{"id":195,"date":"2025-01-22T15:58:08","date_gmt":"2025-01-22T10:13:08","guid":{"rendered":"https:\/\/pokharelsugam.com.np\/bioinformatics\/?p=195"},"modified":"2025-02-03T16:59:37","modified_gmt":"2025-02-03T11:14:37","slug":"fishers-exact-test","status":"publish","type":"post","link":"https:\/\/pokharelsugam.com.np\/bioinformatics\/fishers-exact-test\/","title":{"rendered":"Fisher&#8217;s Exact Test"},"content":{"rendered":"\n<script type=\"text\/javascript\" async \n  src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-MML-AM_CHTML\">\n<\/script>\n\n\n\n<p>This test is used to determine if there are nonrandom associations between two categorical variables in a contingency table usually 2 x 2 tables. It is done when sample sizes are small and the chi-squared test assumptions are violated (expected frequency &gt;= 5).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">When to Use Fisher\u2019s Exact Test?<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>When sample sizes are small (typically when the expected frequency in any cell of the table is less than 5).<\/li>\n\n\n\n<li>When analyzing categorical data in a <strong>2\u00d72 contingency table<\/strong>.<\/li>\n\n\n\n<li>When the chi-square test\u2019s assumptions (large sample size) do not hold.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Steps to Perform Fisher\u2019s Exact Test<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Create a 2\u00d72 contingency table<\/strong> with observed frequencies.<\/li>\n\n\n\n<li><strong>Calculate the probability<\/strong> of obtaining the observed table using the Fisher\u2019s Exact Test formula.<\/li>\n\n\n\n<li><strong>Compute the p-value<\/strong>, which is the sum of probabilities of all tables with equal or lower probability.<\/li>\n\n\n\n<li><strong>Compare the p-value with the significance level (\u03b1, usually 0.05)<\/strong>:\n<ul class=\"wp-block-list\">\n<li>If <strong>p \u2264 \u03b1<\/strong>, reject the null hypothesis (evidence of association).<\/li>\n\n\n\n<li>If <strong>p > \u03b1<\/strong>, fail to reject the null hypothesis (no significant association).<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<h3 class=\"wp-block-heading\">Hypothesis for Fisher\u2019s Exact Test<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Null Hypothesis (H\u2080)<\/strong>: There is no association between the two categorical variables (they are independent).<\/li>\n\n\n\n<li><strong>Alternative Hypothesis (H\u2081)<\/strong>: There is an association between the two categorical variables (they are dependent).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">2\u00d72 Contingency Table Format<\/h3>\n\n\n\n<p>A contingency table is used to summarize the frequency counts of two categorical variables:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>Category B\u2081<\/th><th>Category B\u2082<\/th><th>Total<\/th><\/tr><\/thead><tbody><tr><td><strong>A\u2081<\/strong><\/td><td>a<\/td><td>b<\/td><td>a+b<\/td><\/tr><tr><td><strong>A\u2082<\/strong><\/td><td>c<\/td><td>d<\/td><td>c+d<\/td><\/tr><tr><td><strong>Total<\/strong><\/td><td>a+c<\/td><td>b+d<\/td><td>N<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>a, b, c, and d<\/strong> are observed frequencies in each cell.<\/li>\n\n\n\n<li><strong>N<\/strong> is the total sample size.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Mathematical Formula for Fisher\u2019s Exact Test<\/h3>\n\n\n\n<p class=\"has-medium-font-size\">\\(P= \\frac{^{(a+b)}C_a.^{(c+d)}C_c}{^NC_{(a+c)}} \\) Where C is Combination<\/p>\n\n\n\n<p>Or,<\/p>\n\n\n\n<p class=\"has-medium-font-size\">\\(P= \\frac{(a+b)!.(c+d)!.(a+c)!.(b+d)!}{N!.a!.b!.c!.d!} \\)<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Interpretation of the Formula<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The test calculates the probability of observing the specific arrangement of the table under the assumption that row and column totals are fixed.<\/li>\n\n\n\n<li>The p-value is the sum of probabilities of all tables that have a probability equal to or smaller than the observed table.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Test statistics<\/h3>\n\n\n\n<p>\\(P = P_k +P_{k-1} +P_{k-2} + P_{k-3}+ &#8230;&#8230;&#8230;&#8230;. +P_o\\)<\/p>\n\n\n\n<p>Where<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\\(P\\) : P value of Fisher&#8217;s Exact Test<\/li>\n\n\n\n<li>\\(P_o\\) : is the probability of the observed contingency table (i.e., the exact table we got from the data)<\/li>\n\n\n\n<li>\\(P_{k-1}\\) or  \\(P_{k-2}\\) : Probabilities of more extreme tables than \\(P_o\\)<\/li>\n\n\n\n<li>\\(P_k\\) : Probability of the most extreme table possible.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Example<\/h3>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>Category B\u2081<\/th><th>Category B\u2082<\/th><th>Total<\/th><\/tr><\/thead><tbody><tr><td><strong>A\u2081<\/strong><\/td><td>1<\/td><td>6<\/td><td>7<\/td><\/tr><tr><td><strong>A\u2082<\/strong><\/td><td>4<\/td><td>1<\/td><td>5<\/td><\/tr><tr><td><strong>Total<\/strong><\/td><td>5<\/td><td>7<\/td><td>12<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Hypothesis:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Null: Proportion are equal<\/li>\n\n\n\n<li>Alternate: Proportion are not equal<\/li>\n<\/ul>\n\n\n\n<p><strong>Test Statistics Calculation <\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">\\(P_o =\\frac{(a+b)!.(c+d)!.(a+c)!.(b+d)!}{N!.a!.b!.c!.d!} \\)<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Or, \\(P_o =\\frac{(7)!.(5)!.(5)!.(7)!}{12!.1!.6!.4.1!} = 0.0441  \\)<\/p>\n\n\n\n<p>Now modify value of contingency table keeping sum of all of them constant to get extreme table<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>Category B\u2081<\/th><th>Category B\u2082<\/th><th>Total<\/th><\/tr><\/thead><tbody><tr><td><strong>A\u2081<\/strong><\/td><td>0<\/td><td>7<\/td><td>7<\/td><\/tr><tr><td><strong>A\u2082<\/strong><\/td><td>5<\/td><td>0<\/td><td>5<\/td><\/tr><tr><td><strong>Total<\/strong><\/td><td>5<\/td><td>7<\/td><td>12<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"has-medium-font-size\">\\(P_k =\\frac{(a+b)!.(c+d)!.(a+c)!.(b+d)!}{N!.a!.b!.c!.d!} \\)<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Or, \\(P_k =\\frac{(7)!.(5)!.(5)!.(7)!}{12!.0!.7!.5.0!} = 0.00126  \\)<\/p>\n\n\n\n<p><strong>Final Test Statistics Calculation <\/strong><\/p>\n\n\n\n<p>\\(P = P_o + P_k\\)<\/p>\n\n\n\n<p>Or, \\(P =  0.0441 + 0.00126 \\) = 0.045326<\/p>\n\n\n\n<p><strong>Decision:<\/strong><\/p>\n\n\n\n<p>\\(P(0.045326 &lt; \\alpha (0.05) \\) Hence, do not accept null hypothesis<\/p>\n","protected":false},"excerpt":{"rendered":"<p>This test is used to determine if there are nonrandom associations between two categorical variables in a contingency table usually 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