One-Way ANOVA - Biostatistics

1. An investigator wants to compare mean cholesterol levels across three dietary interventions. She calculates that with 20 subjects per group, she has 80% power to detect a medium effect size. If she decides to add a fourth intervention group while keeping the same total sample size of 60, what will most likely happen to the statistical power?
   Answer: Power will decrease because each group will have fewer subjects despite the same total N When you encounter questions about statistical power and study design changes, focus on how modifications affect your ability to detect true differences between groups. Statistical power depends on several factors: effect size, sample size per group, number of comparisons, and alpha level. In ANOVA designs, what matters most for power is having adequate sample size within each group to detect differences between group means. The correct answer is D because changing from 3 groups to 4 groups while keeping the total sample size at 60 reduces each group from 20 subjects to 15 subjects. This 25% reduction in per-group sample size substantially decreases power. Smaller groups have larger standard errors, making it harder to detect true differences between means. The effect is more pronounced than any benefits from additional comparisons. Answer A incorrectly assumes that more groups automatically provide more information. While additional groups can provide more comparisons, they don't increase power when achieved by splitting the same total sample size. Answer B misses the critical insight that total sample size alone doesn't determine power—the distribution of that sample across groups matters enormously. Answer C incorrectly focuses on error degrees of freedom, which actually increase when you add groups (from 57 to 56 error df), and this minor change wouldn't substantially affect power anyway. Study tip: Remember that in ANOVA designs, power primarily depends on sample size per group, not total sample size. When comparing study designs, always calculate the per-group N first—that's usually where you'll find the answer.
2. A psychologist conducts a one-way ANOVA to compare stress levels among nurses working in four different hospital units. The analysis yields F(3,56) = 2.18, p = 0.101. The hospital administrator wants to conclude that all units have equivalent stress levels. What is the most appropriate response?
   Answer: The conclusion is premature; failure to reject the null hypothesis does not prove equivalence When interpreting ANOVA results, you need to understand the fundamental difference between "failing to reject the null hypothesis" and "proving the null hypothesis is true." This distinction is crucial in statistical inference. The ANOVA result shows F(3,56) = 2.18, p = 0.101. Since p > 0.05, we fail to reject the null hypothesis that all group means are equal. However, this doesn't prove that the groups are actually equivalent—it simply means we don't have sufficient evidence to conclude they're different. The administrator's conclusion commits a logical fallacy by treating "no evidence of difference" as "evidence of no difference." Answer B correctly identifies this reasoning error. Failure to reject the null hypothesis could result from truly equal means, but it could also result from insufficient power, high variability, small effect sizes, or inadequate sample size. Without additional evidence (like equivalence testing), we cannot conclude the groups are equivalent. Answer A makes the classic mistake of interpreting non-significance as proof of equivalence, which is statistically invalid. Answer C suggests the conclusion is correct, which it isn't—confidence intervals around effect sizes don't validate an equivalence claim without proper equivalence testing. Answer D focuses solely on sample size, but the fundamental issue is the logical interpretation error, not necessarily inadequate power. Remember this key principle: in hypothesis testing, non-significant results mean "insufficient evidence for a difference," never "proof of no difference." To demonstrate equivalence, you need specialized equivalence tests, not traditional null hypothesis testing.
3. A researcher conducts a one-way ANOVA comparing memory scores across three age groups: Young (20-30 years), Middle-aged (40-50 years), and Elderly (65-75 years). Before analyzing the data, she should be most concerned about which potential violation of ANOVA assumptions?
