Nonparametric Methods Questions and Answers

Answer: A

P is correct and Q is NOT correct.

A run is a maximal block of identical signs. Splitting the sequence:

(+ +) (−) (+) (− −) (+ + + +) (− − −) (+ +) (−)

gives \(R = 8\) runs, with 9 plus and 7 minus signs, so P is correct.

The alternative 'not random' is two-sided: too few runs (clustering) or too many (alternation) both count against \(H_{0}\). Under \(H_{0}\) the expected number of runs is \(1 + \frac{2 \cdot 9 \cdot 7}{16} \approx 8.9\), and the observed 8 is close to it. The lower tail \(P(R \leq 8) = 0.427\) is far above the 0.025 a two-sided test at level 0.05 allows each side; even 6 runs would not be significant, since \(P(R \leq 6) = 0.108\). So \(H_{0}\) is not rejected at level 0.05, and Q is not correct.

Answer: C

The expected number of runs is 16.76.

Count the letters in the two printed lines: there are 33 people, \(n_{1} = 20\) men and \(n_{2} = 13\) women. Under randomness the number of runs R has mean

\[ \begin{aligned} E(R) &= \frac{2n_{1}n_{2}}{n_{1} + n_{2}} + 1 \\ &= \frac{2 \times 20 \times 13}{33} + 1 \\ &= 15.76 + 1 = 16.76 \end{aligned} \]

(The observed arrangement has 18 runs, close to this expectation, so the runs test would not reject randomness.)

None of 14.92, 15.88 or 18.20 is the expectation for any split of 33 people into two groups, since each would need \(n_{1}n_{2}\) to be a non-integer. Dropping the +1 gives 15.76, which is not offered either.

Answer: C

The asymptotic distribution of \(F_{n}\) is normal.

At a fixed \(x\), \(nF_{n}(x)\) counts the observations not exceeding \(x\), which is binomial with parameters \(n\) and \(F(x)\). By the central limit theorem, \(\sqrt{n}(F_{n}(x) - F(x))\) tends to \(N(0, F(x)(1 - F(x)))\).

The other statements fail. \(F_{n}(x) \to F(x)\) almost surely (and uniformly, by Glivenko-Cantelli), so \(F_{n}\) is consistent. Its variance is \(F(x)(1 - F(x))/n\), which tends to 0, not to a positive constant. And its mean is \(F(x)\), not 1/2.

Answer: B

Kendall's coefficient is \(-1/2\).

With 9 observations and no ties there are \(\binom{9}{2} = 36\) pairs, each concordant or discordant. With 9 concordant pairs, 27 are discordant, so

\[ \begin{aligned} \tau &= \frac{C - D}{\binom{n}{2}} \\ &= \frac{9 - 27}{36} = -\frac{1}{2} \end{aligned} \]

The sign is negative because discordant pairs outnumber concordant ones. +1/2 reverses the sign, and \(\pm 1/4\) come from dividing by 72 or using 9 alone instead of the difference.

Answer: C

The true statement is \(V_{F}(U) \le V_{F}(S)\).

The U-statistic corresponding to S is its average over all permutations of the sample, which is the conditional expectation of S given the order statistics: \(U = E[S \mid X_{(1)}, \ldots, X_{(n)}]\). The order statistics are sufficient for an i.i.d. sample from an unrestricted F.

So U has the same expectation as S, \(\theta(F)\), which rules out A and B, and by the Rao-Blackwell argument (conditional variance formula)

\[ V(S) = V(U) + E[V(S \mid \text{order stats})] \ge V(U) \]

So the U-statistic never has larger variance, and D is wrong.

Answer: B

The Mann-Whitney statistic is 8.

Count, for each \(y\), the number of \(x\)'s that come before it in the combined ordering \(y_{4} \lt x_{1} \lt x_{5} \lt y_{3} \lt x_{2}\) \(\lt y_{1} \lt y_{2} \lt x_{4} \lt x_{3}\):

• \(y_{4}\): 0
• \(y_{3}\): 2 (\(x_{1}, x_{5}\))
• \(y_{1}\): 3
• \(y_{2}\): 3

The total is \(U = 8\). Counting the other way, the number of \(y\)'s before each \(x\) gives \(1 + 1 + 2 + 4 + 4 = 12\), and \(8 + 12 = 5 \times 4 = 20\) as it must. The test statistic is the smaller count, 8. 12 is the complementary count, and 6 and 10 do not arise from either count.

Answer: A

The p-value is 0.5.

This is the sign test. Under \(H_{0}\) the number S of positive \(Z_{i}\) is binomial with \(n = 5\), \(p = \frac{1}{2}\). Since \(H_{1}\) says the median is negative, small values of S are evidence against \(H_{0}\), so the p-value is

\[ \begin{aligned} P(S \le 2) &= \frac{\binom{5}{0} + \binom{5}{1} + \binom{5}{2}}{32} \\ &= \frac{1 + 5 + 10}{32} = 0.5 \end{aligned} \]

The data give no evidence at all against \(H_{0}\). 0.25, 0.10 and 0.05 would require fewer positive differences.

Answer: D

The variance is \(F(x)(1 - F(x))\).

For fixed \(x\), \(nF_{n}(x)\) counts the observations at most \(x\), so it is binomial with parameters \(n\) and \(F(x)\). Hence \(E[F_{n}(x)] = F(x)\) and \(\mathrm{Var}[F_{n}(x)] = F(x)(1 - F(x))/n\).

By the central limit theorem, \(\sqrt{n}(F_{n}(x) - F(x)) \to N(0, F(x)(1 - F(x)))\).

The other variances, \(F^{2}/2\), \((1 - F)^{2}\) and \(F^{2}\), are not the Bernoulli variance \(p(1 - p)\) with \(p = F(x)\).