Rank-Based Tests - Statistics Graduate Level

1. Using the usual chi-square approximation, which result is correct at significance level $0.05$?
   Answer: The statistic is approximately $4.36$ with $2$ degrees of freedom, so the null hypothesis is not rejected When you encounter a Kruskal–Wallis question, your two main tasks are computing the test statistic correctly and identifying the right degrees of freedom for the chi-square approximation. The Kruskal–Wallis statistic is: $H = \frac{12}{N(N+1)} \sum_{i=1}^{k} \frac{R_i^2}{n_i} - 3(N+1)$ Here, $k = 3$ groups, each with $n_i = 3$, giving $N = 9$ total observations. The rank sums are $R_1 = 8$, $R_2 = 15$, $R_3 = 22$. Plugging in: $H = \frac{12}{9(10)} \left(\frac{64}{3} + \frac{225}{3} + \frac{484}{3}\right) - 3(10)$ $= \frac{12}{90} \cdot \frac{773}{3} - 30 = \frac{9276}{270} - 30 = 34.356 - 30 = 4.356$ So $H \approx 4.36$. The degrees of freedom equal $k - 1 = 2$. The chi-square critical value at $\alpha = 0.05$ with 2 df is $5.99$. Since $4.36 < 5.99$, you fail to reject the null hypothesis — confirming answer B. Answer A gets the right degrees of freedom but halves the statistic (a likely arithmetic error or use of a wrong formula). Answer C correctly computes the statistic but uses $df = 3$ instead of $k - 1 = 2$, which is a classic degrees-of-freedom confusion — and then incorrectly concludes rejection. Answer D inflates the statistic to $6.53$, which would cross the critical value, but this figure doesn't arise from the correct formula. As a study tip, always remember: Kruskal–Wallis degrees of freedom are $k - 1$ (number of groups minus one), not $N - 1$. Mixing these up is the most common trap on this type of question.
2. By what factor should the no-tie null variance be multiplied to obtain the standard tie-corrected variance?
   Answer: $1-\dfrac{30}{10^3-10}=\dfrac{32}{33}$ Whenever you encounter a Wilcoxon rank-sum test with ties, your central task is adjusting the null variance using a correction factor that accounts for tied groups. The standard no-tie null variance is $\frac{n_1 n_2 (N+1)}{12}$, and the tie-corrected version multiplies this by a factor of the form $1 - \frac{\sum_j(t_j^3 - t_j)}{N^3 - N}$, where each $t_j$ is the size of a tied group. Here, $N = 10$, one value occurs three times ($t_1 = 3$) and another occurs twice ($t_2 = 2$). Compute the correction sum: $\sum_j(t_j^3 - t_j) = (3^3 - 3) + (2^3 - 2) = (27 - 3) + (8 - 2) = 24 + 6 = 30$ The denominator is $N^3 - N = 1000 - 10 = 990$. The correction factor is therefore: $1 - \frac{30}{990} = 1 - \frac{1}{33} = \frac{32}{33}$ This confirms B is correct. Choice A uses $N^2 - N = 90$ in the denominator instead of $N^3 - N$, which corresponds to no standard formula — a straightforward algebraic error. Choice C computes the numerator as 14 rather than 30; this would result from using $\sum(t_j^2 - t_j)$ (i.e., $6 + 2 + \ldots$) instead of $\sum(t_j^3 - t_j)$, confusing the exponent. Choice D uses 35 in the numerator, which doesn't correspond to any standard calculation of the tied groups given. A reliable memory anchor: the tie correction involves cubed group sizes — $t^3 - t$ — divided by $N^3 - N$. If you remember "cubes on top and bottom," you'll avoid the most common errors here.
3. Which interpretation of the observed statistic is most defensible without assuming that the two distributions differ only by a location shift?