   Answer: Homogeneity of variance across the three age groups When you encounter ANOVA questions, always consider which assumptions are most critical and likely to be violated in the given scenario. ANOVA has four key assumptions: independence, normality, homogeneity of variance, and random sampling. The correct answer is C because homogeneity of variance (homoscedasticity) is the assumption most likely to be problematic when comparing different age groups on cognitive measures like memory. Age groups naturally tend to show different amounts of variability in performance - younger adults typically show more consistent scores clustered around higher values, while elderly participants often display much greater variability due to factors like varying degrees of cognitive decline, health differences, and medication effects. This creates unequal variances across groups, violating the homoscedasticity assumption. A is less concerning because independence within age groups is typically maintained through proper experimental design and random assignment. B (normality) is important but ANOVA is relatively robust to normality violations, especially with reasonable sample sizes. D (random sampling) is a design issue that should be addressed during study planning, but it's not the assumption you'd be "most concerned about" when you're ready to analyze existing data. ANOVA is particularly sensitive to variance inequality when group sizes are unequal, and this violation can seriously affect the validity of your results and increase Type I error rates. Study tip: For biostatistics exams, remember that when comparing groups that naturally differ in variability (age groups, disease severity levels, treatment vs. control), homogeneity of variance is usually your biggest statistical concern.
4. A researcher conducts a one-way ANOVA to compare mean blood pressure across four treatment groups (n = 12 per group). The calculated F-statistic is 2.95. If the critical value at α = 0.05 is 2.84, but the researcher mistakenly uses the critical value for α = 0.01 (which is 4.31) to make their decision, what type of error becomes more likely and what is the immediate consequence for their conclusion?
   Answer: Type II error becomes more likely; they will incorrectly fail to reject the null hypothesis The calculated F-statistic (2.95) exceeds the correct critical value at α = 0.05 (2.84), so the null hypothesis should be rejected. However, by mistakenly using the more stringent critical value for α = 0.01 (4.31), the researcher will fail to reject the null hypothesis since 2.95 < 4.31. This increases the probability of Type II error (failing to reject a false null hypothesis). The immediate consequence is incorrectly concluding no difference exists when one actually does.
5. Based on the study design described above, if the researcher finds a significant one-way ANOVA result (F = 3.89, p = 0.013) but then discovers that the baseline anxiety scores differed significantly between groups, what is the most appropriate analytical approach to address this issue?
   Answer: Perform ANCOVA using baseline anxiety scores as a covariate with change scores as the outcome When you encounter a clinical trial with significant baseline differences between groups, you're dealing with a common threat to internal validity that requires statistical adjustment rather than abandoning your analysis. The discovery of significant baseline differences means your groups weren't truly equivalent at the start, despite randomization. This creates confounding that could explain your treatment effect. ANCOVA (Analysis of Covariance) is specifically designed for this situation - it statistically adjusts for baseline differences by treating them as a covariate, allowing you to isolate the true treatment effect on change scores. Option D correctly applies this principle. Option A is problematic because using only post-treatment scores without accounting for baseline differences would be misleading. If one group started with higher anxiety, their post-treatment scores might still be higher even with effective treatment. Option B misunderstands the issue - this isn't about violating ANOVA assumptions requiring non-parametric tests. Baseline differences don't violate ANOVA's mathematical assumptions; they create a confounding variable that needs statistical control. Option C throws away valuable between-group comparison data unnecessarily. The whole point of a controlled trial is comparing treatments, and paired t-tests within groups can't tell you which treatment is most effective. Remember this pattern: when you discover baseline imbalances in randomized trials, think ANCOVA first. It's the gold standard for adjusting treatment comparisons when groups differ at baseline, preserving your ability to make valid between-group inferences while controlling for confounding variables.
6. A nutrition study compares mean vitamin D levels across three dietary groups: omnivores (n=20, x̄=28.5, s=6.2), vegetarians (n=18, x̄=22.1, s=5.8), and vegans (n=22, x̄=18.3, s=7.1). All units are ng/mL. Before conducting one-way ANOVA, which assumption requires the most careful evaluation based on these descriptive statistics?
   Answer: Independence of observations, because the dietary groups may have overlapping social networks affecting vitamin D intake While all assumptions are important, independence is most critical here because dietary groups often cluster within social networks, families, or communities. People with similar dietary choices may share living situations, shopping habits, geographic locations, or lifestyle factors that affect vitamin D levels, violating the independence assumption. The variance ratios (largest/smallest SD = 7.1/5.8 = 1.22) are actually quite acceptable, and the sample sizes are adequate for CLT.
