75.219 3.253e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

## Call:
## loglm(formula = ~Class + Survived + Sex + Age + Class:Survived + 
##     Class:Sex + Survived:Sex + Class:Age + Survived:Age + Sex:Age + 
##     Class:Survived:Sex + Class:Survived:Age + Class:Sex:Age, 
##     data = Titanic, evaluate = FALSE)
## 
## Statistics:
##                       X^2 df  P(> X^2)
## Likelihood Ratio 1.685426  4 0.7933632
## Pearson               NaN  4       NaN

We can see the backwards stepwise process of sequentially removing terms from the saturated model one-by-one until the model listed under Call: is finally chosen. In this case, we can see that we have all two-way interaction terms, and all three-way interaction terms except that corresponding to Survived, Age and Sex, thus suggesting that these three variable exhibit homogeneous associations amongst each other when conditioning on Class.

Use Forwards selection to fit a log-linear model using BIC (look in the help file for step if you are unsure how to do this). Do you get a different model to that found in part (c)? If so, is this due to the change in model selection criterion or stepwise selection direction?

n <- sum( Titanic )
step( Titanic.I.fit, scope = ~ Class*Survived*Sex*Age, direction="forward",
      test = "Chisq", k = log(n) )

## Start:  AIC=1297.54
## ~Class + Survived + Sex + Age
## 
##                  Df     AIC    LRT  Pr(>Chi)    
## + Survived:Sex    1  870.77 434.47 < 2.2e-16 ***
## + Class:Sex       3  908.03 412.60 < 2.2e-16 ***
## + Class:Survived  3 1139.73 180.90 < 2.2e-16 ***
## + Class:Age       3 1172.30 148.33 < 2.2e-16 ***
## + Sex:Age         1 1281.95  23.28 1.398e-06 ***
## + Survived:Age    1 1285.68  19.56 9.746e-06 ***
## <none>              1297.54                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Step:  AIC=870.77
## ~Class + Survived + Sex + Age + Survived:Sex
## 
##                  Df    AIC    LRT  Pr(>Chi)    
## + Class:Sex       3 481.26 412.60 < 2.2e-16 ***
## + Class:Survived  3 712.96 180.90 < 2.2e-16 ***
## + Class:Age       3 745.53 148.33 < 2.2e-16 ***
## + Sex:Age         1 855.18  23.28 1.398e-06 ***
## + Survived:Age    1 858.90  19.56 9.746e-06 ***
## <none>              870.77                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Step:  AIC=481.26
## ~Class + Survived + Sex + Age + Survived:Sex + Class:Sex
## 
##                  Df    AIC     LRT  Pr(>Chi)    
## + Class:Age       3 356.02 148.327 < 2.2e-16 ***
## + Class:Survived  3 398.27 106.075 < 2.2e-16 ***
## + Sex:Age         1 465.67  23.284 1.398e-06 ***
## + Survived:Age    1 469.39  19.561 9.746e-06 ***
## <none>              481.26                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Step:  AIC=356.02
## ~Class + Survived + Sex + Age + Survived:Sex + Class:Sex + Class:Age
## 
##                  Df    AIC     LRT  Pr(>Chi)    
## + Class:Survived  3 273.03 106.075 < 2.2e-16 ***
## + Survived:Age    1 352.71  11.008 0.0009071 ***
## <none>              356.02                      
## + Sex:Age         1 357.63   6.086 0.0136259 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Step:  AIC=273.03
## ~Class + Survived + Sex + Age + Survived:Sex + Class:Sex + Class:Age + 
##     Class:Survived
## 
##                      Df    AIC    LRT  Pr(>Chi)    
## + Class:Survived:Sex  3 230.94 65.180 4.591e-14 ***
## + Survived:Age        1 255.15 25.583 4.237e-07 ***
## <none>                  273.03                     
## + Sex:Age             1 274.64  6.086   0.01363 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Step:  AIC=230.94
## ~Class + Survived + Sex + Age + Survived:Sex + Class:Sex + Class:Age + 
##     Class:Survived + Class:Survived:Sex
## 
##                Df    AIC    LRT  Pr(>Chi)    
## + Survived:Age  1 213.06 25.583 4.237e-07 ***
## <none>            230.94                     
## + Sex:Age       1 232.56  6.086   0.01363 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Step:  AIC=213.06
## ~Class + Survived + Sex + Age + Survived:Sex + Class:Sex + Class:Age + 
##     Class:Survived + Survived:Age + Class:Survived:Sex
## 
##                      Df    AIC     LRT  Pr(>Chi)    
## + Class:Survived:Age  3 206.94 29.2063 2.027e-06 ***
## <none>                  213.06                      
## + Sex:Age             1 220.62  0.1356    0.7127    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Step:  AIC=206.94
## ~Class + Survived + Sex + Age + Survived:Sex + Class:Sex + Class:Age + 
##     Class:Survived + Survived:Age + Class:Survived:Sex + Class:Survived:Age
## 
##           Df    AIC     LRT Pr(>Chi)
## <none>       206.94                 
## + Sex:Age  1 214.37 0.26868   0.6042

## Call:
## loglm(formula = ~Class + Survived + Sex + Age + Survived:Sex + 
##     Class:Sex + Class:Age + Class:Survived + Survived:Age + Class:Survived:Sex + 
##     Class:Survived:Age, data = Titanic, evaluate = FALSE)
## 
## Statistics:
##                       X^2 df    P(> X^2)
## Likelihood Ratio 22.22167  8 0.004521358
## Pearson               NaN  8         NaN

We can see the forwards stepwise process of sequentially adding terms from the first-order linear model one-by-one until the model listed under Call: is finally chosen. In this case, we can see that we have all two-way interaction terms except that between Sex and Age, and all possible three-way interactions based on this given that the model must be hierarchical. This model suggests that Sex and Age are conditionally independent given Class and Survived. It is likely that this change in model selection choice is due to the change of criteria (i.e. BIC rather than AIC), although the selection method can have an effect.

–>