Biostatistics Quiz: Fishers Exact Test
20 questions · exam conditions
0:00
Fishers Exact TestQuestion 1 of 20

A researcher compares complication rates between two surgical procedures: Procedure A had 1 complication in 4 surgeries, Procedure B had 3 complications in 5 surgeries. For Fisher's exact test, the researcher must enumerate all possible 2×2 tables with the same marginal totals. What constraint determines the range of possible values for the 'Procedure A complications' cell?

The value must be between max(0, total complications - Procedure B surgeries) and min(Procedure A surgeries, total complications)
The value must be between 0 and the smaller of the two procedure sample sizes
The value must be between 0 and the total number of complications observed across both procedures
The value must be between 1 and 4 since these are the observed minimum and maximum for Procedure A
The value can be any integer between 0 and 9 since this represents the total sample size
← Back to quizzes

Biostatistics Quiz

Biostatistics Quiz: Fishers Exact Test

Practice Fishers Exact Test in Biostatistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Fishers Exact Test, giving you a quick way to practice the rules, question types, and explanations that matter most for Biostatistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A researcher compares complication rates between two surgical procedures: Procedure A had 1 complication in 4 surgeries, Procedure B had 3 complications in 5 surgeries. For Fisher's exact test, the researcher must enumerate all possible 2×2 tables with the same marginal totals. What constraint determines the range of possible values for the 'Procedure A complications' cell?

  1. The value must be between max(0, total complications - Procedure B surgeries) and min(Procedure A surgeries, total complications) (correct answer)
  2. The value must be between 0 and the smaller of the two procedure sample sizes
  3. The value must be between 0 and the total number of complications observed across both procedures
  4. The value must be between 1 and 4 since these are the observed minimum and maximum for Procedure A
  5. The value can be any integer between 0 and 9 since this represents the total sample size
Explanation: Fisher's exact test requires you to consider all possible 2×2 tables that could have occurred with the same marginal totals as your observed data. This means keeping the row totals (surgeries per procedure) and column totals (total complications vs. no complications) fixed while allowing the cell values to vary. From the given data: Procedure A has 4 surgeries, Procedure B has 5 surgeries, and there are 4 total complications (1+3) across both procedures. When enumerating possible tables, you need to determine how many complications Procedure A could theoretically have had. The correct range follows two logical constraints. The minimum possible complications for Procedure A is max⁡(0,4−5)=0\max(0, 4-5) = 0 because you can't have negative complications, and if Procedure B had all possible complications, Procedure A would have none. The maximum is min⁡(4,4)=4\min(4, 4) = 4 because Procedure A can't have more complications than either its total surgeries or the total complications observed. Choice B incorrectly focuses on sample sizes rather than the interaction between complications and surgeries. Choice C ignores the constraint that Procedure A can't exceed its own sample size of 4 surgeries. Choice D makes the fundamental error of using the observed data range (1-4) rather than considering all theoretically possible values given the marginal constraints. Remember: In Fisher's exact test, the key insight is that marginal totals are fixed, but internal cell values can vary within logical bounds determined by those margins. Always think about both the minimum and maximum constraints simultaneously.

Question 2

A researcher is comparing treatment outcomes between two small groups using Fisher's exact test. The contingency table shows 3 successes out of 8 patients in Group A and 7 successes out of 9 patients in Group B. What is the primary reason Fisher's exact test is more appropriate than the chi-square test for this analysis?

  1. The expected cell frequencies are too small for the chi-square approximation to be valid (correct answer)
  2. Fisher's exact test provides a more conservative p-value than the chi-square test
  3. The sample sizes are unequal between the two groups being compared
  4. Fisher's exact test accounts for the correlation between treatment assignment and outcome
  5. The chi-square test requires normally distributed data while Fisher's exact test does not
Explanation: When choosing between Fisher's exact test and the chi-square test for analyzing contingency tables, you need to consider whether the assumptions for the chi-square approximation are met. The chi-square test relies on a normal approximation that requires adequate expected frequencies in each cell of the contingency table. Let's examine this data: Group A has 3/8 successes (5/8 failures) and Group B has 7/9 successes (2/9 failures). When you calculate the expected frequencies under the assumption of no association, several cells fall well below the conventional threshold of 5. The chi-square test's normal approximation becomes unreliable when expected frequencies are too small, potentially leading to inaccurate p-values. Fisher's exact test sidesteps this problem by calculating the exact probability distribution rather than relying on approximations, making it the appropriate choice here. Looking at the incorrect options: Option B is wrong because Fisher's exact test isn't inherently more conservative—it simply provides the exact p-value rather than an approximation. Option C misses the point entirely; unequal sample sizes don't determine test choice between these methods—many valid chi-square analyses have unequal groups. Option D reflects a misunderstanding of what Fisher's exact test does; it doesn't account for correlation between variables any differently than the chi-square test—both test for independence in contingency tables. Study tip: Always check expected cell frequencies when deciding between chi-square and Fisher's exact test. If any expected frequency is below 5 (some sources say below 1), or if more than 20% of cells have expected frequencies below 5, choose Fisher's exact test for small contingency tables.

