Question

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,000 and $45,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. What is the planning value for the population standard deviation? How large a sample should be taken if the desired margin of error is

Answer 1

To estimate the value of a bachelor's degree, we can calculate the planning value for the population standard deviation by using upper and lower salary limits and the desired confidence level. The planning value can be used to determine the sample size needed based on the desired margin of error.

Step-by-step explanation:

To calculate the planning value for the population standard deviation, we can use the formula:

Planning value = (Upper limit - Lower limit) / (2 * Z-score * Margin of error)

In this case, the upper limit is $45,000, the lower limit is $30,000, and we want a 95% confidence interval. The Z-score for a 95% confidence interval is approximately 1.96.

Let's assume we want a margin of error of $500:

Planning value = ($45,000 - $30,000) / (2 * 1.96 * $500)

Planning value = $15,000 / (2 * 1.96 * $500)

Planning value ≈ $15,000 / $980 ≈ 15.31

Therefore, the planning value for the population standard deviation is approximately 15.31.

To determine the sample size required, we can use the formula:

Sample size = (Z-score^2 * Population standard deviation^2) / (Margin of error^2)

Using the planning value we calculated earlier:

Sample size = (1.96^2 * 15.31^2) / ($500^2)

Sample size ≈ (3.8416 * 235.1761) / $250000 ≈ 915 / $250000 ≈ 3.66

Since you cannot have a fraction of a sample, you would need to round up to the nearest whole number. Therefore, a sample size of 4 would be required to achieve the desired margin of error.