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

A sample of n = 10 from a normal population has sample variance s^2 = 18. The chi-square(9) points are 2.700 (2.5%) and 19.023 (97.5%). What is the 95% confidence interval for sigma^2?

The interval is (8.52, 60.00). Multiply s squared by n-1 to get 162, then divide by the upper chi-square point 19.023 for the lower limit and by the lower point 2.700 for the upper limit.

  1. A(8.52, 60.00)Correct
  2. B(0.95, 6.67)
  3. C(60.00, 8.52)
  4. D(8.52, 6.67)
  5. (7.67, 54.00)

Explanation

(n-1)s^2 = 9 x 18 = 162. Lower limit = 162/19.023 = 8.52; upper limit = 162/2.700 = 60.00. So the interval is (8.52, 60.00). Swapping the percentage points gives an inverted interval, and dividing 18 directly ignores the factor n-1.

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