Nonparametric Methods Questions and Answers

Q: The recorded air quality index (AQI) at a place in a city for 16 consecutive days for the current year is compared with the historical average AQI. The direction of the recorded AQI is noted as '+' if it is above the historical average and as '−' if it is below the historical average. The dataset is as below:+ + − + − − + + + + − − − + + −The total number of runs, \(R\), is being used to test\(H_0\): The direction of the recorded AQI is randomagainst\(H_1\): The direction of the recorded AQI is not randomConsider the following statements.P: The observed value of total number of runs is 8.Q: Based on the above dataset, \(H_0\) is rejected in favor of \(H_1\) at level 0.05.Then which one of the following is true?[You may use: Under \(H_0\), with 7 negative signs and 9 positive signs, \(P(R \leq 6) = 0.108, P(R \leq 7) = 0.231, P(R \leq 8) = 0.427\)]

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.

Q: The following is an arrangement of men (M) and women (W), lined up to purchase tickets for a rock concert :MWMWMMMWMWMMMWWMMMMWWMWMMMWMMMWWWWhat is the expected value of runs for testing the hypothesis that the arrangement is random ?

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.

Q: Let \(F_{n}(\cdot)\) be empirical cumulative distribution function, based on a random of size \(n\) from a continuous distribution with cumulative distribution function \(F(\cdot)\). Then :

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.

The recorded air quality index (AQI) at a place in a city for 16 consecutive days for the current year is compared with the historical average AQI. The direction of the recorded AQI is noted as '+' if it is above the historical average and as '−' if it is below the historical average. The dataset is as below:

+ + − + − − + + + + − − − + + −

The total number of runs, \(R\), is being used to test
\(H_0\): The direction of the recorded AQI is random
against
\(H_1\): The direction of the recorded AQI is not random

Consider the following statements.

P: The observed value of total number of runs is 8.

Q: Based on the above dataset, \(H_0\) is rejected in favor of \(H_1\) at level 0.05.

Then which one of the following is true?

[You may use: Under \(H_0\), with 7 negative signs and 9 positive signs, \(P(R \leq 6) = 0.108, P(R \leq 7) = 0.231, P(R \leq 8) = 0.427\)]

A.P is correct and Q is NOT correct
B.P is NOT correct and Q is correct
C.Both P and Q are correct
D.Neither P nor Q is correct

Exam Year(s)

IIT JAM - 2026,

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.

Discuss in Forum

The following is an arrangement of men (M) and women (W), lined up to purchase tickets for a rock concert :

MWMWMMMWMWMMMWWMMMMWW
MWMMMWMMMWWW

What is the expected value of runs for testing the hypothesis that the arrangement is random ?

A.14.92
B.15.88
C.16.76
D.18.20

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

Discuss in Forum

Let \(F_{n}(\cdot)\) be empirical cumulative distribution function, based on a random of size \(n\) from a continuous distribution with cumulative distribution function \(F(\cdot)\). Then :

A.\(F_{n}\) is not a consistent estimator for F
B.Variance of \(F_{n}\) converges to a positive constant
C.Asymptotic distribution of \(F_{n}\) is normal
D.Asymptotic mean of \(F_{n}\) is 1/2

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

Nonparametric Methods Questions and Answers

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