Data

For our example, we want to determine whether there is a statistically significant association between smoking and being a professional athlete. Smoking can only be “yes” or “no” and being a professional athlete can only be “yes” or “no”. The two variables of interest are qualitative variables and we collected data on 14 persons.¹

Observed frequencies

Our data are summarized in the contingency table below reporting the number of people in each subgroup:

dat <- data.frame(
  "smoke_no" = c(7, 0),
  "smoke_yes" = c(2, 5),
  row.names = c("Athlete", "Non-athlete"),
  stringsAsFactors = FALSE
)
colnames(dat) <- c("Non-smoker", "Smoker")

dat

##             Non-smoker Smoker
## Athlete              7      2
## Non-athlete          0      5

It is also a good practice to draw a mosaic plot to visually represent the data:

mosaicplot(dat,
  main = "Mosaic plot",
  color = TRUE
)

We can already see from the plot that the proportion of smokers in the sample is higher among non-athletes than athlete. The plot is however not sufficient to conclude that there is such a significant association in the population.

Like many statistical tests, this can be done via a hypothesis test. But before seeing how to do it in R, let’s see the concept of expected frequencies.

Expected frequencies

Remember that the Fisher’s exact test is used when there is at least one cell in the contingency table of the expected frequencies below 5. To retrieve the expected frequencies, use the chisq.test() function together with $expected:

chisq.test(dat)$expected

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

##             Non-smoker Smoker
## Athlete            4.5    4.5
## Non-athlete        2.5    2.5

The contingency table above confirms that we should use the Fisher’s exact test instead of the Chi-square test because there is at least one cell below 5.

Tip: although it is a good practice to check the expected frequencies before deciding between the Chi-square and the Fisher test, it is not a big issue if you forget. As you can see above, when doing the Chi-square test in R (with chisq.test()), a warning such as “Chi-squared approximation may be incorrect” will appear. This warning means that the smallest expected frequencies is lower than 5. Therefore, do not worry if you forgot to check the expected frequencies before applying the appropriate test to your data, R will warn you that you should use the Fisher’s exact test instead of the Chi-square test if that is the case.

(Remember that, as for the Chi-square test of independence, the observations must be independent in order for the Fisher’s exact test to be valid. See more details about the independence assumption in this section.)

Fisher’s exact test in R

To perform the Fisher’s exact test in R, use the fisher.test() function as you would do for the Chi-square test:

test <- fisher.test(dat)
test

## 
##  Fisher's Exact Test for Count Data
## 
## data:  dat
## p-value = 0.02098
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
##  1.449481      Inf
## sample estimates:
## odds ratio 
##        Inf

The most important in the output is the \(p\)-value. You can also retrieve the \(p\)-value with:

test$p.value

## [1] 0.02097902

Note that if your data is not already presented as a contingency table, you can simply use the following code:

fisher.test(table(dat$variable1, dat$variable2))

where dat is the name of your dataset, variable1 and variable2 correspond to the names of the two variables of interest.

Conclusion and interpretation

From the output and from test$p.value we see that the \(p\)-value is less than the significance level of 5%. Like any other statistical test, if the \(p\)-value is less than the significance level, we can reject the null hypothesis. If you are not familiar with \(p\)-values, I invite you to read this section.

\(\Rightarrow\) In our context, rejecting the null hypothesis for the Fisher’s exact test of independence means that there is a significant relationship between the two categorical variables (smoking habits and being an athlete or not). Therefore, knowing the value of one variable helps to predict the value of the other variable.