Example R programs and commands
	20. Chi-squared tests for independence in contingency tables

# All lines preceded by the "#" character are my comments.
# All other left-justified lines are my input.
# All other indented lines are the R program output.

############################################################
# CHI-SQUARED test for independence
#
# Read the contingency table, r=2 rows (X values), c=3 columns (Y values):
data<-scan()
 163     135     71
  86      77     40

tab <- t(matrix(data, nrow=3,ncol=2))  # transpose, as usual
tab         # view the table to make sure it is right
	[,1] [,2] [,3]
   [1,]  163  135   71
   [2,]   86   77   40
   
chisq.test(tab)   # performs a test of independence for matrix input.

	  Pearson's Chi-squared test

  data:  tab
  X-squared = 0.1769, df = 2, p-value = 0.9153

# ==> DO NOT REJECT the null hypothesis that X,Y are independent

# NOTE: the transposed table gives the same result, as X and Y are
# independent if and only if Y and X are independent.
chisq.test(t(tab))

	  Pearson's Chi-squared test

  data:  t(tab)
  X-squared = 0.1769, df = 2, p-value = 0.9153

############################################################
# LOG-LIKELIHOOD computation
# Extract the observed and expected values.
toi<-chisq.test(tab)     # store the whole output data structure
summary(toi)
	     Length Class  Mode
   statistic 1      -none- numeric
   parameter 1      -none- numeric
   p.value   1      -none- numeric
   method    1      -none- character
   data.name 1      -none- character
   observed  6      -none- numeric
   expected  6      -none- numeric
   residuals 6      -none- numeric

# Compute the log-likelihood statistic directly from these parts:
g<-2*sum(toi$observed*log(toi$observed/toi$expected))
g

# Compute the p-value of g assuming it has the Chi-squared density:
nu<-prod(dim(tab)-1)  # this is (nrow-1)*(ncol-1) == (r-1)(c-1)
1-pchisq(g,nu)


############################################################
# YATES correction for CHI-SQUARED test in 2x2 contingency tables
data<-scan()
 163     135 
  86      77 

tab <- matrix(data, nrow=2,ncol=2)  # transpose or not, gives same result
tab                                 # view the table
chisq.test(tab, correct=TRUE)       # with Yates correction (default)
chisq.test(tab, correct=FALSE)      # without Yates correction 


############################################################
# YATES correction for LOG-LIKELIHOOD test in 2x2 contingency tables
data<-scan()
 163     135 
  86      77 

tab <- matrix(data, nrow=2,ncol=2) # transpose or not, gives same result
tab                                # view the table
ycmat <- matrix(0.5*c(-1,1,1,-1),nrow=2,ncol=2)  # correction matrix
tabc <- tab+sign(det(tab))*ycmat    # add or subtract the corrections as needed
tabc                                # view the corrected table

toic<-chisq.test(tabc, correct=FALSE) # Do not use the Chi-sq Yates correction 

# Compute the log-likelihood statistic directly from these parts:
g<-2*sum(toic$observed*log(toic$observed/toic$expected))
g

# Compute the p-value of g assuming it has the Chi-squared density:
nu<-prod(dim(tab)-1)  # this is (nrow-1)*(ncol-1) == (r-1)(c-1)
1-pchisq(g,nu)


############################################################
# HIGHER-DIMENSIONAL contingency tables; test of independence:

# Read the count data for this problem
count<-scan()
   12  34  23         4  47  11
   35  31  11        34  10  18
   12  32   9        18  13  19
   12  12  14         9  33  25 

# Create factor tags:r=rows, c=columns, t=tiers
r <- factor(gl(4, 2*3, 2*3*4, labels=c("r1","r2", "r3", "r4")));  r
c <- factor(gl(3, 1,   2*3*4, labels=c("c1","c2", "c3")));  c
t <- factor(gl(2, 3,   2*3*4, labels=c("t1","t2")));  t

# Cross-tabulation of counts:
xtabs(count~r+c+t)           # all three factors
xtabs(count~r+c)             # RC: sum over tiers
xtabs(count~r+t)             # RT: sum over columns
xtabs(count~c+t)             # CT: sum over rows
xtabs(count~r)       # R: sum over columns and tiers
xtabs(count~c)       # C: sum over rows and tiers
xtabs(count~t)       # T: sum over rows and columns

