Chi-Square Test for Association

Chi Square Test

We use the Chi square test for testing association in two way contingency tables.
Contingency tables: Outcomes are categorized into rows and columns.
This can be used to test the differences between several groups.
The null hypothesis is that there is not differences between the groups.
The alternative is that there is difference between the groups.
Is there Independence or Association?

Smoking Example

Using this in-built dataset, we shall test the association between smoking and exercise.

Test the hypothesis whether the students smoking habit is independent of their exercise level at 0.05 significance level.

library(MASS)       # load the MASS package 
tbl = table(survey$Smoke, survey$Exer) 
tbl                 # the contingency table

##        
##         Freq None Some
##   Heavy    7    1    3
##   Never   87   18   84
##   Occas   12    3    4
##   Regul    9    1    7

# Notice the small cell sizes

class(survey$Smoke)

## [1] "factor"

# Sort the factors

levels(survey$Smoke)=c('Never','Occas','Regul','Heavy')
levels(survey$Exer)=c('None','Some','Freq')

Chi Square Test for Independence

chisq.test(tbl)

## Warning in chisq.test(tbl): Chi-squared approximation may be incorrect

## 
##  Pearson's Chi-squared test
## 
## data:  tbl
## X-squared = 5, df = 6, p-value = 0.5

We have applied the chisq.test() function to the contingency table tbl, and found the p-value to be 0.4828.
We fail to reject the null hypothesis.
The warning message found in the solution above is due to the small cell values in the contingency table.
To avoid such warning, we could combine the second and third columns of tbl.

# Dont throw out the raw data,make a 'derived variable' instead.
survey$Exer2 <- survey$Exer

levels(survey$Exer2) <- list(Rare = c('None','Some'),Freq = 'Freq')

chisq.test(survey$Smoke,survey$Exer2)

## Warning in chisq.test(survey$Smoke, survey$Exer2): Chi-squared approximation may
## be incorrect

## 
##  Pearson's Chi-squared test
## 
## data:  survey$Smoke and survey$Exer2
## X-squared = 5, df = 3, p-value = 0.2

`prob.table()`

A useful command associated with the Chi-Square Test is prob.table(), whicn converts count data to proportions.

### OVerall Proportions
prop.table(tbl)

##        
##            Freq    None    Some
##   Heavy 0.02966 0.00424 0.01271
##   Never 0.36864 0.07627 0.35593
##   Occas 0.05085 0.01271 0.01695
##   Regul 0.03814 0.00424 0.02966

### Proportion of Row Variable
prop.table(tbl,1)

##        
##           Freq   None   Some
##   Heavy 0.6364 0.0909 0.2727
##   Never 0.4603 0.0952 0.4444
##   Occas 0.6316 0.1579 0.2105
##   Regul 0.5294 0.0588 0.4118

### Proportion of Column Variable
prop.table(tbl,2)

##        
##           Freq   None   Some
##   Heavy 0.0609 0.0435 0.0306
##   Never 0.7565 0.7826 0.8571
##   Occas 0.1043 0.1304 0.0408
##   Regul 0.0783 0.0435 0.0714