   Answer: The estimated probability that a randomly selected $X$ is less than a randomly selected $Y$, with half credit for ties, is $0.75$ Whenever you see a question about the Mann–Whitney statistic, anchor yourself to its core probabilistic interpretation: $U$ estimates $P(X < Y)$ (with ties split), not a location shift or a median difference, unless you explicitly impose a location-shift assumption. With $n_1 = 8$ and $n_2 = 10$, the total number of pairs is $8 \times 10 = 80$. Dividing the observed statistic by the total pairs gives $\hat{\theta} = U/(n_1 n_2) = 60/80 = 0.75$. This quantity estimates $\theta = P(X < Y) + \tfrac{1}{2}P(X = Y)$, which is the probability that a randomly drawn $X$ falls below a randomly drawn $Y$, with half-credit awarded when they tie. No distributional assumptions beyond independence are required for this interpretation — making A the most defensible choice. B is wrong on two levels: it claims to exclude ties entirely (which changes the estimand and isn't what the formula computes), and it misidentifies the denominator as $n_1$ times "pairs counted" rather than $n_1 n_2$. C is a conceptual error. The ratio $U/(n_1 n_2) = 0.75$ is a probability estimate, not a shift in original measurement units. Interpreting it as a median difference requires the location-shift model — precisely the assumption the question tells you not to make. D reverses the direction. Since $\hat{\theta} = 0.75 > 0.5$, it is $Y$ that tends to exceed $X$, meaning $X$ is stochastically smaller, not larger. The estimated probability that $Y < X$ would be $1 - 0.75 = 0.25$. Strategy tip: Always check direction and denominator. On exam questions about Mann–Whitney, confirm who is being compared to whom and verify that you divide by $n_1 n_2$, not a subset of pairs.
4. What are the positive signed-rank sum $W^+$ and the two-sided statistic $T=\min(W^+,W^-)$?
   Answer: $W^+=6.5$ and $T=3.5$, using midranks $2.5$ for the tied absolute value $3$ When applying the Wilcoxon signed-rank test, your first step is always to drop zero differences, then rank the remaining absolute values, using midranks for ties. Only after ranking do you separate positives from negatives. Here, discarding the zero leaves four differences: $-2, 3, -3, 5$. Their absolute values are $2, 3, 3, 5$, which rank as follows: $|2| \to 1$, then $|3|$ and $|3|$ tie for ranks 2 and 3, so each gets midrank $2.5$, and $|5| \to 4$. Now attach signs: the negative differences ($-2$ and $-3$) carry ranks $1$ and $2.5$, while the positive differences ($3$ and $5$) carry ranks $2.5$ and $4$. So $W^+ = 2.5 + 4 = 6.5$ and $W^- = 1 + 2.5 = 3.5$, giving $T = \min(6.5, 3.5) = 3.5$. This confirms answer A. Answer B breaks the tie by awarding the higher integer rank $3$ to the positive difference — but ties must be averaged into midranks, not resolved by sign. Answer C makes the same integer-rank mistake and compounds it by giving both higher ranks to positive differences, inflating $W^+$ further. Answer D incorrectly retains the zero difference and assigns it rank 1, violating the fundamental rule that zeros are discarded before ranking. Your study tip: memorize the three rules in order — drop zeros, midrank ties, then sum by sign. Exam distractors almost always tempt you to skip one of these steps or apply them out of order.
5. If the t test requires a total sample size of approximately $100$, what total sample size would the Wilcoxon test require for comparable asymptotic power under this model?
   Answer: Approximately $105$, because the required size is $100/0.955$ after rounding Whenever you see a question involving asymptotic relative efficiency (ARE), your first instinct should be to think carefully about what the efficiency ratio actually tells you — specifically, which test is in the numerator and which is in the denominator. The ARE of the Wilcoxon relative to the t-test is defined here as $e_{W,t} = \lim n_t / n_W \approx 3/\pi \approx 0.955$. This ratio tells you how many t-test observations are needed per Wilcoxon observation to achieve equivalent power. Since $e_{W,t} < 1$, the Wilcoxon actually requires more observations than the t-test under normality — it is the slightly less efficient procedure in this setting. To find the Wilcoxon sample size, you rearrange: $n_W = n_t / e_{W,t} = 100 / 0.955 \approx 104.7 \approx 105$. That confirms C is correct. Choice A flips the ratio, multiplying rather than dividing: $100 \times 0.955 = 95.5$. This would mean the Wilcoxon needs fewer observations, but that contradicts the fact that ARE < 1 implies the Wilcoxon is less efficient under normality. Choice B is tempting because both tests are valid under normality, but "valid" doesn't mean "identically powerful at the same sample size" — the ARE being less than 1 (even slightly) means a real, quantifiable difference exists asymptotically. Choice D invents a heuristic about "rank conversion losing observations" that has no grounding in ARE theory; the ~5% extra cost comes from the efficiency formula, not data loss. The key study tip: always track the direction of the ARE ratio. If $e_{A,B} = n_B/n_A$, then $n_A = n_B / e_{A,B}$. Writing it out algebraically before plugging in numbers will prevent the flip error that makes A so tempting.
