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Part BCSIR NET June 2023runs-distribution

Runs distribution

Let and be independent random samples from continuous CDFs F and G. For the Wald–Wolfowitz run test of vs , let R be the total number of runs in the combined ordered arrangement. Which of the following is true?

  1. A. = 6) = 28/286, = 9) = 28/143.
  2. B. = 6) = 21/286, = 9) = 15/286.
  3. C. = 6) = 21/286, = 9) = 28/143.✓
  4. D. = 6) = 21/286, = 9) = 15/286.

Solution

With m = 7, n = 9 and C(16,7) = 11440 arrangements: P(R = 2k) = 2·C(m−1,k−1)·C(n−1,k−1)/C(16,7) gives P(R=6) = 2·15·28/11440 = 21/286; P(R = 2k+1) = [C(m−1,k)C(n−1,k−1) + C(m−1,k−1)C(n−1,k)]/C(16,7) gives P(R=9) = (15·56 + 20·70)/11440 = 28/143.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: Any interval of the form [X̄ − (S/√n)t, ∞) with a 90% quantile t is a 90% confidence interval

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