Given the information, we need to calculate the expected value and the standard error of the sample mean for a random sample of 46 CEO compensations.

1. Expected Value (Mean of the Sample Mean):

The expected value of the sample mean is equal to the population mean. Since the mean of the population is given as:

$\mu = 11.81 \text{ million}$

The expected value of the sample mean is also:

$E(\bar{X}) = 11.81 \text{ million}$

2. Standard Error of the Sample Mean (SE):

The standard error of the sample mean is calculated using the formula:

$SE(\bar{X}) = \frac{\sigma}{\sqrt{n}}$

Where:

• $\sigma = 11.19 \text{ million}$ (population standard deviation),
• $n = 46$ (sample size).

Substituting the values:

$SE(\bar{X}) = \frac{11.19}{\sqrt{46}} \approx \frac{11.19}{6.78} \approx 1.65 \text{ million}$

Final Answer:

• Expected Value of the Sample Mean: $11.81 \text{ million}$
• Standard Error of the Sample Mean: $1.65 \text{ million}$

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Would you like more details or have any questions?

Here are 5 related questions:

1. How does the sample size affect the standard error of the sample mean?
2. What is the significance of the standard error in the context of sampling?
3. How would the expected value change if the sample size were increased?
4. Why is the standard deviation of the population used instead of the sample in this calculation?
5. How does the central limit theorem relate to the sample mean in large populations?

Tip: Increasing the sample size reduces the standard error, leading to more precise estimates of the population mean.