Test of independence for contingency tables

Below is the calculation using direct formula.

        obs = rbind(c(762,327,468), c(484, 239, 477))    # observed counts
        n = sum(obs)                                     # total sample size
        row = rowSums(obs)                               # row total
        col = colSums(obs)                               # col total
        ept = outer(row, col, "*")/n                     # expected counts
        Pearson = (obs-ept)/sqrt(ept)                    # Pearson residual
        Pearson

##           [,1]       [,2]      [,3]
## [1,]  2.198856  0.4113702 -2.843240
## [2,] -2.504669 -0.4685829  3.238673

        Pearson/sqrt(outer(1-row/n, 1-col/n, "*"))       # Standardized residual

##           [,1]       [,2]      [,3]
## [1,]  4.502054  0.6994517 -5.315946
## [2,] -4.502054 -0.6994517  5.315946

In R, you can use as.table() function to tell R that the matrix represents a contingency table of counts. Then the functions developed for the contingency tables can be applied.

 dimnames(obs) <- list(Gender=c("Female","Male"), 
                      Party=c("Democrat","Independent","Republican")) 
obs.tab = as.table(obs)
obs.tab

##         Party
## Gender   Democrat Independent Republican
##   Female      762         327        468
##   Male        484         239        477

margin.table(obs.tab,1)

## Gender
## Female   Male 
##   1557   1200

margin.table(obs.tab,2)

## Party
##    Democrat Independent  Republican 
##        1246         566         945

addmargins(obs.tab)

##         Party
## Gender   Democrat Independent Republican  Sum
##   Female      762         327        468 1557
##   Male        484         239        477 1200
##   Sum        1246         566        945 2757

prop.table(obs.tab,1)

##         Party
## Gender    Democrat Independent Republican
##   Female 0.4894027   0.2100193  0.3005780
##   Male   0.4033333   0.1991667  0.3975000

prop.table(obs.tab,2)

##         Party
## Gender    Democrat Independent Republican
##   Female 0.6115570   0.5777385  0.4952381
##   Male   0.3884430   0.4222615  0.5047619

prop.table(obs.tab)

##         Party
## Gender     Democrat Independent Republican
##   Female 0.27638738  0.11860718 0.16974973
##   Male   0.17555314  0.08668843 0.17301415

obs.chisq = chisq.test(obs.tab)   # Chi-squared Test for Independence
obs.chisq

## 
##  Pearson's Chi-squared test
## 
## data:  obs.tab
## X-squared = 30.07, df = 2, p-value = 2.954e-07

names(obs.chisq)

## [1] "statistic" "parameter" "p.value"   "method"    "data.name" "observed" 
## [7] "expected"  "residuals" "stdres"

obs.chisq$statistic

## X-squared 
##  30.07015

obs.chisq$observed

##         Party
## Gender   Democrat Independent Republican
##   Female      762         327        468
##   Male        484         239        477

obs.chisq$expected

##         Party
## Gender   Democrat Independent Republican
##   Female 703.6714    319.6453   533.6834
##   Male   542.3286    246.3547   411.3166

with(obs.chisq, sum((observed - expected)^2/expected))

## [1] 30.07015

# The output residuals from  chisq.test are Pearson residuals
obs.chisq$residuals

##         Party
## Gender     Democrat Independent Republican
##   Female  2.1988558   0.4113702 -2.8432397
##   Male   -2.5046695  -0.4685829  3.2386734

# You can verify the output residuals by
with(obs.chisq, (observed - expected)/sqrt(expected))

##         Party
## Gender     Democrat Independent Republican
##   Female  2.1988558   0.4113702 -2.8432397
##   Male   -2.5046695  -0.4685829  3.2386734

# To    To get the standardized residuals, we also need marginal proportions.
n <- sum(obs.tab)
n.gender <- margin.table(obs.tab, 1)
n.party  <- margin.table(obs.tab, 2)
stand.res = obs.chisq$residual/sqrt((1-n.gender/n) %o% (1-n.party/n))
stand.res

##         Party
## Gender     Democrat Independent Republican
##   Female  4.5020535   0.6994517 -5.3159455
##   Male   -4.5020535  -0.6994517  5.3159455

For the trend test,

Total = c(17114, 14502, 793, 127, 38)
present = c(48, 38, 5, 1, 1)
prop.trend.test(present, Total, c(0, 0.5, 1.5, 4.0, 7))

## 
##  Chi-squared Test for Trend in Proportions
## 
## data:  present out of Total ,
##  using scores: 0 0.5 1.5 4 7
## X-squared = 6.5701, df = 1, p-value = 0.01037