The issues surrounding the levels and structure of executive compensation have g

Question

The issues surrounding the levels and structure of executive compensation have gained added prominence in the wake of the financial crisis that erupted in the fall of 2008. Based on the 2006 compensation data obtained from the Securities and Exchange Commission (SEC) website, it was determined that the mean and the standard deviation of compensation for the 546 highest paid CEOs in publicly traded U.S. companies are $11.27 million and $10.68 million, respectively. An analyst randomly chooses 31 CEO compensations for 2006. Use Table 1. a. Is it necessary to apply the finite population correction factor? Yes No b. Is the sampling distribution of the sample mean approximately normally distributed? Yes No c. Calculate the expected value and the standard error of the sample mean. (Round "expected value" to 2 decimal places and "standard deviation" to 4 decimal places.) Expected value Standard error d. What is the probability that the sample mean is more than $15 million? (Round "z" value to 2 decimal places, and final answer to 4 decimal places.) Probability

Explanation / Answer

a. Is it necessary to apply the finite population correction factor?

YES, as the sample (31) is a significant percentage of the population (546).
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b. Is the sampling distribution of the sample mean approximately normally distributed?

YES, as the sample size is large enough, 31 > 30.

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c. Calculate the expected value and the standard error of the sample mean. (Round "expected value" to 2 decimal places and "standard deviation" to 4 decimal places.)

The expected value is

u = 11.27 [ANSWER, EXPECTED VALUE]

and the standard error is

sigma(X) = sigma/sqrt(n) [sqrt[(N-n)/(N-1)]]

sigma(X) = 10.68/sqrt(31)*(sqrt((546-31)/(546-1))) = 1.864643318 [ANSWER, STANDARD ERROR]

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d. What is the probability that the sample mean is more than $15 million?

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    15      
u = mean =    11.27      
          
s = standard deviation =    1.864643318      
          
Thus,          
          
z = (x - u) / s =    2      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2   ) =    0.0228 [ANSWER]

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