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

Two independent samples from normal populations with known standard deviations give: sample 1 has n = 36, mean 52, sigma = 6; sample 2 has n = 25, mean 47, sigma = 5. Using z = 1.96, which is the 95% confidence interval for the difference in population means (mu1 - mu2)?

The standard error of the difference is sqrt(1 + 1) = 1.414, so the margin is 1.96 x 1.414 = 2.77. The interval is 5 ± 2.77, about (2.2, 7.8), which is closest to the option (2.1, 7.9).

  1. A(2.5, 7.5)
  2. B(2.1, 7.9)Correct
  3. C(1.9, 8.1)
  4. D(2.6, 7.4)
  5. (3.0, 7.0)

Explanation

Standard error = sqrt(36/36 + 25/25) = sqrt(2) = 1.4142. Margin = 1.96 x 1.4142 = 2.772. The difference is 5, so the interval is 5 ± 2.77 = (2.23, 7.77). The nearest option is (2.1, 7.9), but check: the margin is 2.77, so the interval is (2.23, 7.77), which is closest to option B. Option A uses a margin of 2.5, which is too small.

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