IAI Actuarial Core Principles · Actuarial Statistics · Confidence intervals and prediction intervals
A sample of 16 claim amounts (in Rs thousand) from a normal population has mean 120 and sample variance 64. Using t(15) 97.5% point = 2.131, what is the 95% confidence interval for the population mean?
The interval is (115.74, 124.26). The sample standard deviation is 8, giving a standard error of 2, and the t value of 2.131 with 15 degrees of freedom gives a margin of 4.262 around the mean of 120.
- A(115.74, 124.26)Correct
- B(116.08, 123.92)
- C(112.00, 128.00)
- D(102.95, 137.05)
- (113.53, 126.47)
Explanation
S = 8, so S/sqrt(16) = 2. Margin = 2.131 x 2 = 4.262. Interval = 120 +/- 4.262 = (115.74, 124.26). Using 1.96 gives (116.08, 123.92), wrongly ignoring the estimation of sigma.
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