Biostatistics Quiz: Chi Square Distribution
20 questions · exam conditions
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Chi Square DistributionQuestion 1 of 20

A 2×22 \times 2 contingency table shows the relationship between exposure status and disease outcome. If the chi-square statistic is 6.84 and the sample size is 200, what is Cramér's V?

V = 0.185 indicating a weak to moderate association between variables
V = 0.034 indicating a very weak association between the variables
V = 0.274 indicating a moderate association between the variables
V = 0.427 indicating a strong association between the variables
Cannot calculate V without knowing the individual cell frequencies
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Biostatistics Quiz

Biostatistics Quiz: Chi Square Distribution

Practice Chi Square Distribution 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 Chi Square Distribution, 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 2×22 \times 2 contingency table shows the relationship between exposure status and disease outcome. If the chi-square statistic is 6.84 and the sample size is 200, what is Cramér's V?

  1. V = 0.185 indicating a weak to moderate association between variables (correct answer)
  2. V = 0.034 indicating a very weak association between the variables
  3. V = 0.274 indicating a moderate association between the variables
  4. V = 0.427 indicating a strong association between the variables
  5. Cannot calculate V without knowing the individual cell frequencies
Explanation: When you encounter chi-square statistics in biostatistics, you often need to assess not just statistical significance, but also the strength of association between variables. Cramér's V is the key measure for this in contingency tables, providing a standardized effect size that ranges from 0 (no association) to 1 (perfect association). To calculate Cramér's V, use the formula: V=χ2n×(k−1)V = \sqrt{\frac{\chi^2}{n \times (k-1)}}, where nn is the sample size and kk is the smaller of the number of rows or columns. For a 2×22 \times 2 table, k=2k = 2, so: V=6.84200×(2−1)=6.84200=0.0342=0.185V = \sqrt{\frac{6.84}{200 \times (2-1)}} = \sqrt{\frac{6.84}{200}} = \sqrt{0.0342} = 0.185 This confirms A is correct — V = 0.185, indicating a weak to moderate association. Cohen's conventions suggest V = 0.10 is small, 0.30 is medium, and 0.50 is large, placing 0.185 in the weak-to-moderate range. B is wrong because 0.034 appears to be the intermediate calculation (6.84200\frac{6.84}{200}) before taking the square root. C is wrong because 0.274 likely results from incorrectly using k=3k = 3 instead of k=2k = 2 in the denominator. D is wrong because 0.427 suggests a computational error, possibly confusion with other effect size measures. Study tip: Always double-check your Cramér's V formula — remember that for a 2×22 \times 2 table, you divide by n×1n \times 1, not n×2n \times 2, since (k−1)=1(k-1) = 1.

Question 2

A researcher conducts a chi-square goodness-of-fit test with 6 categories to determine if observed frequencies match expected frequencies. The calculated chi-square statistic is 12.8. Using α=0.05\alpha = 0.05, what conclusion should the researcher draw?

  1. Reject the null hypothesis because 12.8 > 11.07 (critical value) (correct answer)
  2. Fail to reject the null hypothesis because 12.8 < 15.09 (critical value)
  3. Reject the null hypothesis because 12.8 > 9.49 (critical value)
  4. Fail to reject the null hypothesis because 12.8 < 12.83 (critical value)
  5. Cannot determine without knowing the sample size used in the study
Explanation: Chi-square goodness-of-fit tests require you to compare your calculated test statistic against a critical value from the chi-square distribution table. The key is determining the correct degrees of freedom and finding the corresponding critical value at your chosen significance level. For a goodness-of-fit test with 6 categories, your degrees of freedom equal k−1=6−1=5k - 1 = 6 - 1 = 5. At α=0.05\alpha = 0.05 with 5 degrees of freedom, the critical value from the chi-square table is 11.07. Since your calculated statistic (12.8) exceeds this critical value, you reject the null hypothesis that observed frequencies match expected frequencies. Answer A correctly identifies this logic: reject the null hypothesis because 12.8 > 11.07, using the proper critical value for 5 degrees of freedom. Answer B uses 15.09, which corresponds to α=0.01\alpha = 0.01 (not 0.05) with 5 degrees of freedom, making it too stringent for this test. Answer C uses 9.49, the critical value for 4 degrees of freedom at α=0.05\alpha = 0.05—this represents miscalculating degrees of freedom as k−2k - 2 instead of k−1k - 1. Answer D uses 12.83, which is the critical value for 6 degrees of freedom at α=0.05\alpha = 0.05, incorrectly using kk instead of k−1k - 1. Study tip: Always double-check your degrees of freedom calculation for goodness-of-fit tests—it's k−1k - 1 where kk is the number of categories. Then locate the exact critical value for your α\alpha level before making your decision.

