IAI Actuarial Core Principles · Actuarial Statistics · Confidence intervals and prediction intervals
A sample of 16 claim sizes from a normal population has mean 200 and sample variance 400 (divisor n-1). Using t(15) with 97.5th percentile 2.131, what is the 95% confidence interval for the population mean?
The interval is (189.35, 210.65). With unknown variance and a small sample, use the t distribution with 15 degrees of freedom: 200 plus or minus 2.131 times 20 over 4, which is a margin of 10.655.
- A(189.35, 210.65)Correct
- B(190.20, 209.80)
- C(160.76, 239.24)
- D(180.00, 220.00)
- (194.67, 205.33)
Explanation
s = 20, so the standard error is 20/4 = 5. The margin is 2.131 × 5 = 10.655, giving (189.345, 210.655), i.e. (189.35, 210.65). Using z = 1.96 would give (190.20, 209.80), which is wrong because σ is unknown and n is small.
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