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IAI Actuarial Core Principles · Actuarial Statistics · Confidence intervals and prediction intervals

A random sample of 25 claim sizes (in Rs thousand) from a normal distribution has sample mean 120 and sample standard deviation 10. Using t(24) 97.5% point 2.064, which is a 95% prediction interval for the next claim size, to the nearest Rs thousand?

Use mean plus or minus t times s times sqrt(1+1/n): 2.064 x 10.198 is about 21, giving (99, 141). The narrower interval (116,124) is only for the mean and ignores the variability of a new claim.

  1. A(99, 141)Correct
  2. B(116, 124)
  3. C(100, 140)
  4. D(102, 138)
  5. (109, 131)

Explanation

Standard error = 10 x sqrt(1+1/25) = 10 x 1.0198 = 10.198. Margin = 2.064 x 10.198 = 21.05. Interval = (98.95, 141.05), i.e. (99, 141). Option (116,124) is the confidence interval for the mean, ignoring the new observation's variability.

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