# 3-way Chi squared test of independence
summary(xtabs(count~r+c+t))

# NOTE: The number of degrees of freedom in this 3-D case is the sum 
# of the degrees of freedom for all 3-way and 2-way interactions:
nr <- 4; nc<-3; nt<-2;  # number of rows, columns, tiers, respectively
dof <- (nr-1)*(nc-1)*(nt-1)+(nr-1)*(nc-1)+(nr-1)*(nt-1)+(nc-1)*(nt-1)


# 2-way Chi squared test of partial independence: C versus T
summary(xtabs(count~c+t))  # dof=(nc-1)*(nt-1)

# 2-way Chi squared test of partial independence: R versus T
summary(xtabs(count~r+t))  # dof=(nr-1)*(nt-1)

# 2-way Chi squared test of partial independence: R versus C
summary(xtabs(count~r+c))  # dof=(nr-1)*(nc-1)

# 1-way Chi squared test of no dependence on R:
chisq.test(xtabs(count~r))  # dof= nr-1

# 1-way Chi squared test of no dependence on C:
chisq.test(xtabs(count~c))  # dof= nc-1

# 1-way Chi squared test of no dependence on T:
chisq.test(xtabs(count~t))  # dof= nt-1


#### Output and interpretation (at significance level alpha=0.05):
   # 3-way Chi squared test of independence
   > summary(xtabs(count~r+c+t))
   Call: xtabs(formula = count ~ r + c + t)
   Number of cases in table: 478
   Number of factors: 3
   Test for independence of all factors:
	   Chisq = 102.17, df = 17, p-value = 3.514e-14
#
# ==> reject H0 in favor of 
#            HA: r,c,t are NOT mutually independent
#

   > # 2-way Chi squared test of partial independence: C versus T
   > summary(xtabs(count~c+t))
   Call: xtabs(formula = count ~ c + t)
   Number of cases in table: 478
   Number of factors: 2
   Test for independence of all factors:
	   Chisq = 2.3704, df = 2, p-value = 0.3057
#
# ==> do not reject H0: c,t are mutually independent
#

   >
   > # 2-way Chi squared test of partial independence: R versus T
   > summary(xtabs(count~r+t))
   Call: xtabs(formula = count ~ r + t)
   Number of cases in table: 478
   Number of factors: 2
   Test for independence of all factors:
	   Chisq = 10.057, df = 3, p-value = 0.01809
#
# ==> reject H0 in favor of 
#            HA: r,t are NOT mutually independent
#

   > # 2-way Chi squared test of partial independence: R versus C
   > summary(xtabs(count~r+c))
   Call: xtabs(formula = count ~ r + c)
   Number of cases in table: 478
   Number of factors: 2
   Test for independence of all factors:
	   Chisq = 58.67, df = 6, p-value = 8.363e-11
#
# ==> reject H0 in favor of 
#            HA: r,c are NOT mutually independent
#


   > # 1-way Chi squared test of no dependence on R:
   > chisq.test(xtabs(count~r))  # dof= nr-1

	   Chi-squared test for given probabilities

   data:  xtabs(count ~ r)
   X-squared = 8.3264, df = 3, p-value = 0.03973
#
# ==> reject H0 in favor of 
#            HA: count is NOT independent of R
#

   > # 1-way Chi squared test of no dependence on C:
   > chisq.test(xtabs(count~c))  # dof= nc-1

	   Chi-squared test for given probabilities

   data:  xtabs(count ~ c)
   X-squared = 26.226, df = 2, p-value = 2.019e-06
#
# ==> reject H0 in favor of 
#            HA: count is NOT independent of C
#


   > # 1-way Chi squared test of no dependence on T:
   > chisq.test(xtabs(count~t))  # dof= nt-1

	   Chi-squared test for given probabilities

   data:  xtabs(count ~ t)
   X-squared = 0.033473, df = 1, p-value = 0.8548
#
# ==> DO NOT reject H0: count is independent of T
#