Question 3

A clinical trial compares two treatments with the following results: Treatment A had 2 cures out of 5 patients, Treatment B had 0 cures out of 4 patients. When applying Fisher's exact test, what assumption is made about the marginal totals during the calculation?

  1. The marginal totals are treated as fixed constants given by the experimental design (correct answer)
  2. The marginal totals are assumed to follow a binomial distribution with unknown parameters
  3. The marginal totals are considered random variables that must be estimated from the data
  4. The marginal totals are assumed to be equal between groups for valid comparison
  5. The marginal totals are treated as normally distributed approximations to the true population values
Explanation: Fisher's exact test is used when you have small sample sizes in contingency tables and need to calculate exact p-values rather than relying on normal approximations. The key insight is understanding what assumptions the test makes about your data structure. Fisher's exact test treats the marginal totals (row and column sums) as fixed constants determined by your experimental design. In your example, you decided to treat 5 patients with Treatment A and 4 patients with Treatment B - these totals are fixed by design. Similarly, the total number of cures (2) is treated as fixed when calculating the exact probability distribution. The test then asks: "Given these fixed marginal totals, what's the probability of observing this particular arrangement or one more extreme?" Looking at the incorrect options: B) suggests marginal totals follow a binomial distribution, but Fisher's test doesn't model the totals as random - it conditions on them being fixed. C) claims marginal totals are random variables to be estimated, which contradicts Fisher's conditioning approach where totals are treated as known constants from your study design. D) assumes equal marginal totals between groups, but Fisher's test works perfectly fine with unequal group sizes (like your 5 vs 4 patients). The correct answer is A because Fisher's exact test fundamentally assumes marginal totals are fixed by your experimental design choices. Study tip: Remember that "exact" in Fisher's exact test refers to conditioning on fixed marginal totals - this is what allows the test to provide exact p-values for small samples without normal approximations.

Question 4

In a 2×2 contingency table analysis using Fisher's exact test, a researcher obtains a one-tailed p-value of 0.032. If the alternative hypothesis is non-directional, what would be the appropriate two-tailed p-value for this test?

  1. The two-tailed p-value could be 0.064, 0.032, or somewhere between, depending on the table symmetry (correct answer)
  2. The two-tailed p-value is exactly 0.064 by doubling the one-tailed result
  3. The two-tailed p-value is exactly 0.032 since it includes both tails by definition
  4. The two-tailed p-value is 0.016 since each tail contributes half the probability
  5. The two-tailed p-value cannot be determined without recalculating from the original data
Explanation: When you encounter Fisher's exact test questions, remember that the relationship between one-tailed and two-tailed p-values isn't as straightforward as with symmetric distributions like the normal or t-distribution. Fisher's exact test calculates p-values by summing probabilities of all tables as extreme or more extreme than the observed table. For a one-tailed test, you sum probabilities in one direction. For a two-tailed test, you must sum probabilities in both directions—but here's the key insight: the distribution of possible table configurations is often asymmetric. The correct answer is A because the two-tailed p-value depends on the specific contingency table structure. If the table is perfectly symmetric, doubling gives you the exact two-tailed p-value (2×0.032=0.0642 \times 0.032 = 0.064). However, most real tables are asymmetric, meaning extreme tables in one tail might not have corresponding tables of equal probability in the other tail. This results in a two-tailed p-value that could be 0.064, 0.032, or anywhere between. Option B incorrectly assumes perfect symmetry—this "doubling rule" only applies to symmetric distributions. Option C misunderstands the calculation entirely; a one-tailed result of 0.032 cannot simultaneously be the two-tailed result. Option D incorrectly reverses the relationship, suggesting you divide rather than account for both tails. Study tip: For Fisher's exact test, always remember that the exact two-tailed p-value requires examining the actual table structure. Unlike parametric tests, you can't simply double the one-tailed result due to the discrete, often asymmetric nature of contingency table probability distributions.

Question 5

A pharmaceutical company tests a new drug in a small trial with 6 patients receiving the drug (3 improved, 3 did not improve) and 4 patients receiving placebo (0 improved, 4 did not improve). When would Fisher's exact test be preferred over other tests of association for these data?