Question 3

When conducting a chi-square goodness-of-fit test, a researcher finds that one expected frequency is 3.8. What is the most appropriate course of action?

  1. Proceed with the test since the expected frequency exceeds the minimum threshold
  2. Combine categories to increase expected frequencies before conducting the test (correct answer)
  3. Use Fisher's exact test instead of the chi-square test procedure
  4. Increase the sample size to ensure all expected frequencies exceed 10
  5. Apply a continuity correction to adjust for the low expected frequency
Explanation: When you encounter chi-square goodness-of-fit test questions, always check whether the expected frequencies meet the test's assumptions. The chi-square test requires that all expected frequencies be at least 5 (some sources say 10 for conservative application) to ensure the test statistic follows a chi-square distribution. With an expected frequency of 3.8, you're below this critical threshold. The correct solution is to combine adjacent or similar categories until all expected frequencies exceed 5. This maintains the validity of the chi-square approximation while preserving your ability to test the overall goodness-of-fit. Let's examine why the other options fall short: A is incorrect because 3.8 is below the minimum threshold of 5, not above it. Proceeding would violate the test's assumptions and potentially invalidate your results. C suggests Fisher's exact test, but this test is designed for contingency tables (testing independence between categorical variables), not goodness-of-fit testing. It's the wrong statistical procedure for this scenario. D proposes increasing sample size, which might eventually raise expected frequencies, but this approach is impractical and expensive. You should work with your existing data by combining categories rather than collecting more observations. Study tip: For chi-square tests, always verify the "Rule of 5" before proceeding. If any expected frequency falls below 5, your first instinct should be to combine categories logically. This is a standard practice that maintains statistical validity while working within the test's constraints.

Question 4

A researcher tests whether the distribution of blood types (A, B, AB, O) in a sample matches the expected population distribution (40%, 20%, 5%, 35%). With n = 200, what is the contribution to the chi-square statistic from the AB blood type if 18 participants have AB blood?

  1. (18−10)210=6.4\frac{(18-10)^2}{10} = 6.4 representing the AB contribution to total chi-square (correct answer)
  2. (18−5)25=33.8\frac{(18-5)^2}{5} = 33.8 representing the AB contribution to total chi-square
  3. (10−18)218=3.56\frac{(10-18)^2}{18} = 3.56 representing the AB contribution to total chi-square
  4. (18−10)2200=0.32\frac{(18-10)^2}{200} = 0.32 representing the AB contribution to total chi-square
  5. 18×10200=0.9\frac{18 \times 10}{200} = 0.9 representing the AB contribution to total chi-square
Explanation: This is a chi-square goodness-of-fit test, which compares observed frequencies to expected frequencies. When you encounter these problems, remember that each category contributes one term to the overall chi-square statistic using the formula: (O−E)2E\frac{(O-E)^2}{E}, where O is observed and E is expected. For the AB blood type, you need to calculate the expected frequency first. With n = 200 participants and an expected population percentage of 5% for AB blood type, the expected frequency is: 200 × 0.05 = 10 people. The observed frequency is 18 people with AB blood type. The contribution to chi-square is: (18−10)210=6410=6.4\frac{(18-10)^2}{10} = \frac{64}{10} = 6.4. This makes answer A correct. Looking at the wrong answers: Answer B uses 5 as the denominator instead of 10 – this incorrectly uses the percentage (5%) rather than the actual expected frequency (10 people). Answer C flips the formula by putting the expected frequency in the numerator and observed in the denominator, which is backwards. Answer D uses the total sample size (200) in the denominator, which completely misapplies the chi-square formula. Study tip: Always convert percentages to actual expected frequencies by multiplying by the sample size, and remember the chi-square formula puts expected frequency in the denominator, not observed frequency. Practice identifying which number represents what in word problems – this is where most errors occur.

Question 5

When performing a chi-square test of independence on a 2×32 \times 3 contingency table, a researcher calculates χ2=7.2\chi^2 = 7.2. If the critical value at α=0.05\alpha = 0.05 is 5.99, what additional information is needed to determine the strength of association?

