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IAI Actuarial Core Principles · Actuarial Statistics · Hypothesis testing and goodness of fit

A sequence of 12 claims is classified as above (A) or below (B) the median: A A B B B A B A A A B B. How many runs does the sequence contain, and what is the expected number of runs under randomness with n1 = 6 and n2 = 6, given E[R] = 2n1n2/(n1+n2) + 1?

Grouping consecutive identical letters gives six runs: AA, BBB, A, B, AAA, BB. With six observations in each category the expected number of runs under randomness is 2×36/12 + 1 = 7, so the observed count is 6 and the expected is 7.

  1. A7 runs observed; expected 7Correct
  2. B6 runs observed; expected 7
  3. C7 runs observed; expected 6
  4. D6 runs observed; expected 6
  5. 8 runs observed; expected 7

Explanation

Runs: AA, BBB, A, B, AAA, BB gives 6 runs. Recount: AA(1), BBB(2), A(3), B(4), AAA(5), BB(6). So 6 runs. Expected = 2*36/12 + 1 = 7. So the correct option is 6 observed, expected 7.

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