  1. When any expected cell frequency under independence would be less than 5 (correct answer)
  2. When the total sample size is less than 20 patients regardless of cell frequencies
  3. When one of the groups shows no events in either outcome category
  4. When the research hypothesis predicts a specific direction of treatment effect
  5. When the outcome variable has only two possible categories to choose from
Explanation: When analyzing categorical data in small samples, you need to decide between chi-square tests and Fisher's exact test based on whether the chi-square approximation is reliable. The key principle is that chi-square tests assume large sample sizes where expected frequencies follow a normal distribution. Fisher's exact test is preferred when the chi-square test's assumptions are violated, specifically when expected cell frequencies are too small. The standard rule is that if any expected cell frequency under the null hypothesis of independence is less than 5, you should use Fisher's exact test instead of chi-square. In this drug trial, with such small sample sizes (6 and 4 patients), calculating expected frequencies would likely yield values below 5 in several cells, making Fisher's exact test the appropriate choice. Looking at the incorrect options: Choice B is wrong because the decision isn't based solely on total sample size—it's about expected cell frequencies specifically. A study with 15 patients might still use chi-square if expected frequencies are adequate. Choice C is incorrect because having zero events in a category doesn't automatically require Fisher's exact test; it depends on whether expected frequencies meet the threshold. Choice D misses the point entirely—the choice between these tests is about statistical assumptions, not about directional versus non-directional hypotheses. Remember this rule: When you see small sample categorical data analysis questions, always think about expected cell frequencies first. If any expected frequency would be less than 5, Fisher's exact test is your go-to method for reliable results.

Question 6

In Fisher's exact test, the null hypothesis of independence is tested by examining all possible 2×2 tables with the same marginal totals as the observed data. What key assumption about the data collection process makes this approach valid?

  1. The marginal totals represent fixed constraints from the study design rather than random outcomes (correct answer)
  2. The observations within each cell of the table are independent and identically distributed
  3. The probability of success is constant across all individuals in the study population
  4. The sample was randomly selected from a much larger population of similar individuals
  5. The outcome variable follows a normal distribution within each treatment group
Explanation: Fisher's exact test is fundamentally based on the hypergeometric distribution, which models sampling without replacement from a finite population. Understanding when this approach is valid requires recognizing what makes the marginal totals "fixed" versus "random." The key insight is that Fisher's exact test assumes you're conditioning on the marginal totals - treating them as if they were predetermined by your study design rather than random outcomes. This happens in scenarios like case-control studies where you deliberately choose to sample a specific number of cases and controls, or in randomized experiments where you assign a fixed number of subjects to each treatment group. When marginal totals are truly fixed by design, the hypergeometric distribution correctly models all possible ways the data could have been distributed within those constraints, making answer A correct. Answer B describes a general assumption for most statistical tests but doesn't explain why we can condition on marginal totals specifically. Answer C refers to the assumption of constant probability, which is more relevant to binomial models than to the conditioning logic of Fisher's exact test. Answer D describes random sampling assumptions that apply broadly to inference but doesn't address the specific mathematical foundation that makes Fisher's exact test valid. Study tip: When you encounter Fisher's exact test questions, always ask yourself: "Were the row and column totals determined by the researcher's design choices, or did they emerge randomly from the data?" This distinction determines whether the hypergeometric conditioning approach is appropriate.

Question 7

A medical researcher analyzes the relationship between a new screening test result (positive/negative) and disease status (present/absent) in 15 patients. The study finds 4 true positives, 2 false positives, 1 false negative, and 8 true negatives. Why might Fisher's exact test be necessary for this analysis?

  1. The expected frequency for false negatives under independence is approximately 3.33, which is less than 5 (correct answer)
  2. The total sample size of 15 patients is below the minimum requirement for chi-square testing
  3. The screening test shows high specificity, making the normal approximation inappropriate
  4. The study design used a case-control approach rather than a cohort study design
  5. The presence of both false positives and false negatives creates dependency in the data
Explanation: When analyzing categorical data in contingency tables, you need to check whether the assumptions for chi-square testing are met. The key assumption is that expected frequencies under the null hypothesis of independence should be at least 5 in each cell. Let's construct the 2×2 table from the given data:
  • True positives: 4, False positives: 2 (Total positive tests: 6)
  • False negatives: 1, True negatives: 8 (Total negative tests: 9)
  • Disease present: 5, Disease absent: 10 (Total: 15)
Under independence, the expected frequency for each cell equals (row total × column total) ÷ grand total. For false negatives: (9 × 5) ÷ 15 = 3.0. Since this is less than 5, the chi-square test's normal approximation breaks down, making Fisher's exact test necessary. Answer A correctly identifies this violation. Answer B is incorrect because there's no absolute minimum sample size requirement for chi-square testing—it's about expected frequencies, not total sample size. Answer C misses the point entirely; high specificity doesn't determine test choice, and the specificity here is actually moderate (8/10 = 80%). Answer D is wrong because the choice between Fisher's exact test and chi-square depends on expected frequencies, not study design type. Study tip: Always calculate expected frequencies for all cells in a contingency table before choosing your statistical test. If any expected frequency is below 5, use Fisher's exact test instead of chi-square. This is a fundamental rule in categorical data analysis.

Question 8

In a Fisher's exact test comparing treatment outcomes, a statistician calculates that 12 different 2×2 tables are possible with the given marginal totals. The observed table has a probability of 0.08 under the null hypothesis. For a two-tailed test with α = 0.05, which tables should be included in the p-value calculation?