7. A pharmaceutical company tests four different formulations of a drug on separate groups of patients. The ANOVA results show F(3,28) = 2.45 with p = 0.083. Given that the company had initially hypothesized that Formulation C would be superior to all others, what is the most appropriate interpretation?
   Answer: There is insufficient evidence to conclude any differences exist among the formulations at α = 0.05 When you encounter ANOVA results in biostatistics, focus on what the F-statistic and p-value actually tell you about group differences. ANOVA tests the null hypothesis that all group means are equal against the alternative that at least one group differs from the others. Here, F(3,28) = 2.45 with p = 0.083 means there's an 8.3% probability of observing these differences (or greater) if all four formulations were truly equivalent. Since p = 0.083 > 0.05, you fail to reject the null hypothesis at the standard α = 0.05 level. This means there's insufficient statistical evidence to conclude that any meaningful differences exist among the four formulations. Answer B correctly interprets this result. The p-value exceeds the conventional significance threshold, so you cannot conclude differences exist. Answer A is wrong because you cannot claim Formulation C is superior when the overall ANOVA isn't even significant. ANOVA only tells you whether groups differ collectively—it doesn't identify which specific group is best. Answer C incorrectly suggests the results show a "trend" for Formulation C. The company's initial hypothesis about Formulation C is irrelevant to interpreting the ANOVA results, which test overall group differences, not specific formulations. Answer D commits the error of post-hoc significance level adjustment. You cannot change your α level after seeing the results to make them "significant"—this inflates Type I error rates and represents poor statistical practice. Remember: stick to your predetermined α level and interpret ANOVA results for what they test—overall group differences, not specific group superiority.
8. A researcher compares the effectiveness of three physical therapy techniques on pain reduction scores. The ANOVA table shows MSB = 45.6 and MSW = 12.3. If there are 10 patients in each group, what assumption is most critical to verify before interpreting these results?
   Answer: The three treatment groups have equal variances in their pain scores When you encounter ANOVA problems in biostatistics, you're dealing with a statistical test that compares means across multiple groups. ANOVA has several key assumptions, but their relative importance depends on factors like sample size and the robustness of the test to violations. The correct answer is B because equal variances (homoscedasticity) is the most critical assumption to verify here. With only 10 patients per group, ANOVA is particularly sensitive to variance differences between groups. Unequal variances can dramatically inflate Type I error rates and make your F-statistic unreliable. The fact that you're given MSB and MSW values suggests the focus is on variance-related calculations, reinforcing this assumption's importance. Let's examine why the other options are less critical: A is incorrect because ANOVA is remarkably robust to normality violations, especially with equal sample sizes of 10 per group. The Central Limit Theorem helps here, and F-tests can tolerate non-normal data quite well. C is wrong because random assignment affects external validity and causal inference, but it's not a statistical assumption required for ANOVA calculations themselves. D is incorrect because while measurement scale matters, most pain scores in research already meet interval/ratio requirements, and this is typically established during study design rather than being verified post-hoc. Study tip: Remember that with smaller sample sizes (like n=10 per group), focus on the equal variances assumption first. Use Levene's test or examine residual plots before interpreting your ANOVA results. ANOVA forgives normality violations more easily than variance violations.
9. In a one-way ANOVA with 3 groups and 45 total observations, SSTotal = 890.5 and SSWithin = 623.8. If the calculated F-statistic is 8.94, what is the most likely error in the analysis?
