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

In a test of whether residual signs from a fitted model are random, there are n1 = 10 positive and n2 = 10 negative residuals, with 6 runs observed. Under randomness, the number of runs has mean 11 and variance 4.74 (approx.) using the normal approximation. What is the approximate z statistic (no continuity correction) and conclusion at 5% two-sided?

The standard deviation of the run count is about 2.18, so z = (6 − 11)/2.18 ≈ −2.30. This lies beyond −1.96, so randomness is rejected at 5%. Too few runs indicates clustering of residual signs, suggesting positive serial correlation.

  1. Az = -2.30; reject randomnessCorrect
  2. Bz = -1.15; do not reject randomness
  3. Cz = -2.30; do not reject randomness
  4. Dz = +2.30; reject randomness
  5. z = -5.00; reject randomness

Explanation

Variance 4.74 gives sd = 2.177. z = (6-11)/2.177 = -2.30. Since |z| > 1.96 we reject randomness; too few runs suggests positive serial correlation. Check variance: 2n1n2(2n1n2-n1-n2)/((n1+n2)^2(n1+n2-1)) = 200*180/(400*19)=4.74.

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