  1. The sample size is needed to calculate Cramér's V coefficient (correct answer)
  2. The individual cell frequencies are needed to calculate the phi coefficient
  3. The marginal totals are needed to verify the chi-square calculation
  4. No additional information needed since chi-square indicates association strength
  5. The degrees of freedom must be recalculated to ensure proper interpretation
Explanation: When you encounter chi-square tests with questions about association strength, remember that statistical significance and effect size are two different concepts. A significant chi-square tells you an association exists, but not how strong it is. Answer A is correct because Cramér's V is the appropriate measure of association strength for a 2×32 \times 3 contingency table. The formula is V=χ2n(k−1)V = \sqrt{\frac{\chi^2}{n(k-1)}}, where nn is the sample size and kk is the smaller of the number of rows or columns. Since you have χ2=7.2\chi^2 = 7.2 and know the table dimensions, you only need the sample size to calculate this standardized measure that ranges from 0 to 1. Answer B is incorrect because the phi coefficient is only appropriate for 2×22 \times 2 tables, not 2×32 \times 3 tables. Using phi here would be methodologically wrong. Answer C is incorrect because marginal totals aren't needed to determine association strength. While they might help verify calculations, they don't directly measure how strong the relationship is between your variables. Answer D is incorrect because the chi-square statistic alone doesn't indicate association strength. A χ2=7.2\chi^2 = 7.2 could represent a weak association in a large sample or a stronger association in a smaller sample. The raw chi-square value increases with sample size, making it unreliable for measuring effect size. Study tip: Always distinguish between statistical significance (does an association exist?) and effect size (how strong is it?). For contingency tables larger than 2×22 \times 2, Cramér's V is your go-to measure, and it always requires sample size.

Question 6

A researcher conducts a chi-square goodness-of-fit test to determine if a die is fair. After rolling 300 times, each face shows approximately 50 times. If the calculated chi-square statistic is 2.1, what can be concluded?

  1. The die is significantly biased since the test statistic exceeds 1.0
  2. The die appears fair since 2.1 < 11.07 (critical value for α = 0.05) (correct answer)
  3. The test is inconclusive due to the small sample size of 300 rolls
  4. The die is fair since each face appeared exactly 50 times as expected
  5. Additional rolls are needed since the chi-square value is too small
Explanation: When you encounter chi-square goodness-of-fit tests, you're examining whether observed data matches an expected distribution. The key is comparing your calculated test statistic to the critical value at your chosen significance level. In this problem, you have a fair die hypothesis where each face should appear 3006=50\frac{300}{6} = 50 times. With a calculated chi-square statistic of 2.1, you need to compare this to the critical value. For a chi-square test with 5 degrees of freedom (6 categories - 1) at α = 0.05, the critical value is 11.07. Since 2.1 < 11.07, you fail to reject the null hypothesis that the die is fair. Answer A incorrectly suggests that any test statistic above 1.0 indicates significance. Chi-square critical values depend on degrees of freedom and significance level—there's no universal threshold of 1.0. Answer C wrongly claims 300 rolls is a small sample size. This is actually quite adequate for a chi-square test, where you typically need expected frequencies of at least 5 per category (here you have 50). Answer D makes the error of assuming that because observed values are close to expected values, the die must be fair. Statistical tests require proper comparison to critical values, not just eyeballing the data. Remember: in hypothesis testing, never conclude significance just by looking at the test statistic alone. Always compare it to the appropriate critical value for your chosen α level and degrees of freedom. Chi-square tests are particularly sensitive to these comparisons.

Question 7

In comparing readmission rates across 3 hospitals, a researcher finds χ2=12.4\chi^2 = 12.4 with df = 2. If this result is significant at α=0.01\alpha = 0.01, what is the most appropriate next step?

  1. Calculate standardized residuals to identify which hospitals differ from expected (correct answer)
  2. Perform pairwise chi-square tests between each pair of hospitals
  3. Increase the sample size to confirm the significant finding
  4. Apply a Bonferroni correction to adjust for multiple comparisons
  5. Report only the overall test result since post-hoc analysis is inappropriate
Explanation: When you obtain a significant chi-square test with multiple categories, you've only established that the groups differ overall — but you don't know which specific groups are driving that difference. The next logical step is to examine the pattern of deviations to understand where the significant differences lie. Standardized residuals (option A) are the appropriate follow-up tool because they tell you which cells in your contingency table contribute most to the overall chi-square statistic. These residuals show how many standard deviations each observed frequency is from its expected frequency, helping you identify which hospitals have unusually high or low readmission rates. Values greater than ±2 typically indicate meaningful deviations. Option B is problematic because pairwise chi-square tests would require you to collapse your 3×2 table into multiple 2×2 tables, losing information and creating multiple testing issues. Option C misses the point entirely — you already have a significant result at a stringent α level (0.01), so the finding is robust. Option D reflects a misunderstanding of when Bonferroni corrections apply; you're not conducting multiple hypothesis tests here, but rather exploring the components of a single omnibus test. Think of the chi-square test as asking "Is there a difference?" while standardized residuals answer "Where is the difference?" Always follow a significant chi-square test by examining residuals to understand the pattern — this transforms a simple yes/no answer into actionable insight about which groups are actually problematic.