   Answer: No error is apparent; the calculations appear consistent with the given values When you encounter one-way ANOVA problems, always verify the calculations systematically by checking each component against the given information. Let's work through this step-by-step. With 3 groups and 45 total observations, we have: - Between groups df = k-1 = 3-1 = 2 - Within groups df = N-k = 45-3 = 42 - Total df = N-1 = 45-1 = 44 First, calculate SSBetween: $SSBetween = SSTotal - SSWithin = 890.5 - 623.8 = 266.7$ Next, find the mean squares: - $MSBetween = \frac{SSBetween}{df_{between}} = \frac{266.7}{2} = 133.35$ - $MSWithin = \frac{SSWithin}{df_{within}} = \frac{623.8}{42} = 14.85$ Finally, calculate F: $F = \frac{MSBetween}{MSWithin} = \frac{133.35}{14.85} = 8.98$ This matches the given F-statistic of 8.94 (allowing for rounding differences), indicating no error exists in the analysis. Answer choice A is incorrect because the within groups df is correctly 42 (N-k = 45-3). Answer B is wrong since SSBetween does equal SSTotal - SSWithin. Answer C is incorrect because the mean square values and F-statistic calculation are proper. Answer D is wrong because total df correctly equals N-1 = 44. Study tip: Always double-check ANOVA calculations by working backwards from the F-statistic. If your recalculated F matches the given value, the analysis is likely correct. Practice identifying when "no error" might be the right answer rather than assuming something must be wrong.
10. A researcher conducts a one-way ANOVA to compare mean pain scores across four treatment groups and obtains SSTotal = 2400, SSBetween = 720, with a total sample size of 48 (n=12 per group). If the researcher wants to calculate the proportion of total variance explained by the treatment factor, what is eta-squared (η²) and how should it be interpreted?
    Answer: η² = 0.30; treatment accounts for 30% of variance in pain scores, indicating a large effect size Eta-squared (η²) = SSBetween/SSTotal = 720/2400 = 0.30. This indicates that 30% of the total variance in pain scores is explained by the treatment factor. According to Cohen's guidelines for effect sizes, η² = 0.30 represents a large effect size (small ≈ 0.01, medium ≈ 0.06, large ≈ 0.14 and above). This suggests the treatment has a substantial practical impact on pain scores.
11. A researcher performs a one-way ANOVA comparing mean cholesterol levels across three diet groups and obtains a significant F-statistic (F = 4.82, p = 0.015). The group means are: Diet A = 180 mg/dL, Diet B = 165 mg/dL, Diet C = 195 mg/dL. If the researcher concludes that 'Diet B is significantly better than Diet C,' what error has been made?
    Answer: The error is making pairwise comparisons without post-hoc testing after a significant omnibus test A significant omnibus F-test only indicates that at least one group differs from the others, but does not specify which pairs are significantly different. To make claims about specific pairwise comparisons (like Diet B vs. Diet C), post-hoc tests (such as Tukey's HSD, Bonferroni, or Scheffe) must be conducted to control for multiple comparisons and maintain the overall Type I error rate.
12. A researcher conducts a one-way ANOVA to compare the mean blood pressure reduction among four different antihypertensive medications. The F-statistic is 3.24 with 3 degrees of freedom in the numerator and 36 degrees of freedom in the denominator. If the critical value at α = 0.05 is 2.87, which conclusion is most appropriate?
    Answer: Reject the null hypothesis; at least one medication mean differs significantly from the others When you encounter a one-way ANOVA question, you're dealing with hypothesis testing that compares means across multiple groups. The null hypothesis states that all group means are equal, while the alternative hypothesis states that at least one mean differs from the others. The decision rule is straightforward: compare your calculated F-statistic to the critical value. Here, your F-statistic is 3.24 and the critical value is 2.87. Since 3.24 > 2.87, you reject the null hypothesis. This means there's sufficient evidence that at least one medication produces a different mean blood pressure reduction than the others. Option A is incorrect because when your test statistic exceeds the critical value, you have sufficient evidence to reject the null hypothesis, not accept it. Option C makes a critical error in interpretation—rejecting the null hypothesis in ANOVA only tells you that at least one mean differs, not that all means are different from each other. You'd need post-hoc tests to determine which specific pairs differ. Option D is also wrong because ANOVA doesn't tell you exactly how many means differ, only that at least one does. Option B correctly states that you reject the null hypothesis and that at least one medication mean differs significantly from the others—this is precisely what a significant ANOVA result indicates. Remember: ANOVA is like a screening test. A significant result tells you differences exist somewhere among your groups, but you need additional pairwise comparisons to pinpoint exactly where those differences lie.