  1. All tables with probability ≤ 0.08, regardless of which marginal cell they favor (correct answer)
  2. The 5% of tables with the smallest probabilities under the null hypothesis
  3. All tables in the same direction as the observed table plus the single most extreme opposite table
  4. Tables with probabilities ≤ 0.025 in each tail of the distribution separately
  5. The observed table plus any tables with probabilities exactly equal to 0.08
Explanation: Fisher's exact test questions often trip up students because they require understanding how two-tailed p-values are calculated when dealing with discrete probability distributions rather than continuous ones. In Fisher's exact test, you calculate the p-value by summing probabilities of all tables that are "as extreme or more extreme" than your observed table. The key insight is that "extreme" means having a probability under the null hypothesis that's as small or smaller than what you observed, regardless of which direction the deviation goes. Since your observed table has probability 0.08, you include all possible tables with probability ≤ 0.08 in your p-value calculation. Answer A is correct because it captures this principle exactly - you sum probabilities of all tables with probability ≤ 0.08, whether they show effects in the same direction as your observed table or the opposite direction. Answer B misunderstands how the α level relates to table selection. You don't simply take the 5% most extreme tables; you take all tables as extreme as or more extreme than what you actually observed. Answer C describes a flawed hybrid approach that doesn't properly account for the discrete nature of the distribution. You can't just add "the single most extreme opposite table." Answer D incorrectly tries to split the discrete distribution into symmetric tails with exactly 0.025 probability each, which isn't possible with discrete data where only specific probability values exist. Remember: In Fisher's exact test, "two-tailed" means including all tables with probabilities as small or smaller than your observed table's probability, regardless of direction.

Question 9

When performing Fisher's exact test on a 2×2 contingency table, the hypergeometric probability for any specific table with cell counts a, b, c, d and marginal totals R₁, R₂, C₁, C₂, and total n is calculated using a specific formula. What does this probability represent in the context of the test?

  1. The probability of observing exactly that table configuration given the null hypothesis of independence and fixed marginal totals (correct answer)
  2. The probability that the null hypothesis of independence is true given the observed table configuration
  3. The probability of rejecting the null hypothesis when it is actually true for that table
  4. The probability of obtaining a more extreme table than the one specified in the calculation
  5. The probability that the alternative hypothesis of association is true for that specific table arrangement
Explanation: When you encounter Fisher's exact test questions, remember that this test evaluates independence in 2×2 tables when sample sizes are small, using the hypergeometric distribution with fixed marginal totals. The hypergeometric probability formula in Fisher's exact test calculates the exact probability of observing a specific table configuration under very precise conditions: assuming the null hypothesis of independence is true AND the marginal totals are fixed. This probability represents how likely you are to see exactly those cell counts (a, b, c, d) by random chance alone, given those constraints. Answer A correctly captures this interpretation. Answer B confuses the direction of conditional probability - this would be a Bayesian posterior probability, but Fisher's exact test calculates the probability of data given the hypothesis, not the hypothesis given the data. Answer C describes a Type I error probability (alpha level), which is conceptually different from the probability of any single table configuration. Answer D describes what you'd use to calculate the p-value - you sum probabilities of tables as extreme or more extreme than observed, but the hypergeometric probability itself is just for that specific table. The key distinction is between the probability of one specific table (what the hypergeometric formula gives you) versus the p-value of the test (which sums multiple table probabilities). Remember: in Fisher's exact test, you're calculating precise probabilities for exact table configurations under the null hypothesis, then summing the relevant ones to get your final p-value for hypothesis testing.

Question 10

A clinical researcher tests whether a rare side effect occurs equally in two treatment groups. Group 1 (n=7) had 2 patients with the side effect, while Group 2 (n=5) had 0 patients with the side effect. In Fisher's exact test, how many different 2×2 tables share the same marginal totals as this observed data?

  1. 3 tables, with Group 1 side effects ranging from 0 to 2 patients (correct answer)
  2. 6 tables, corresponding to all possible combinations of the cell values
  3. 8 tables, equal to the number of ways to distribute 2 side effects among 12 patients
  4. 12 tables, equal to the total sample size across both groups
  5. 5 tables, with Group 2 side effects ranging from 0 to 2 patients
Explanation: When you encounter Fisher's exact test questions, focus on understanding how the test constrains the possible outcomes by fixing the marginal totals (row and column sums). In this problem, the observed 2×2 table has fixed margins: Group 1 has 7 patients total, Group 2 has 5 patients total, and exactly 2 patients across both groups experienced the side effect. Fisher's exact test asks: given these fixed totals, how many different ways could the data be arranged? Since the margins are fixed, if Group 1 has kk patients with the side effect, then Group 2 must have (2−k)(2-k) patients with the side effect. The constraint is that kk cannot exceed the smaller of: (1) Group 1's size (7), or (2) the total number of side effects (2). Therefore, kk can only be 0, 1, or 2, giving us exactly 3 possible tables. Answer A is correct. Looking at the wrong answers: B suggests 6 tables from "all possible combinations," but this ignores that Fisher's exact test fixes the marginal totals, severely limiting the possibilities. C proposes 8 tables based on distributing 2 side effects among 12 patients, but this calculation doesn't account for the group structure and fixed group sizes. D simply uses the total sample size (12) as the number of tables, which has no logical connection to how contingency tables work. Study tip: Remember that Fisher's exact test always fixes both row and column totals. Once you identify these constraints, the number of possible tables is determined by how the smallest total can be distributed while respecting all margins.

