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A journalist wants to determine the average annual salary of CEOs in the S&P 1,500. He does not have time to survey all 1,500 CEOs but wants to be 95% confident that his estimate is within $50,000 of the true mean. The journalist takes a preliminary sample and estimates that the standard deviation is approximately $449,300. What is the minimum number of CEOs that the journalist must survey to be within $50,000 of the true average annual salary? Remember that the z-value associated with a 95% confidence interval is 1.96.

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Answer:

Atleast 315

Step-by-step explanation:

Given that a journalist wants to determine the average annual salary of CEOs in the S&P 1,500. To save time a sample is taken.

s = sample std dev = 449300

Sample size = n

Margin of error = 50000

Z critical = 1.96

Hence 1.96*std error = 50000

std error = 25510.20

Std deviation/sqrt n = 25510.20

Simplify to get

[tex]\sqrt{n} =\frac{449300}{25510.20} =17.752\\n=315\\[/tex]

So sample size should be atleast 315

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