FRM Part I · FRM Exam Part I · Hypothesis Testing
A analyst computes a 95% confidence interval for a mean return using 25 observations, with unknown population variance and normally distributed returns. The sample mean is 2.0%, and the sample standard deviation is 5.0%. The t critical value with 24 degrees of freedom at 95% two-sided is 2.064. What is the interval?
The interval is approximately -0.06% to 4.06%. With unknown variance and 25 observations, use the t value of 2.064 times the standard error of 1.0%, giving a margin of about 2.06% around the 2.0% mean. The normal value of 1.96 would understate the width.
- A-0.06% to 4.06%Correct
- B0.04% to 3.96%
- C-0.58% to 4.58%
- D-8.28% to 12.28%
Explanation
SE = 5.0/sqrt(25) = 1.0%. Margin = 2.064 x 1.0 = 2.064%. Interval = 2.0 ± 2.064 = -0.064% to 4.064%, i.e. about -0.06% to 4.06%. Using z = 1.96 gives 0.04% to 3.96%, which understates the width for a small sample.
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