Question 11

In a small randomized trial comparing two pain medications, 4 out of 6 patients improved with Drug A, while 1 out of 5 patients improved with Drug B. A statistician performs Fisher's exact test and finds three possible table configurations with the same marginal totals. If these tables have probabilities 0.12, 0.30, and 0.58 under the null hypothesis, which represents the observed data?

  1. Cannot be determined without knowing which table corresponds to the actual cell counts observed (correct answer)
  2. The table with probability 0.30, representing the middle value among the three possibilities
  3. The table with probability 0.12, representing the most extreme departure from independence
  4. The table with probability 0.58, representing the most likely configuration under the null hypothesis
  5. All three tables equally represent the observed data since they have identical marginal totals
Explanation: When you encounter Fisher's exact test questions, remember that this test evaluates all possible ways to arrange data while keeping the row and column totals (marginal totals) fixed. The key insight here is distinguishing between what the test calculates versus what actually happened in your study. The question describes a completed study with specific observed results: 4/6 patients improved with Drug A and 1/5 improved with Drug B. These are the actual data collected. Fisher's exact test then considers this observed configuration alongside all other possible arrangements that maintain the same marginal totals, calculating probabilities for each under the null hypothesis of no treatment difference. Answer A is correct because the question doesn't tell you which of the three probability values (0.12, 0.30, or 0.58) corresponds to the specific table containing 4 successes in the Drug A group and 1 success in the Drug B group. You know the observed data, but you don't know which probability matches those exact cell counts. Answer B incorrectly assumes the middle probability represents observed data. Answer C wrongly suggests the most extreme table (lowest probability) must be the observed one - but Fisher's exact test considers all possible tables regardless of which was actually observed. Answer D incorrectly identifies the most likely table under the null hypothesis as the observed data. Study tip: In Fisher's exact test problems, always distinguish between the observed data (what actually happened in your study) and the theoretical tables the test examines (all possible arrangements with the same margins).

Question 12

In a Fisher's exact test with a 2×2 table, if one cell has a count of 0, what effect does this have on the calculation and interpretation of the test?

  1. The test remains valid, but the number of possible tables with the same marginal totals is reduced (correct answer)
  2. The test becomes invalid because hypergeometric probabilities cannot be calculated with zero counts
  3. The test automatically yields a p-value of 1.0 indicating no evidence against independence
  4. The test requires a continuity correction to account for the discrete nature of the zero count
  5. The test must be modified to use the exact binomial distribution instead of hypergeometric
Explanation: Fisher's exact test evaluates independence in 2×2 contingency tables by calculating the exact probability of observing your data (or more extreme) given fixed marginal totals. The test uses the hypergeometric distribution to enumerate all possible tables with the same row and column sums, then determines where your observed table falls in this distribution. When one cell contains zero, the test remains completely valid. Zero is a legitimate count that simply represents no observations in that category combination. The hypergeometric formula P=(a+ba)(c+dc)(na+c)P = \frac{\binom{a+b}{a}\binom{c+d}{c}}{\binom{n}{a+c}} handles zeros perfectly well since you're essentially calculating combinations where some groups have zero members. Answer A correctly identifies that while the test stays valid, having a zero reduces the number of possible table configurations. With tighter constraints from the zero, fewer arrangements of the data are mathematically possible given the fixed marginals. Answer B is wrong because hypergeometric probabilities calculate perfectly fine with zero counts - zero is just another integer in the combinatorial formulas. Answer C incorrectly assumes zeros automatically eliminate evidence against independence. In fact, extreme values (including zeros) often provide strong evidence for rejecting independence, potentially yielding very small p-values. Answer D confuses Fisher's exact test with chi-square tests - Fisher's exact test doesn't use continuity corrections because it calculates exact probabilities rather than relying on asymptotic approximations. Remember: Fisher's exact test is called "exact" precisely because it handles small counts and zeros without approximations, making it ideal when chi-square assumptions fail.

Question 13

A epidemiologist studies the association between exposure to a chemical (yes/no) and development of a rare disease (yes/no) in a small community of 12 people. The study finds 3 exposed individuals with disease, 2 exposed individuals without disease, 1 unexposed individual with disease, and 6 unexposed individuals without disease. What assumption of Fisher's exact test might be questionable in this scenario?