Question 8

When testing the null hypothesis that treatment response rates are equal across 4 treatment groups, a researcher obtains p = 0.03. If 2 groups have response rates of 60% and 2 groups have rates of 45%, what conclusion is most appropriate?

  1. Reject H0H_0; evidence suggests response rates differ significantly across groups (correct answer)
  2. Fail to reject H0H_0; the 15% difference is not practically significant
  3. Reject H0H_0 but conclude the difference is not clinically meaningful
  4. The test is invalid because only 2 distinct response rates exist
  5. Additional data are needed to distinguish between the statistical hypotheses
Explanation: When you encounter hypothesis testing questions involving multiple groups, you're dealing with tests like ANOVA or chi-square that compare several groups simultaneously. The key is understanding what a significant p-value tells you and what it doesn't tell you about the nature of the differences. Here, the researcher obtained p = 0.03, which is less than the conventional α = 0.05 significance level. This means you reject the null hypothesis that all treatment response rates are equal. The statistical test has detected a significant difference somewhere among the four groups, and the evidence suggests the observed differences (60% vs 45%) are unlikely due to chance alone. Choice A correctly interprets this result - you reject H0H_0 because the evidence indicates significant differences exist across groups. Choice B incorrectly fails to reject H0H_0 despite p < 0.05, and confuses practical significance with statistical significance. The 15% difference might actually be quite meaningful clinically. Choice C makes a contradictory statement by accepting statistical significance but dismissing clinical meaning without any clinical context provided. You can't determine clinical meaningfulness from the statistical results alone. Choice D reflects a fundamental misunderstanding of hypothesis testing. Having only two distinct response rates doesn't invalidate the test - what matters is whether the overall pattern of differences across all four groups is statistically significant. Remember: when p < α, always reject H0H_0. Don't let seemingly small effect sizes fool you into ignoring statistically significant results, especially when clinical context isn't provided to judge practical importance.

Question 9

A study examines the association between smoking status (never, former, current) and lung disease severity (mild, moderate, severe). If the contingency table shows strong evidence against independence (p < 0.001), which statement about the residuals is most likely true?

  1. All standardized residuals will be approximately equal in magnitude across cells
  2. The largest standardized residuals will likely correspond to current smokers with severe disease (correct answer)
  3. Standardized residuals will all be positive since the association is significant
  4. The residuals in the diagonal cells will be systematically larger than off-diagonal cells
  5. Residual analysis is not meaningful when the overall test is highly significant
Explanation: When analyzing contingency tables with significant associations, you need to understand what standardized residuals reveal about the patterns driving that association. Standardized residuals measure how much each observed cell count deviates from what you'd expect if the variables were independent, scaled by the standard error. With strong evidence against independence (p < 0.001) between smoking status and lung disease severity, the significant association is driven by cells where observed counts dramatically differ from expected counts. Given the biological relationship, you'd expect current smokers to have disproportionately high rates of severe disease compared to what independence would predict. This creates large positive standardized residuals in cells representing current smokers with severe disease, as the observed counts exceed expected counts substantially. Answer A is incorrect because significant associations arise precisely from unequal residuals—some cells show strong departures from independence while others may be close to expected values. Answer C misunderstands residual interpretation; residuals can be positive (observed > expected) or negative (observed < expected) regardless of significance. A significant chi-square test doesn't determine residual signs. Answer D incorrectly assumes this contingency table has meaningful diagonal cells—unlike correlation matrices, contingency tables for categorical variables don't have special diagonal interpretations. Study tip: When you see significant associations in contingency tables, focus on the biological or logical relationships between categories. The largest residuals typically occur where these relationships are strongest, helping you identify which specific combinations drive the overall association.

Question 10

A researcher plans to test whether the proportions of patients experiencing side effects are equal across 6 different medications. If the overall side effect rate is expected to be 30% and the researcher wants 80% power to detect a meaningful difference, approximately how many patients should be recruited per medication group?

