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

A random sample of 10 observations from a normal distribution has sample variance 18 (divisor n-1). Given that the chi-square(9) distribution has 2.5th percentile 2.700 and 97.5th percentile 19.023, what is the 95% confidence interval for the population variance σ²?

The interval is (8.52, 60.00). The pivot (n-1)S²/σ² follows chi-square with 9 degrees of freedom; with 162 as the numerator, divide by 19.023 for the lower limit and by 2.700 for the upper limit.

  1. A(8.52, 60.00)Correct
  2. B(6.00, 43.33)
  3. C(0.85, 6.00)
  4. D(7.67, 54.00)
  5. (9.00, 63.00)

Explanation

The pivot is (n-1)S²/σ² ~ χ²(9), with (n-1)S² = 162. The lower limit is 162/19.023 = 8.52 and the upper limit is 162/2.700 = 60.00. Using 10×18 = 180 with the wrong multiplier gives different limits (9.46, 66.67), and mixing up the percentiles reverses the roles.

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