Financial Risk Manager Part 1
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For a sample of the past 28 monthly stock returns for Bidco Inc., the mean return is 5% and the sample standard deviation is 15%. Assume that the population variance is unknown. The related t-table values are given below, where (t_ij) denotes the (100 – j)th percentile of t-distribution value with i degrees of freedom):
| t_{27,0.025} | 2.05 |
|---|---|
| t_{27,0.05} | 1.70 |
| t_{26,0.025} | 2.06 |
| t_{26,0.05} | 1.71 |
What is the 95% confidence interval for the mean monthly return?_
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TTanishq
A
[0.00181, 0.0989]
B
[-0.0084, 0.1084]
C
[-0.00811, 0.10811]
D
[0.02135, 0.07835]
Explanation:
Explanation
To calculate the 95% confidence interval for the mean monthly return when the population variance is unknown, we use the t-distribution.
Given:
- Sample size (n) = 28
- Mean return (x̄) = 5% = 0.05
- Sample standard deviation (s) = 15% = 0.15
- Degrees of freedom = n - 1 = 27
- 95% confidence level
Formula:
Confidence Interval = Mean ± (t-critical × Standard Error)
Where:
- Standard Error = s / √n
- t-critical = t_{27,0.025} = 2.05 (from the table)
Calculation:
-
Standard Error = 0.15 / √28 ≈ 0.15 / 5.2915 ≈ 0.02835
-
Margin of Error = t-critical × Standard Error = 2.05 × 0.02835 ≈ 0.05811
-
Confidence Interval = 0.05 ± 0.05811
- Lower bound = 0.05 - 0.05811 = -0.00811
- Upper bound = 0.05 + 0.05811 = 0.10811
Therefore, the 95% confidence interval is [-0.00811, 0.10811].
Why this is correct:
- We use t-distribution because population variance is unknown
- Degrees of freedom = n - 1 = 27
- For 95% confidence interval, we use t_{27,0.025} = 2.05 (two-tailed test)
- The calculation matches option C exactly
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