  1. At least 45 patients per group based on standard chi-square power calculations
  2. At least 60 patients per group based on standard chi-square power calculations
  3. At least 85 patients per group based on standard chi-square power calculations
  4. At least 120 patients per group based on standard chi-square power calculations
  5. Cannot determine sample size without specifying the effect size to detect (correct answer)
Explanation: When you encounter questions about comparing proportions across multiple groups, you're dealing with chi-square test power calculations. This scenario involves testing equality of side effect rates across 6 medications, which requires careful sample size planning. For a chi-square test comparing proportions across 6 groups with an expected overall rate of 30% and desired power of 80%, the standard approach uses the formula that accounts for the number of groups (k=6), expected proportion (p=0.30), effect size, and power requirements. The calculation involves: n=(Zα+Zβ)2×p(1−p)×design effect(effect size)2n = \frac{(Z_\alpha + Z_\beta)^2 \times p(1-p) \times \text{design effect}}{(\text{effect size})^2} With multiple groups, the design effect increases substantially. For 6 groups at 80% power detecting meaningful differences (typically defined as differences that would change clinical practice), standard biostatistics references indicate approximately 85 patients per group are needed. Option A (45 patients) severely underestimates the requirements and would yield insufficient power, likely around 60%. Option B (60 patients) is closer but still underpowered, providing roughly 70% power. Option D (120 patients) overestimates and represents a more conservative approach typically used when effect sizes are smaller than "meaningful differences." Option C correctly reflects the standard chi-square power calculation for this scenario, balancing adequate power with practical recruitment constraints. Study tip: For multi-group proportion comparisons, sample size requirements increase significantly with the number of groups. Always verify that power calculations account for multiple comparisons and the specific effect size you want to detect.

Question 11

A quality control study examines whether defect rates vary across 5 production lines. After collecting data from 1000 items (200 per line), the chi-square test yields χ2=15.8\chi^2 = 15.8 with p = 0.003. What does this result indicate about the production process?

  1. All production lines have significantly different defect rates from each other
  2. At least one production line has a defect rate different from the others (correct answer)
  3. The overall defect rate is significantly different from the expected rate
  4. Production line 1 has the highest defect rate based on the test statistic
  5. The production process meets quality standards since p < 0.05
Explanation: When you encounter a chi-square goodness-of-fit test comparing multiple groups, you're testing whether the observed frequencies differ significantly from what you'd expect if all groups were equal. The null hypothesis states that defect rates are the same across all production lines, while the alternative hypothesis suggests at least one line differs. With χ2=15.8\chi^2 = 15.8 and p = 0.003, you reject the null hypothesis at typical significance levels (α = 0.05). This significant result tells you that the defect rates aren't uniform across all five production lines—at least one line has a meaningfully different defect rate. Answer B correctly captures this interpretation. Answer A is wrong because the chi-square test doesn't tell you that all lines differ from each other—only that there's significant variation somewhere among the groups. You'd need post-hoc pairwise comparisons to determine which specific lines differ. Answer C misinterprets what's being tested. You're not comparing the overall defect rate to some external expected rate; you're comparing the rates among the five production lines to see if they're equal to each other. Answer D makes an impossible claim. The chi-square test statistic alone doesn't identify which specific production line has the highest defect rate—it only detects that significant differences exist somewhere in the data. Study tip: Remember that chi-square tests detect any significant variation among groups but don't pinpoint exactly where those differences lie. The phrase "at least one" is your clue that you're dealing with an omnibus test result.

Question 12

A chi-square test of homogeneity compares smoking rates across 3 different cities. Each city contributes 150 participants, and overall 40% are smokers. If the test statistic is χ2=8.7\chi^2 = 8.7, what can be concluded at α=0.01\alpha = 0.01?

  1. Reject H0H_0; smoking rates differ significantly across the three cities
  2. Fail to reject H0H_0; insufficient evidence that smoking rates differ across cities (correct answer)
  3. Reject H0H_0; the overall smoking rate is significantly different from 40%
  4. Fail to reject H0H_0; the sample size is too small for reliable conclusions
  5. Cannot determine without knowing the individual city smoking rates
Explanation: Chi-square tests of homogeneity compare whether proportions are the same across different groups. Here, you're testing whether smoking rates are homogeneous (the same) across three cities. The null hypothesis states that smoking rates are equal across all cities, while the alternative hypothesis claims they differ. To determine significance, you need to compare your test statistic (χ2=8.7\chi^2 = 8.7) to the critical value. With 3 cities, you have df=3−1=2df = 3-1 = 2 degrees of freedom. At α=0.01\alpha = 0.01 with df=2df = 2, the critical value is χ0.01,22=9.21\chi^2_{0.01,2} = 9.21. Since your calculated value (8.7) is less than 9.21, you fail to reject the null hypothesis. There's insufficient evidence that smoking rates differ significantly across the three cities. Answer A is wrong because 8.7 < 9.21, so you don't reject H0H_0. Answer C misunderstands the test entirely—this isn't testing whether the overall rate differs from 40%, but whether rates are consistent across cities. Answer D incorrectly suggests sample size issues when you have 450 total participants (150 per city), which is adequate for chi-square analysis. The correct answer is B: you fail to reject H0H_0 due to insufficient evidence of differences across cities. Study tip: Always identify your degrees of freedom first (number of groups minus 1), then compare your test statistic to the critical value at your chosen alpha level. If calculated < critical, you fail to reject H0H_0.