  1. The assumption that observations are independent may be violated due to shared community exposure sources (correct answer)
  2. The assumption that disease status follows a binomial distribution may not hold for rare diseases
  3. The assumption that exposure and disease are categorical variables is clearly violated
  4. The assumption that marginal totals should be approximately equal is not satisfied
  5. The assumption that the sample size exceeds 10 observations is barely met
Explanation: When analyzing statistical tests like Fisher's exact test, you need to consider whether the underlying assumptions are met in your specific study context. Fisher's exact test assumes that observations are independent - meaning one person's exposure or disease status doesn't influence another's. In this community study, the independence assumption is highly questionable because all 12 people live in the same small community. They likely share environmental factors, water sources, air quality, and other exposure pathways that could affect both their chemical exposure and disease risk. If the chemical comes from a contaminated local well or factory, for example, exposure patterns aren't random but clustered by geography and shared resources. This violates the independence assumption that Fisher's exact test requires. Looking at the wrong answers: B is incorrect because Fisher's exact test actually works well for rare diseases and doesn't assume any particular disease prevalence. The binomial distribution assumption isn't problematic here. C is wrong because exposure (yes/no) and disease (yes/no) are clearly categorical variables, which is exactly what Fisher's exact test is designed for. D misunderstands the test - Fisher's exact test doesn't require approximately equal marginal totals. In fact, it's specifically useful when sample sizes are small and unbalanced, like in this 2×2 table with 5 exposed vs. 7 unexposed individuals. Study tip: When evaluating statistical test assumptions, always consider the real-world context of data collection. Geographic clustering, shared environments, or related individuals often violate independence assumptions, which is a common issue in epidemiological studies.

Question 14

A biostatistician performs Fisher's exact test on treatment data and obtains a one-tailed p-value of 0.028. The research question asks whether the treatments differ in either direction (non-directional alternative). If the distribution of possible tables under the null hypothesis is symmetric, what would be the two-tailed p-value?

  1. 0.056, assuming perfect symmetry allows doubling the one-tailed p-value (correct answer)
  2. 0.028, because the one-tailed p-value already accounts for both directions
  3. 0.014, representing half the one-tailed p-value for each direction tested
  4. Cannot be determined because Fisher's exact test is inherently one-tailed
  5. 0.972, representing one minus the one-tailed p-value for the complement
Explanation: When you encounter Fisher's exact test questions involving one-tailed versus two-tailed p-values, focus on understanding how the test distributes probability across possible outcomes under the null hypothesis. Fisher's exact test calculates the probability of observing your data or more extreme results, given the null hypothesis of no association. A one-tailed test examines extremeness in only one direction, while a two-tailed test considers both directions. When the distribution of possible tables under the null hypothesis is symmetric, the probability mass is evenly distributed on both sides. Since you obtained a one-tailed p-value of 0.028 and the distribution is symmetric, this represents the probability of observing results this extreme or more in one specific direction. For a non-directional (two-tailed) alternative hypothesis, you need to account for equally extreme results in the opposite direction. Under perfect symmetry, this opposite tail also contributes 0.028, making the two-tailed p-value 0.028+0.028=0.0560.028 + 0.028 = 0.056. Answer A correctly identifies this doubling relationship under symmetric conditions. Answer B incorrectly assumes the one-tailed p-value already accounts for both directions—it doesn't. Answer C mistakenly halves the p-value, confusing the conversion direction. Answer D is wrong because Fisher's exact test can absolutely be adapted for two-tailed hypotheses. Study tip: Remember that converting from one-tailed to two-tailed p-values under symmetry requires doubling, not halving. The key phrase "symmetric distribution" signals that this doubling relationship holds perfectly.

Question 15

A researcher wants to test the association between gender and disease status in a small cohort study. The data shows 4 diseased males out of 8 males total, and 1 diseased female out of 6 females total. Which probability distribution underlies the calculation of Fisher's exact test for this scenario?

  1. Hypergeometric distribution conditioning on the marginal totals of the contingency table (correct answer)
  2. Binomial distribution with success probability estimated from the combined sample
  3. Chi-square distribution with degrees of freedom equal to the number of cells minus one
  4. Normal distribution approximation to the sampling distribution of the odds ratio
  5. Poisson distribution modeling the rare occurrence of disease in small populations
Explanation: When analyzing associations in small contingency tables, you need to understand which probability distribution governs the test statistic. Fisher's exact test is specifically designed for situations where sample sizes are too small for chi-square approximations to be reliable. Fisher's exact test calculates the probability of observing your data (or more extreme results) by treating the marginal totals as fixed. In your example, you condition on having exactly 8 males, 6 females, 5 total diseased individuals, and 9 total healthy individuals. Given these fixed margins, the test asks: "What's the probability of seeing exactly 4 diseased males?" This follows the hypergeometric distribution, which models sampling without replacement from a finite population with two categories. Choice A correctly identifies this hypergeometric distribution conditioning on marginal totals. Choice B is wrong because the binomial distribution assumes independent trials with constant probability, but Fisher's exact test fixes the total number of successes (diseased individuals), making trials dependent. Choice C confuses Fisher's exact test with Pearson's chi-square test - while chi-square tests do use the chi-square distribution, Fisher's exact test uses exact probabilities, not asymptotic approximations. Choice D is incorrect because Fisher's exact test doesn't rely on normal approximations at all; it calculates exact probabilities using discrete distributions. Study tip: Remember that Fisher's exact test = hypergeometric distribution with fixed margins. When you see "exact test" in small samples, think "hypergeometric conditioning on totals," not approximations or asymptotic distributions.