13. A medical researcher compares pain relief scores across three analgesic medications using ANOVA. The between-groups sum of squares is 156.8, within-groups sum of squares is 284.2, with 15 patients per medication group. What percentage of the total variance in pain relief scores is explained by the medication type?
    Answer: 35.6% When you encounter ANOVA results asking about "variance explained," you're being tested on the coefficient of determination (R²) - a fundamental measure of how much variability in your outcome variable is attributable to group differences. To calculate R², you need the ratio of between-groups sum of squares to total sum of squares. First, find the total sum of squares: $SS_{total} = SS_{between} + SS_{within} = 156.8 + 284.2 = 441$ Then calculate R²: $R² = \frac{SS_{between}}{SS_{total}} = \frac{156.8}{441} = 0.356 = 35.6\%$ This means 35.6% of the variance in pain relief scores is explained by medication type, making A correct. Let's examine why the other answers are wrong. B (55.2%) would result from incorrectly calculating $\frac{284.2}{515}$ - perhaps from adding the sums of squares incorrectly. C (64.4%) might come from using $\frac{284.2}{441}$, which would be calculating the percentage of variance NOT explained by the medication (the error variance). D (44.8%) could result from various computational errors, possibly involving the sample size information unnecessarily. Study tip: Remember that R² in ANOVA always equals $\frac{SS_{between}}{SS_{total}}$, and percentages of variance explained typically range from 0-100%. Be suspicious of answers requiring complex calculations involving sample sizes when the question provides sum of squares directly - ANOVA variance components depend on sums of squares, not sample sizes.
14. A pharmacologist tests four different dosages of a new medication on blood glucose levels. The ANOVA results show F(3,36) = 5.72, p = 0.003. The effect size (eta-squared) is 0.32. How should this result be interpreted in terms of practical significance?
    Answer: Practical significance cannot be determined from eta-squared alone; clinical context is needed When you encounter questions about practical vs. statistical significance, remember that statistical significance tells you if an effect likely exists, while practical significance tells you if that effect matters in real-world applications. This ANOVA shows clear statistical significance (p = 0.003), meaning the dosage differences are unlikely due to chance. The eta-squared of 0.32 indicates a large effect size by conventional standards (small = 0.01, medium = 0.06, large = 0.14). However, practical significance depends entirely on clinical context that isn't provided here. Answer D is correct because determining practical significance requires knowing what constitutes a clinically meaningful change in blood glucose. A 32% variance explanation could represent either a modest change across a wide range of values or a dramatic change across a narrow range. Without knowing baseline glucose levels, the magnitude of actual changes, or clinical thresholds for meaningful improvement, you cannot assess practical significance. Answer A incorrectly assumes that a large statistical effect size automatically equals practical significance. Answer B makes the opposite error, calling 0.32 a "low effect size" when it's actually quite large by statistical conventions. Answer C wrongly suggests practical significance exists independent of statistical significance, when both are needed for meaningful clinical conclusions. Study tip: In biostatistics, always distinguish between statistical measures (p-values, effect sizes) and clinical relevance. Large effect sizes suggest important findings worth investigating, but practical significance always requires domain knowledge about what constitutes meaningful change in the specific clinical context.
15. In a one-way ANOVA with unequal group sizes (n₁ = 8, n₂ = 12, n₃ = 15), a researcher finds F(2,32) = 3.89, p = 0.031. When checking assumptions, she discovers that the group with the smallest sample size has the largest variance. What is the most likely consequence for the ANOVA results?