Question 13

In a 3×53 \times 5 contingency table testing independence, one cell has an observed frequency of 12 and an expected frequency of 8. What is the standardized residual for this cell?

  1. 12−88=42.83=1.41\frac{12-8}{\sqrt{8}} = \frac{4}{2.83} = 1.41 (correct answer)
  2. 12−88=48=0.50\frac{12-8}{8} = \frac{4}{8} = 0.50
  3. (12−8)28=168=2.0\frac{(12-8)^2}{8} = \frac{16}{8} = 2.0
  4. 12−812=43.46=1.16\frac{12-8}{\sqrt{12}} = \frac{4}{3.46} = 1.16
  5. 12−820=44.47=0.89\frac{12-8}{\sqrt{20}} = \frac{4}{4.47} = 0.89
Explanation: When you encounter a contingency table question asking for standardized residuals, you're working with a measure that tells you how far an observed frequency deviates from what you'd expect under independence, expressed in standard deviation units. The standardized residual formula is: observed−expectedexpected\frac{\text{observed} - \text{expected}}{\sqrt{\text{expected}}}. This standardizes the raw difference by dividing by the standard deviation of the expected frequency, which equals expected\sqrt{\text{expected}} under the assumption of independence. For this cell with observed = 12 and expected = 8, the calculation is 12−88=42.83=1.41\frac{12-8}{\sqrt{8}} = \frac{4}{2.83} = 1.41, making A correct. Option B calculates 12−88=0.50\frac{12-8}{8} = 0.50, which gives you the raw residual divided by the expected frequency—this isn't standardized and has no clear statistical interpretation. Option C computes (12−8)28=2.0\frac{(12-8)^2}{8} = 2.0, which is actually the contribution of this cell to the overall chi-square test statistic, not a standardized residual. Option D uses 12−812=1.16\frac{12-8}{\sqrt{12}} = 1.16, incorrectly using the square root of the observed frequency rather than the expected frequency as the denominator. Remember that standardized residuals follow approximately a standard normal distribution when the null hypothesis of independence is true. Values greater than 2 in absolute value suggest that cell contributes significantly to any departure from independence. Always use expected\sqrt{\text{expected}} as your denominator—this is the key detail that distinguishes standardized residuals from other residual measures.

Question 14

A clinical trial randomizes 240 patients to three treatment groups (80 per group). Researchers want to test if the distribution of treatment outcomes differs across groups.

Based on the passage above, if there are 4 possible outcomes for each patient, what are the degrees of freedom for the appropriate chi-square test?

  1. df = 6 degrees based on (groups - 1) × (outcomes - 1) calculation (correct answer)
  2. df = 11 degrees based on (total cells - 1) calculation method
  3. df = 9 degrees based on (groups × outcomes - 1) calculation method
  4. df = 239 degrees based on (total sample size - 1) calculation
  5. df = 3 degrees based on (number of groups - 1) calculation only
Explanation: When you encounter a chi-square test involving multiple groups and multiple outcomes, you're dealing with a test of independence or homogeneity. The key is recognizing this as a contingency table analysis where you're testing whether the distribution of outcomes is the same across all treatment groups. For chi-square tests with contingency tables, the degrees of freedom formula is always (r−1)×(c−1)(r-1) \times (c-1) where r is the number of rows and c is the number of columns. In this scenario, you have 3 treatment groups and 4 possible outcomes, creating a 3×4 contingency table. Therefore: df=(3−1)×(4−1)=2×3=6df = (3-1) \times (4-1) = 2 \times 3 = 6 Choice A correctly applies this standard formula, giving 6 degrees of freedom. Choice B incorrectly uses total cells minus 1 (3×4-1=11). This approach ignores the constraint structure of contingency tables, where both row and column totals are fixed, requiring you to subtract degrees of freedom for both dimensions. Choice C mistakenly calculates groups times outcomes minus 1 (3×4-1=9). This treats the problem as if you only lose one degree of freedom total, when actually you lose one for each dimension of the table. Choice D uses total sample size minus 1 (240-1=239), which would be appropriate for a one-sample chi-square goodness-of-fit test, not a multi-group comparison. Remember: For any contingency table chi-square test, always use (rows-1)×(columns-1). The "minus 1" accounts for the constraints in both dimensions of your table.

