Part BCSIR NET June 2023runs-distribution

Runs distribution

Let $X_{1}, \dots, X_{7}$ and $Y_{1}, \dots, Y_{9}$ be independent random samples from continuous CDFs F and G. For the Wald–Wolfowitz run test of $H_{0}: F = G$ vs $H_{1}: F \ne G$ , let R be the total number of runs in the combined ordered arrangement. Which of the following is true?

A. $P_{H_{0}}(R$ = 6) = 28/286, $P_{H_{0}}(R$ = 9) = 28/143.
B. $P_{H_{0}}(R$ = 6) = 21/286, $P_{H_{0}}(R$ = 9) = 15/286.
C. $P_{H_{0}}(R$ = 6) = 21/286, $P_{H_{0}}(R$ = 9) = 28/143.✓ Correct
D. $P_{H_{0}}(R$ = 6) = 21/286, $P_{H_{0}}(R$ = 9) = 15/286.

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

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Runs distribution

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