IAI Actuarial Core Principles · Actuarial Statistics · Confidence intervals and prediction intervals
For paired data on 9 policyholders, the differences (after minus before) have mean 4.0 and sample standard deviation 3.0. The t value with 8 degrees of freedom at 95% is 2.306. What is the 95% confidence interval for the mean difference?
The standard error is 3/√9 = 1, so the interval is 4 ± 2.306 × 1, which is (1.69, 6.31).
- A(1.69, 6.31)Correct
- B(0.00, 8.00)
- C(2.00, 6.00)
- D(1.23, 6.77)
- (3.00, 5.00)
Explanation
SE = 3/√9 = 1. Margin = 2.306 × 1 = 2.306. Interval = 4 ± 2.306 = (1.69, 6.31). Using z = 1.96 would give (2.04, 5.96), which is wrong for a small sample.
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