IAI Actuarial Core Principles · Actuarial Statistics · Random sampling and sampling distributions
A sample of 16 observations from a normal distribution has sample mean 42 and sample standard deviation 8 (divisor n-1). The upper 2.5% point of t with 15 degrees of freedom is 2.131. What is the 95% confidence interval for the mean?
The 95% interval is 42 plus or minus 2.131 x (8/4) = 42 +/- 4.262, giving (37.74, 46.26), using the t distribution with 15 degrees of freedom.
- A(37.74, 46.26)Correct
- B(38.09, 45.91)
- C(39.00, 45.00)
- D(34.00, 50.00)
- (36.11, 47.89)
Explanation
Standard error = 8/4 = 2. Margin = 2.131 x 2 = 4.262. Interval = 42 +/- 4.262 = (37.74, 46.26). Using 1.96 would wrongly give (38.08, 45.92).
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