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

A 95% confidence interval for a normal mean, based on a sample of 25 with unknown variance, is (92.0, 108.0). Using t(24) 97.5% point = 2.064, what is the sample standard deviation (nearest value)?

The sample standard deviation is about 19.4. The half-width is 8, which equals 2.064 times s divided by the square root of 25, so s = 8 times 5 divided by 2.064, giving roughly 19.4.

  1. A19.4Correct
  2. B9.7
  3. C15.5
  4. D7.8
  5. 38.8

Explanation

Half-width = 8 = 2.064 x s/5. So s = 8 x 5 / 2.064 = 19.38, about 19.4. Option 9.7 forgets that the width is 16 not 8 in one step of confusion; 38.8 uses the full width.

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