Question 16

In Fisher's exact test, the p-value is calculated by summing probabilities of all possible tables that are 'as extreme or more extreme' than the observed table. For a two-tailed test, which criterion determines whether a table is considered 'as extreme'?

  1. Tables with the same or smaller probability as the observed table under the null hypothesis (correct answer)
  2. Tables with test statistics greater than or equal to the observed test statistic
  3. Tables with the same marginal totals but different cell arrangements than observed
  4. Tables with the same degrees of freedom and similar expected frequencies as observed
  5. Tables with cell counts that deviate from independence by the same amount as observed
Explanation: Fisher's exact test is used when you have a 2×2 contingency table and want to test for independence between two categorical variables. Unlike chi-square tests that use approximations, Fisher's exact test calculates the exact probability of observing your data under the null hypothesis of independence. The key insight is that Fisher's exact test uses the hypergeometric distribution to calculate probabilities. Under the null hypothesis, the marginal totals are fixed, and you're asking: "Given these row and column totals, what's the probability of seeing this particular arrangement of cells, or something even more unlikely?" Answer A is correct because Fisher's exact test defines "as extreme or more extreme" based on probability rankings. You calculate the probability of your observed table, then sum the probabilities of all possible tables (with the same marginal totals) that have equal or smaller probabilities. This gives you the exact p-value. Answer B is wrong because Fisher's exact test doesn't use a traditional test statistic like chi-square or t-tests do. It works directly with probabilities from the hypergeometric distribution. Answer C misunderstands the concept—you do consider all tables with the same marginal totals, but only those that are as or more extreme (less probable) than what you observed. Answer D confuses Fisher's exact test with chi-square tests, which do involve expected frequencies and degrees of freedom. Fisher's exact test bypasses these concepts entirely. Remember: Fisher's exact test ranks tables by their probability under the null hypothesis. The p-value includes all arrangements that are as unlikely or more unlikely than your observed data.

Question 17

A researcher performs Fisher's exact test on a 2×2 table and finds that only 3 possible tables (including the observed one) have the same marginal totals. If the observed table has a probability of 0.15 under the null hypothesis, and the other two tables have probabilities of 0.05 and 0.30, what is the one-tailed p-value for the most extreme direction?

  1. 0.20, representing the sum of probabilities for tables as extreme or more extreme (correct answer)
  2. 0.15, representing the probability of observing exactly the data that was seen
  3. 0.50, representing the probability of the most likely table configuration
  4. 0.05, representing the probability of the most extreme possible table
  5. 0.35, representing the sum of the two most extreme table probabilities
Explanation: When you encounter Fisher's exact test questions, remember that this test calculates the exact probability of observing data as extreme or more extreme than what was actually seen, given fixed marginal totals. Fisher's exact test works by identifying all possible tables with the same marginal totals as your observed data, then calculating each table's probability under the null hypothesis. The p-value is the sum of probabilities for tables that are "as extreme or more extreme" in the direction of interest. In this problem, you have three possible tables with probabilities 0.15 (observed), 0.05, and 0.30. For a one-tailed test, you need to determine which direction represents "more extreme" than your observed table, then sum the probabilities accordingly. Since the question asks for "the most extreme direction," you're looking at the tail that includes the table with probability 0.05 (the smallest probability, hence most extreme). The one-tailed p-value would be 0.15 + 0.05 = 0.20, making choice A correct. Choice B (0.15) incorrectly gives only the probability of the exact observed table, not the cumulative probability for the tail. Choice C (0.50) mistakenly suggests this relates to the most likely configuration, but p-values measure extremeness, not likelihood. Choice D (0.05) only includes the most extreme table while ignoring the observed table's contribution to the tail. Remember: Fisher's exact test p-values always represent cumulative probabilities in the tail(s) of interest, never just single table probabilities. Always sum the probabilities for tables as extreme or more extreme than observed.

Question 18

A quality control study examines defect rates in products from two production lines. Line A produced 3 defective items out of 9 total, while Line B produced 1 defective item out of 6 total. When applying Fisher's exact test, what is the key difference between this application and a chi-square test of independence for the same data?