6. Using a two-sided exact Wilcoxon rank-sum test that doubles the smaller one-sided tail probability, what is the p-value?
   Answer: $0.20$, because two allocations have rank sum at most $7$, the one-sided probability is $2/20=0.10$, and doubling gives $0.20$ Whenever you see an exact Wilcoxon rank-sum test, your first move should be to enumerate all equally likely rank allocations, identify the tail probabilities, then apply the two-sided rule carefully. With two samples of size 3 drawn from ranks 1–6, the total number of ways to assign 3 ranks to the first sample is $\binom{6}{3} = 20$, so each allocation has probability $1/20$ under $H_0$. Now list the allocations with the smallest possible rank sums: $\{1,2,3\}$ gives $W=6$, and $\{1,2,4\}$ gives $W=7$. No other allocation produces a rank sum of 7 or below. That means exactly 2 allocations fall in the lower tail ($W \leq 7$), giving a one-sided tail probability of $2/20 = 0.10$. For a two-sided test that doubles the smaller tail probability, the p-value is $2 \times 0.10 = \boxed{0.20}$, confirming C is correct. Choice A is wrong because it counts only the single allocation strictly less than 7 (i.e., $W=6$), giving $1/20 = 0.05$. The two-sided rule requires including the observed value itself in the tail, so $W=7$ must be counted too. Choice B correctly identifies 2 allocations and computes $2/20 = 0.10$, but stops there — it forgets to double for the two-sided test, reporting the one-sided p-value instead. Choice D invents a count of 4 allocations by misapplying the "both tails" logic, conflating doubling the probability with doubling the count of allocations. Study tip: In exact nonparametric tests, always include the observed statistic in the tail count, then double that one-sided probability for a two-sided result — don't double the count of allocations.
7. What does the Friedman test conclude using its usual chi-square approximation at significance level $0.05$?
   Answer: The statistic is $6.50$ with $2$ degrees of freedom, so the treatment effect is significant When you see a Friedman test question, focus on two things: the correct formula and the correct degrees of freedom. The Friedman statistic is: $\chi^2_F = \frac{12}{bk(k+1)} \sum R_j^2 - 3b(k+1)$ where $b$ is the number of blocks, $k$ is the number of treatments, and $R_j$ are the rank sums. Here, $b = 4$, $k = 3$, and the rank sums are $5, 7, 12$, so $\sum R_j^2 = 25 + 49 + 144 = 218$. Plugging in: $\frac{12}{4 \cdot 3 \cdot 4}(218) - 3(4)(4) = \frac{12}{48}(218) - 48 = 54.5 - 48 = 6.50$. The degrees of freedom equal $k - 1 = 3 - 1 = 2$. The critical value for $\chi^2$ with 2 df at $\alpha = 0.05$ is $5.991$, and since $6.50 > 5.991$, the treatment effect is significant — confirming B. Choice A gets the degrees of freedom right but reports a statistic of $4.50$, which likely comes from an arithmetic error in computing $\sum R_j^2$. Choice C reports the correct statistic of $6.50$ but uses $3$ degrees of freedom instead of $k-1=2$, leading to an incorrect non-significant conclusion. Choice D uses both the wrong statistic and wrong degrees of freedom. Study tip: Always remember the Friedman degrees of freedom are $k - 1$ (treatments minus one), not $b - 1$ and not $k$. Confusing df with the number of blocks or forgetting to subtract 1 is the most common trap on this type of question.
