Financial Risk Manager Part 1

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The mean return of a sample of 28 BB+ corporate bonds is 7.5%, and the sample's standard deviation is 14%. Assuming that the population is normally distributed and the population variance is unknown, what is the 95% confidence interval for the population mean?

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TTanishq

[2.77%; 12.23%]

[2.07%; 12.93%]

[2.93%; 11.43%]

[4.12%; 13.3%]

Explanation:

Explanation

Since the population variance is unknown and the population is normally distributed, and the sample size is less than 30, we use a t-statistic. The t-statistic for a 95% confidence interval and 27 degrees of freedom (df = n-1 = 28-1 = 27) is 2.052.

Step 1: Calculate the Standard Error

\text{Standard Error} = \frac{\text{Sample Standard Deviation}}{\sqrt{\text{Sample Size}}} = \frac{14}{\sqrt{28}} = \frac{14}{5.2915} \approx 2.646

Step 2: Calculate the Margin of Error

\text{Margin of Error} = t_{\alpha/2, df} \times \text{Standard Error} = 2.052 \times 2.646 \approx 5.43

Step 3: Calculate the Confidence Interval

\text{Lower Bound} = \text{Sample Mean} - \text{Margin of Error} = 7.5 - 5.43 = 2.07\%

\text{Upper Bound} = \text{Sample Mean} + \text{Margin of Error} = 7.5 + 5.43 = 12.93\%

Therefore, the 95% confidence interval is [2.07%; 12.93%].

Key Points:

t-distribution is used when population variance is unknown and sample size is small (n ≤ 30)
Degrees of freedom = n - 1 = 27
t-critical value for 95% confidence with 27 df is 2.052
Using z-statistic instead would give incorrect results for small samples with unknown population variance

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