Fisher's Exact Test

In the odds ratio topic note we noted that the odds ratio could help us determine whether the odds were greater in one group versus another. We can test the strength of association of the group and event with Fisher's Exact Test. Fisher's Exact Test has the following hypotheses:

\(H_0\) or null hypothesis: there is no association between the group and event (Odds ratio = 1)
\(H_a\) or alternative hypothesis: there is an association between the group and event (Odds ratio != 1)

We can calculate the probability given the following contingency table with:

	group 1	group 2
Event	a	b
No Event	c	d

\[p = \frac{(a+b)!(c+d)!(a+c)!(b+d)!}{a!b!c!d!(a+b+c+d)!}\]

Explanation of Terms

\(a\) number of members in group 1 with event
\(b\) number of members in group 2 with event
\(c\) number of members in group 1 without event
\(d\) number of members in group 2 without event

Let's assess the relationship between pathiet sex and losing patients to follow up:

library(tidyverse)
# load meta data
meta <- read.table("./data/gbm_cptac_2021/data_clinical_patient.txt",
                   header = T,
                   sep="\t")

# What is the frequency distribution of losing males/females to follow up
table <- as.data.frame.matrix(
  table(meta$SEX,meta$LOST_TO_FOLLOW_UP)
)

table

       No Yes
Female 33  10
Male   44  10

Before we continue, we need to make this table match the contingency table above. With the rows being the event and the columns being the group:

# Reorder so that we are assessing the odds ratio of losing patients to follow up
table <- as.data.frame.matrix(
  table(meta$SEX,meta$LOST_TO_FOLLOW_UP)
) %>%
  select(c(Yes,No)) %>%
  t()

table

    Female Male
Yes     10   10
No      33   44

Now let's conduct our hypothesis test:

#apply our test
fisher.test(table)

    Fisher's Exact Test for Count Data

data:  table
p-value = 0.6191
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
 0.4395952 4.0274603
sample estimates:
odds ratio 
   1.32933

Explanation of Results

Here we note:

our p-value is above 0.05 and thus not strong enough to reject the null (a.k.a the true odds ratio is equal to 1)
the 95% confidence interval reveals that our true odds ratio is somewhere between 0.4395952 and 4.0274603
our odds ratio that patient sex is associated with losing the patient to follow up is about 1.3

Fisher's Exact Test

Fisher's Exact Test

References