    Answer: The F-test will be more conservative, reducing the chance of Type I error below the nominal α level When you encounter ANOVA questions involving assumption violations, focus on how heteroscedasticity (unequal variances) combined with unequal sample sizes affects the F-test's behavior. The key insight here is understanding the relationship between sample size and variance in ANOVA. When the smallest group (n₁ = 8) has the largest variance, this creates a specific pattern that makes the F-test more conservative. The F-statistic becomes less sensitive to detecting true differences because the high-variance group contributes less weight to the overall analysis due to its small sample size. This reduces the actual Type I error rate below the nominal α level (typically 0.05), meaning you're less likely to falsely reject the null hypothesis than your significance level suggests. Option A is correct because this variance-sample size pattern consistently produces conservative results. Option B represents the opposite scenario—if the largest group had the largest variance, the test would become liberal and inflate Type I error rates. Option C incorrectly suggests that unequal sample sizes somehow automatically fix heteroscedasticity; they don't—the combination actually creates the problem. Option D wrongly assumes that adequate total sample size (35 participants) guarantees robustness regardless of assumption violations, but ANOVA's robustness depends heavily on the specific pattern of variance and sample size relationships. Study tip: Remember the critical pattern: small n + large variance = conservative test (Type I error below α); large n + large variance = liberal test (Type I error above α). This relationship appears frequently on biostatistics exams when testing ANOVA assumption knowledge.
16. In a one-way ANOVA comparing mean recovery times across three surgical techniques, the sum of squares between groups (SSB) is 240, and the sum of squares within groups (SSW) is 360. If there are 8 patients per group, what is the mean square error (MSE) and what does it primarily represent in this context?
    Answer: MSE = 17.14; it represents the average variability in recovery times within each surgical technique MSE = SSW/(n-k) where n is total sample size and k is number of groups. With 8 patients per group across 3 groups: n = 24, k = 3. MSE = 360/(24-3) = 360/21 = 17.14. MSE represents the pooled within-group variance, which is the average variability of observations around their respective group means - essentially the 'noise' or error variance within each treatment group.
17. A researcher studying the effect of three different exercise programs on weight loss finds the following group variances: Program A: s² = 12.4, Program B: s² = 18.7, Program C: s² = 9.2. Before proceeding with ANOVA, what should be the researcher's primary concern?
    Answer: The ratio of largest to smallest variance exceeds 2:1, suggesting possible heteroscedasticity When you encounter ANOVA questions involving multiple group variances, you need to evaluate whether the assumption of homoscedasticity (equal variances) is reasonably met before proceeding with the analysis. Let's examine these variances systematically. The largest variance is Program B at 18.7, and the smallest is Program C at 9.2. The ratio of largest to smallest variance is $\frac{18.7}{9.2} = 2.03$, which exceeds the commonly used 2:1 rule of thumb for ANOVA. This suggests potential heteroscedasticity (unequal variances), which violates a key ANOVA assumption and can lead to unreliable results. Choice A is incorrect because the concern isn't about sample size adequacy for variance estimation—we're given actual variance values to work with. Choice C is too vague and doesn't provide the specific guidance needed; while variances should be "similar," it doesn't tell you how to quantify "similar enough." Choice D suggests conducting a formal test, but this is often unnecessary when you can apply the practical 2:1 rule, and formal tests like Levene's test can be overly sensitive with large samples. Choice B correctly identifies the specific problem: the variance ratio exceeds 2:1, indicating possible heteroscedasticity. This rule of thumb is widely used because when the largest variance is more than twice the smallest, ANOVA results become less reliable, especially with unequal sample sizes. Study tip: Remember the 2:1 variance ratio rule for ANOVA. Calculate largest variance ÷ smallest variance—if it exceeds 2.0, consider this a red flag for heteroscedasticity before proceeding with standard ANOVA.
18. In a one-way ANOVA comparing four exercise interventions on weight loss, the researcher reports: 'The F-test was significant (p = 0.032), and post-hoc analysis revealed that high-intensity interval training (HIIT) produced significantly greater weight loss than both moderate cardio and strength training.' What crucial information is missing from this report that affects interpretation?
