Fisher’s Exact test

What does this test do?

• decides if two (or more) proportions differ significantly
• Example I:
  • in group A, 20 out of 40 animals survived (proportion = 0.5)
  • in group B, 5 out of 25 animals survived (proportion = 0.2)
  • is there a statistically significant difference between these proportions?
• Example II (enrichment analysis, differentially expressed genes):
  • in a test group, 10 of 20 genes are up-regulated (proportion = 0.5)
  • in a control group, 15 of 20 genes are up-regulated (proportion = 0.75)
  • is there a statistically significant difference between these proportions?
  • (or, are the up-regulated genes enriched in the control group)
• can also be seen as test of independence:
  • is the survival rate independent of the group membership? (example I)
  • is the gene regulation independent of the group membership? (example II)
• the test works only for relativly low numbers (high computational workload)
• for larger numbers, use the prop.test() function (next chapter)
• \(H_0\) : \(p_1 = p_2\) (equal proportions, independence)
• a small p-value means that the null is rejected, i.e. that the proportions differ significantly
• function: fisher.test()
• entries should be nonnegative integers

Running the test

x <- matrix(c(2,10,20,13), ncol=2, byrow=T)
x

##      [,1] [,2]
## [1,]    2   10
## [2,]   20   13

fisher.test(x)

## 
##  Fisher's Exact Test for Count Data
## 
## data:  x
## p-value = 0.01655
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
##  0.01256709 0.78740829
## sample estimates:
## odds ratio 
##  0.1360765

Reading the output

The point estimate for the odds ratio is calculated as \(\frac{2 \cdot 13}{10 \cdot 20} = 0.13\) in this case. The confidence interval refers to the odds ratio, i.e. the given interval includes the true value with 95% chance. For this example, \(H_0\) is rejected, i.e. the proportions are significantly different (no independence).