Question 15

A pharmaceutical company tests if adverse event rates are homogeneous across 5 clinical sites. With 120 patients per site and an overall adverse event rate of 25%, what is the expected frequency of adverse events at each site under the null hypothesis?

  1. 30 adverse events per site based on site sample size and overall rate (correct answer)
  2. 25 adverse events per site based on the given percentage value directly
  3. 150 adverse events per site based on total sample calculations
  4. 75 adverse events per site based on half the site sample size
  5. Cannot determine without knowing the observed adverse event frequencies
Explanation: When testing for homogeneity of adverse event rates across multiple sites, you're examining whether the proportion of adverse events is consistent across all locations. Under the null hypothesis, you assume that all sites have the same underlying adverse event rate, which means each site should show the expected frequency based on the overall rate. To find the expected frequency at each site, you multiply the site's sample size by the overall adverse event rate: 120 patients×0.25=30 adverse events per site120 \text{ patients} \times 0.25 = 30 \text{ adverse events per site}. This calculation gives you what you'd expect to see at each site if the null hypothesis (homogeneous rates) is true. Choice A correctly applies this formula, calculating 30 adverse events per site based on the site sample size and overall rate. Choice B incorrectly uses 25, which confuses the percentage (25%) with the actual count—this represents a common error of using the rate itself rather than applying it to the sample size. Choice C suggests 150 adverse events per site, which would require multiplying by the total sample across all sites (600 patients), not the individual site size. Choice D uses 75, which represents half the site sample size but ignores the adverse event rate entirely. Remember that expected frequencies in chi-square tests always equal the sample size times the hypothesized proportion. Don't confuse percentages with counts, and always apply rates to the appropriate sample size—in this case, individual site sizes, not the total study population.

Question 16

Consider a chi-square test where H0H_0: the data follow a specified multinomial distribution. If the test statistic follows a chi-square distribution with 8 degrees of freedom under the null hypothesis, how many categories are being tested?

  1. 9 categories since degrees of freedom equal categories minus one
  2. 8 categories since degrees of freedom equal the number of categories
  3. 10 categories since one additional constraint reduces degrees of freedom
  4. 7 categories since degrees of freedom equal categories minus two
  5. Cannot determine without knowing if parameters were estimated from data (correct answer)
Explanation: When you encounter chi-square goodness-of-fit tests, remember that degrees of freedom depend on both the number of categories and the constraints imposed by the hypothesis structure. For a chi-square goodness-of-fit test with a specified multinomial distribution, the degrees of freedom formula is df=k−1df = k - 1, where kk is the number of categories. The "-1" comes from the constraint that all observed frequencies must sum to the total sample size, which reduces the degrees of freedom by one. Given that df=8df = 8, we can solve: 8=k−18 = k - 1, so k=9k = 9 categories. Looking at the incorrect options: Answer B incorrectly assumes degrees of freedom equal the number of categories, ignoring the fundamental constraint that reduces df by 1. Answer C suggests 10 categories with an "additional constraint," but there's only one standard constraint (frequencies summing to n) in basic goodness-of-fit tests. Answer D proposes subtracting 2 from the category count, which would apply to contingency table tests of independence where you lose degrees of freedom for both row and column totals, not goodness-of-fit tests. The key distinction is recognizing this as a goodness-of-fit test (testing against a "specified" distribution) rather than a test of independence between variables. In goodness-of-fit tests, you always lose exactly one degree of freedom due to the sum constraint. Study tip: For biostatistics exams, memorize that goodness-of-fit tests use df=categories−1df = categories - 1, while independence tests use df=(rows−1)(columns−1)df = (rows-1)(columns-1). The word "specified" in the null hypothesis signals a goodness-of-fit scenario.

Question 17

In a chi-square test of independence examining the relationship between gender (2 levels) and treatment response (4 levels), what is the expected frequency for the cell corresponding to males with complete response if there are 80 males total, 120 participants with complete response, and 300 total participants?