  1. Fisher's exact test provides the exact p-value while chi-square uses an asymptotic approximation that may be inaccurate (correct answer)
  2. Fisher's exact test assumes equal production rates between lines while chi-square allows for different rates
  3. Fisher's exact test is one-tailed by default while chi-square test is always two-tailed
  4. Fisher's exact test requires the defect rate to be less than 50% while chi-square has no such restriction
  5. Fisher's exact test conditions on production line assignment while chi-square treats it as random
Explanation: When you encounter questions comparing statistical tests for categorical data with small sample sizes, focus on the fundamental difference between exact methods and approximation methods. Fisher's exact test calculates the precise probability by enumerating all possible outcomes under the null hypothesis using the hypergeometric distribution. With your data (3/9 defects for Line A, 1/6 for Line B), Fisher's test determines the exact p-value by considering every possible arrangement of defects that could occur given these marginal totals. The chi-square test, conversely, relies on an asymptotic approximation that assumes your data follows a chi-square distribution. This approximation becomes unreliable with small expected cell counts (the rule of thumb is expected counts ≥ 5 in each cell). In your example, some expected cell counts would be quite small, making the chi-square approximation potentially inaccurate while Fisher's exact test remains valid regardless of sample size. Answer A correctly identifies this key distinction. Answer B is backwards—neither test assumes equal production rates; both test for independence between variables. Answer C mischaracterizes the tests' sidedness; both can be conducted as one- or two-tailed depending on your hypothesis. Answer D incorrectly suggests Fisher's test has restrictions on the proportion values, when in fact it works for any proportions. Study tip: Remember that "exact" tests calculate precise probabilities while "asymptotic" tests use approximations that require sufficient sample sizes. When you see small sample categorical data, think Fisher's exact test for accuracy.

Question 19

A researcher applies Fisher's exact test to determine if treatment response differs between males and females. The data shows 5 responding males out of 8 total males, and 1 responding female out of 4 total females. If the researcher reports a 'mid-p' value instead of the standard Fisher's exact p-value, what modification is being made to the calculation?

  1. Half the probability of the observed table is added to the sum of more extreme table probabilities (correct answer)
  2. The p-value is calculated using the median probability among all possible table configurations
  3. The critical value is adjusted to α/2 to account for the two-tailed nature of the test
  4. The hypergeometric probabilities are replaced with their normal approximation counterparts
  5. The marginal totals are treated as random rather than fixed during the probability calculation
Explanation: When you encounter questions about Fisher's exact test modifications, focus on understanding how different p-value calculations handle the probability of the observed data itself. Fisher's exact test calculates p-values by summing probabilities of tables as extreme or more extreme than the observed table. The standard approach includes the full probability of the observed table in this sum. However, the mid-p modification makes a specific adjustment: it adds only half the probability of the observed table to the sum of probabilities from more extreme tables. This approach addresses the inherent conservatism of Fisher's exact test, which tends to produce p-values that are too large because it fully counts the observed outcome. Choice A correctly describes this modification - half the observed table's probability is added to the more extreme probabilities. Choice B incorrectly suggests using median probabilities across table configurations, which isn't how mid-p works. Choice C confuses the mid-p adjustment with multiple comparison corrections or two-tailed test adjustments - the mid-p method doesn't involve adjusting alpha levels. Choice D incorrectly references normal approximations, but mid-p still uses the exact hypergeometric distribution; it just modifies how the observed table's probability is incorporated. Study tip: Remember that "mid-p" literally means using the "middle" or half of the observed probability. When you see mid-p mentioned with any exact test, think "half the observed probability plus all the more extreme probabilities" - this creates less conservative p-values than the standard exact approach.

Question 20

A biostatistician is reviewing a colleague's analysis that used Fisher's exact test on a dataset with 50 patients per group and found a p-value of 0.048. The colleague argues that Fisher's exact test was necessary because 'it provides the exact p-value rather than an approximation.' What is the most appropriate critique of this reasoning?

  1. Fisher's exact test becomes computationally intensive with large samples and provides essentially the same result as the chi-square test
  2. With 50 patients per group, the chi-square test approximation would be highly accurate, making Fisher's exact test unnecessarily conservative
  3. Fisher's exact test is only valid for small samples where cell counts are less than 10, so it was incorrectly applied here
  4. The 'exact' nature of Fisher's test refers to the probability calculation method, but both tests would yield very similar p-values with this sample size (correct answer)
Explanation: With large samples (50 per group), both Fisher's exact test and chi-square test would yield very similar p-values. The 'exact' in Fisher's exact test refers to using the exact hypergeometric distribution rather than asymptotic approximations, but this advantage becomes negligible with large samples. Choice A mentions computational intensity but doesn't address the core statistical reasoning. Choice B incorrectly characterizes Fisher's test as 'conservative' in large samples. Choice C is wrong - Fisher's exact test is valid for any sample size, though unnecessary for large samples.