    Answer: The specific post-hoc test used and whether it controls for multiple comparisons appropriately When interpreting ANOVA results with follow-up analyses, you must critically evaluate whether the statistical procedures properly control for inflated Type I error rates that occur with multiple testing. The correct answer is D because when researchers conduct multiple pairwise comparisons after ANOVA, they're performing numerous statistical tests simultaneously. Without knowing which post-hoc test was used and whether it appropriately adjusts for multiple comparisons, you can't properly interpret the reported p-values. Different post-hoc tests (Tukey's HSD, Bonferroni, Dunnett's, etc.) have varying levels of conservativeness in controlling family-wise error rates. Some methods like Fisher's LSD don't adequately control for multiple comparisons, potentially leading to false discoveries. Option A is incorrect because while confidence intervals provide valuable information about effect magnitude and precision, their absence doesn't fundamentally compromise the validity of the statistical conclusions being drawn. Option B is wrong because assumption violations, while important for ANOVA validity, don't directly affect the interpretation of post-hoc test results that have already been conducted. Option C is incorrect because effect size measures address practical significance rather than statistical validity—the missing information about multiple comparison control is more fundamental to interpreting whether the reported significant differences are statistically reliable. Study tip: Always ask "How many comparisons?" when reviewing ANOVA follow-up analyses. If the post-hoc method isn't specified or doesn't control for multiple testing, treat the results with skepticism. This is a frequent oversight in research reporting that biostatisticians must recognize.
19. An ANOVA comparing four different teaching methods yields SSTotal = 850, SSBetween = 340, and SSWithin = 510. With 8 students per method, the researcher concludes that the teaching methods differ significantly (p < 0.05). However, a colleague notes an error in the calculation. What is the most likely problem?
    Answer: SSBetween + SSWithin should equal SSTotal, but 340 + 510 = 850, so the calculations appear correct When you encounter ANOVA problems, always start by checking the fundamental relationship: $SS_{Total} = SS_{Between} + SS_{Within}$. This partitioning of variance is the foundation of analysis of variance. Let's verify the calculations: $SS_{Between} + SS_{Within} = 340 + 510 = 850$, which exactly equals $SS_{Total} = 850$. This confirms the sum of squares calculations are mathematically correct. However, the question states there's an error, making this a tricky problem that tests your understanding of what can go wrong beyond basic arithmetic. Looking at the answer choices: Choice A correctly identifies that the calculations appear arithmetically sound, which is the most likely "problem" the colleague noticed - there's actually no computational error in the given values. Choice B incorrectly suggests 31 total degrees of freedom, but with 32 students total (8 per group × 4 groups), there are actually 31 total degrees of freedom, making this calculation correct. Choice C contains a logical error - it states that 850 exceeds $SS_{Total}$, but 850 equals $SS_{Total}$, not exceeds it. Choice D assumes an F-statistic error, but we can't determine this from the given information since the F-statistic could be calculated correctly from these sum of squares values. The key insight is that this question tests whether you'll second-guess correct calculations when told there's an error. Sometimes in statistics, what appears to be an error actually isn't - always verify the fundamental relationships first before assuming computational mistakes.
20. A pharmaceutical company tests four different dosages of a new medication on blood glucose reduction. The ANOVA table shows F = 3.45 with p = 0.028. However, upon reviewing the raw data, the researcher notices that the highest dosage group has considerably more variable responses (SD = 45 mg/dL) compared to the other groups (SD = 12-18 mg/dL). What is the most appropriate next step?
    Answer: Investigate the homogeneity of variance assumption and consider data transformation or Welch's ANOVA The large difference in standard deviations suggests violation of the homogeneity of variance assumption (homoscedasticity) required for one-way ANOVA. The highest group's SD (45) is more than double the others, indicating heteroscedasticity. The appropriate next step is to formally test this assumption (e.g., Levene's test) and consider remedial measures like data transformation or using Welch's ANOVA, which doesn't assume equal variances.

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