  1. 32.0 participants in this cell based on independence assumption (correct answer)
  2. 40.0 participants in this cell based on independence assumption
  3. 24.0 participants in this cell based on independence assumption
  4. 48.0 participants in this cell based on independence assumption
  5. Cannot calculate without knowing all observed cell frequencies
Explanation: When you encounter chi-square test questions, you're dealing with expected frequencies under the assumption of independence. The key formula to remember is: Expected frequency = (Row total × Column total) ÷ Grand total. In this 2×4 contingency table, you need the expected frequency for males with complete response. You have all the necessary information: 80 males (row total), 120 participants with complete response (column total), and 300 total participants (grand total). Applying the formula: Expected frequency = (80 × 120) ÷ 300 = 9,600 ÷ 300 = 32.0. This represents what you'd expect to see in that cell if gender and treatment response were completely independent. Looking at the wrong answers: Choice B (40.0) likely comes from incorrectly using 100 instead of 120 for the complete response total, or making an arithmetic error. Choice C (24.0) might result from using 90 instead of 120 for the column total, possibly confusing it with another response category. Choice D (48.0) could come from incorrectly multiplying 80 × 120 ÷ 200, perhaps miscalculating the grand total. The correct answer is A: 32.0 participants. Study tip: Always write out the expected frequency formula and double-check your row totals, column totals, and grand total before calculating. Chi-square problems often include extra numbers that aren't needed for the specific cell you're calculating, so identify exactly which totals apply to your target cell.

Question 18

A clinical trial randomizes patients to three treatment groups (A, B, C) and measures response as complete response, partial response, or no response. The study protocol specifies testing for treatment differences using a chi-square test at α=0.05\alpha = 0.05.

After data collection, the researchers discover that treatment C had very few patients due to recruitment difficulties. The contingency table shows several cells with expected frequencies below 3. What is the most statistically sound approach?

  1. Combine treatments B and C into a single group and perform a 2×3 chi-square test comparing the combined group to treatment A
  2. Collapse the response categories by combining partial and no response, then perform a 3×2 chi-square test
  3. Exclude treatment C entirely and perform a 2×3 chi-square test comparing only treatments A and B
  4. Use Monte Carlo simulation to generate the exact p-value for the 3×3 chi-square test with low expected frequencies (correct answer)
Explanation: When expected frequencies are too low for standard chi-square assumptions but you want to preserve the original study design and all collected data, Monte Carlo simulation provides exact p-values without requiring asymptotic assumptions. Choice A inappropriately combines treatment groups, potentially masking important differences. Choice B collapses meaningful outcome categories. Choice C wastes collected data and may introduce bias by excluding a treatment arm post-hoc.

Question 19

A researcher conducts a chi-square test of independence on a 3×4 contingency table and obtains χ2=14.7\chi^2 = 14.7 with p=0.065p = 0.065. The study had 80% power to detect a medium effect size. What is the most appropriate interpretation of this result?

  1. The result suggests no association exists between the variables, and the study had adequate power to detect meaningful effects
  2. The result is inconclusive; a larger sample size is needed to definitively determine whether an association exists
  3. The result suggests a weak but potentially meaningful association that approaches statistical significance with adequate power (correct answer)
  4. The result indicates insufficient power to detect the observed effect size, requiring post-hoc power analysis for proper interpretation
Explanation: With p = 0.065, the result approaches significance and suggests a potential association. The chi-square value of 14.7 with df = 6 indicates a moderate effect size, and the study had adequate power (80%) to detect medium effects. Choice A incorrectly concludes no association exists. Choice B suggests the study was underpowered, which contradicts the stated 80% power. Choice D incorrectly suggests inadequate power when 80% power was achieved.

Question 20

A biostatistician analyzes adverse events in a clinical trial using a chi-square test of independence. The analysis reveals χ2=11.2\chi^2 = 11.2 with 6 degrees of freedom (p=0.082p = 0.082). Post-hoc examination of standardized residuals shows one cell with a residual of -2.8. What does this specific finding indicate?

  1. This cell contributed significantly to the overall chi-square statistic and shows fewer adverse events than expected under independence (correct answer)
  2. This cell shows more adverse events than expected and represents a safety signal requiring immediate investigation
  3. This cell contributed minimally to the overall chi-square statistic due to the negative residual value
  4. This cell indicates a data collection error since standardized residuals should not exceed ±2.0 in clinical trial data
Explanation: A standardized residual of -2.8 indicates this cell contributed substantially to the chi-square statistic (|2.8| > 2.0) and had fewer observed cases than expected (negative value). Standardized residuals with absolute values > 2.0 are considered significant contributors. Choice B incorrectly interprets the negative sign. Choice C is wrong because large absolute values indicate major contributions. Choice D incorrectly suggests standardized residuals have inherent limits in clinical trials.