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

Independent random samples of sizes 40 and 50 are taken from two normal populations with known standard deviations 6 and 8 respectively. The sample means are 52 and 47. Which of the following is the 95% confidence interval for the difference in population means (first minus second)? Use z = 1.96.

The interval is 5 ± 1.96 × √(36/40 + 64/50), which gives roughly 5 ± 2.89, so about (2.1, 7.9).

  1. A(2.3, 7.7)
  2. B(1.9, 8.1)Correct
  3. C(0.8, 9.2)
  4. D(2.9, 7.1)
  5. (3.5, 6.5)

Explanation

Difference = 5. Variance of difference = 36/40 + 64/50 = 0.9 + 1.28 = 2.18, so SE = 1.4765. Margin = 1.96 × 1.4765 = 2.894. Interval = (2.11, 7.89). Computed precisely this is about (2.1, 7.9), which is closest to option 1 only if rounded, so check: option 0 (2.3, 7.7) uses a margin of 2.7, which is wrong.

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