8. Which statement about the resulting test is most accurate?
   Answer: The test is exact under exchangeability implied by independence, because the permutations preserve the observed ties When you see a question about permutation tests and rank correlations, the key concept to anchor on is what makes a permutation test exact: the exchangeability of observations under the null hypothesis. Under independence, the joint distribution of $(X_i, Y_i)$ satisfies exchangeability — meaning any permutation of the $Y$ values across the fixed $X$ values is equally likely. This holds regardless of whether ties exist. When the researcher permutes the observed $Y$ values (including their tied values), each permutation is equally probable under the null, so the resulting permutation distribution is exactly the correct null distribution. The observed Spearman statistic is compared against a distribution generated by the same data structure, ties included. This is precisely why D is correct: the test is exact under the exchangeability that independence guarantees, and the midrank-handling of ties is preserved consistently across every permutation. A contains a widespread misconception — ties don't invalidate a permutation test. Ties only cause problems for the asymptotic approximation (using the standard normal or $t$-table), not for the permutation distribution itself, which is constructed empirically from the data. B is subtly wrong because exactness is a property of the procedure under the null hypothesis, not of the statistic equaling zero. A test being "exact when the statistic is zero" is incoherent — exactness refers to the Type I error rate, not the observed value. C incorrectly ties the validity of a nonparametric permutation test to a parametric distributional assumption. The permutation test requires no normality whatsoever. Study tip: On questions about nonparametric tests, always ask yourself: what does the null hypothesis guarantee? Exchangeability under independence is the engine that makes permutation tests exact — distributional assumptions are irrelevant.
9. Which condition most directly justifies the usual exact null distribution for this test?
   Answer: The differences are independent and arise from a continuous distribution symmetric about zero When you see a question about the Wilcoxon signed-rank test's null distribution, you should immediately think about what mathematical properties make that specific distribution valid — not just what assumptions are "nice to have." The signed-rank test works by ranking the absolute differences $|D_i|$ and attaching the original signs. Under the null hypothesis, the exact null distribution is derived by assuming that every possible assignment of $\pm$ signs to the ranks is equally likely. This equal-likelihood property holds if and only if the differences are symmetric about zero: symmetry implies $P(D_i > 0) = P(D_i < 0)$ and that positive and negative differences of the same magnitude are equally probable. Additionally, independence ensures the joint distribution factors, and continuity eliminates ties (which would complicate the exact distribution). Together, these three properties — independence, continuity, and symmetry about zero — are precisely what B states, making it the direct justification. Choice A fails because median zero alone doesn't guarantee symmetry. A strongly asymmetric distribution can have median zero, but then positive and negative signs on equal-magnitude ranks won't be equally likely, invalidating the exact permutation argument. Choice C is a distractor that sounds like a classical paired-test assumption. Equal marginal variances are relevant for the paired *t*-test's validity, not for the signed-rank null distribution, which makes no variance assumptions. Choice D is wrong because normality of absolute differences is neither required nor sufficient. The signed-rank test is nonparametric precisely to avoid distributional shape assumptions like normality; unequal sign probabilities would directly violate symmetry. Your study tip: distinguish between assumptions needed for a test statistic's null distribution versus assumptions for its power or efficiency. Exam questions often conflate these — symmetry about zero is the load-bearing assumption for the signed-rank null distribution specifically.
10. What is the Hodges–Lehmann estimate of the location shift?
    Answer: $1.75$, obtained as the median of all six Walsh averages including the three self-pairs Whenever you see a question involving the Hodges–Lehmann estimator, remember its defining rule: compute all Walsh averages $(D_i + D_j)/2$ for every pair where $i \le j$, including the "self-pairs" where $i = j$, then take the median of the full set. With observed differences $D_1 = -1,\ D_2 = 2,\ D_3 = 4$, you form all six Walsh averages: - Self-pairs: $\frac{-1+(-1)}{2} = -1.0$, $\frac{2+2}{2} = 2.0$, $\frac{4+4}{2} = 4.0$ - Cross-pairs: $\frac{-1+2}{2} = 0.5$, $\frac{-1+4}{2} = 1.5$, $\frac{2+4}{2} = 3.0$ Sorting these six values: $-1.0,\ 0.5,\ 1.5,\ 2.0,\ 3.0,\ 4.0$. The median of six values is the average of the 3rd and 4th: $\frac{1.5 + 2.0}{2} = 1.75$. So D is correct. Choice A is wrong because it discards the self-pairs, leaving only three cross-pair values $\{0.5, 1.5, 3.0\}$ whose median is $1.50$ — but the self-pairs are explicitly part of the estimator's definition. Choice B restricts computation to positive differences only, which has no basis in the procedure and ignores $D_1 = -1$ entirely. Choice C simply returns the sample median of the raw data ($2.00$), which is a different estimator altogether — the Hodges–Lehmann estimator exists precisely because it has better robustness properties than the sample median. Your study tip: memorize that $i \le j$ (not $i < j$) is the indexing convention. That single inequality is what forces you to include the $n$ self-pairs, giving you $n(n+1)/2$ total Walsh averages — here $3(4)/2 = 6$.

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