FRM Part I · FRM Exam Part I · Hypothesis Testing
From 31 daily observations of a normally distributed return series, the sample standard deviation is 0.20 (s² = 0.04). For 30 degrees of freedom, the chi-square values are 43.773 (5% upper tail) and 18.493 (95% upper tail). Which is the 90% confidence interval for the population variance?
The 90% confidence interval for variance is 0.0274 to 0.0649. It is found by dividing (n−1)s² = 1.2 by the upper chi-square value 43.773 for the lower bound and by the lower value 18.493 for the upper bound.
- A0.0274 to 0.0649Correct
- B0.0283 to 0.0671
- C0.1370 to 0.3245
- D0.1656 to 0.2547
Explanation
The interval is [(n−1)s²/χ²upper, (n−1)s²/χ²lower] with (n−1)s² = 30 × 0.04 = 1.2. Lower = 1.2/43.773 = 0.0274; upper = 1.2/18.493 = 0.0649. Using df of 31 gives 0.0283 to 0.0671. Using s instead of s² gives 0.137 to 0.3245. The last option is the interval for standard deviation, not variance.
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