Fisher’s Exact Test Calculator

Fisher's Exact Test

Row Factor Name (Groups)

Column Factor Name (Outcomes)

2 × 2 Contingency Table Frequencies Enter non-negative integer cell counts

Cell a (Group 1, Outcome 1)

Cell b (Group 1, Outcome 2)

Cell c (Group 2, Outcome 1)

Cell d (Group 2, Outcome 2)

All 4 cells must contain valid non-negative integers.

Hypothesis Direction / Tail Type

Significance Level (α) Default rejection threshold is 0.05 (5%)

Significance level α must be strictly between 0 and 1 (e.g. 0.05).

Exact Categorical Association

What Is Fisher's Exact Test?

Fisher's Exact Test is a non-parametric statistical significance test used to determine whether there are non-random associations between two categorical variables in a $2 \times 2$ contingency table.

Unlike the standard asymptotic Chi-Square Test of Independence, which relies on large-sample approximations, Fisher's test computes the exact hypergeometric probability of obtaining the observed frequencies (or more extreme) under fixed marginal totals.

When to Use It

Small Sample Sizes ($N < 30$): When total sample size is small or cell counts are sparse.
Violation of Chi-Square Assumptions: When expected cell frequencies ($E_{ij}$) fall below 5 in any cell.
Binary Clinical Outcomes: Comparing rare adverse drug events or surgical success rates between treatment and control arms.
Exact Odds Ratio Evaluation: Assessing whether the odds of success differ significantly from 1 ($OR \neq 1.0$).

The Hypergeometric Formula

$$P(a) = \frac{\binom{a + b}{a} \binom{c + d}{c}}{\binom{N}{a + c}} = \frac{(a + b)! (c + d)! (a + c)! (b + d)!}{a! \, b! \, c! \, d! \, N!}$$ $$\text{Odds Ratio } (OR) = \frac{a \cdot d}{b \cdot c}$$ $$\text{Two-Tailed } p = \sum_{P(k) \le P(a_{\text{obs}})} P(k)$$

Where $N = a + b + c + d$, and the summation aggregates probabilities of all possible table configurations having fixed marginal totals that are at least as extreme as the observed data.

Worked Example

Scenario: 16 subjects are evaluated: Treated ($a=8$ recovered, $b=2$ unrecovered); Control ($c=1$ recovered, $d=5$ unrecovered) at $\alpha=0.05$.

Step 1: Compute Observed Table Probability:
Marginals: $R_1 = 10, R_2 = 6, C_1 = 9, C_2 = 7, N = 16$.
$P(a=8) = \frac{10! \, 6! \, 9! \, 7!}{8! \, 2! \, 1! \, 5! \, 16!} = \frac{270}{11440} \approx 0.0236$.

Step 2: Aggregate Two-Tailed Probabilities ($P(k) \le 0.0236$):
Valid range for $a$: $3 \le a \le 9$.
$P(3) = \frac{120}{11440} \approx 0.0105, \quad P(8) = 0.0236, \quad P(9) = \frac{10}{11440} \approx 0.0009$.
$p = P(3) + P(8) + P(9) = \frac{400}{11440} = 0.0350$.

Step 3: Calculate Odds Ratio:
$OR = \frac{8 \times 5}{2 \times 1} = 20.0000 \quad (95\% \text{ CI: } [1.42, 282.45])$.

Conclusion: Since $p = 0.0350 \le 0.05$, reject $H_0$. Recovery odds are significantly higher in the treated cohort.

Exact Probability Computation

Fisher's Exact Test Results

Computed Exact P-Value

p = 0.0350

Statistically Significant Association (Reject H₀ at α = 0.05)

Exact P-Value 0.0350

Odds Ratio (OR) 20.0000

Relative Risk (RR) 4.8000

P(Observed Table) 0.0236

Total Sample (N) 16

Valid Configurations 7

Hypergeometric Table Permutations & Probabilities

Configuration	Cell a	Cell b	Cell c	Cell d	Exact P(Configuration)	Cumulative Probability	Extreme Tail? (p ≤ P_obs)

Fisher’s Exact Test Calculator