IAI Actuarial Core Principles · Actuarial Statistics · Confidence intervals and prediction intervals
A random sample of 16 observations from a normal distribution with unknown variance has sample mean 50 and sample standard deviation 8. Using t(15) 97.5% point = 2.131, what is the 95% confidence interval for the population mean?
The interval is (45.74, 54.26). With unknown variance the t distribution on 15 degrees of freedom applies, so the margin is 2.131 times the standard error of 2, which is 4.262, around the sample mean of 50.
- A(45.74, 54.26)Correct
- B(46.08, 53.92)
- C(44.32, 55.68)
- D(45.00, 55.00)
- (47.37, 52.63)
Explanation
Standard error = 8/sqrt(16) = 2. Margin = 2.131 x 2 = 4.262. Interval = 50 ± 4.262 = (45.74, 54.26). Using 1.96 would give (46.08, 53.92), wrongly ignoring that the variance is estimated.
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