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 (to the nearest whole number)?



How large a sample should be taken if the desired margin of error is as shown below (to the nearest whole number)?
a. $500?
b. $200?
c. $100?

To find the planning value for the population standard deviation, we need to use the range provided for the starting salaries. Since the range is given as $30,000 to $45,000, we can calculate the estimate for the standard deviation by taking the difference between the two values and dividing it by 4 (since it is a 95% confidence interval).

The range is $45,000 - $30,000 = $15,000
The estimate for the standard deviation is $15,000 / 4 = $3,750

So, the planning value for the population standard deviation is $3,750 (to the nearest whole number).

To calculate the sample size required for the desired margin of error, we need to know the formula:
n = (Z * σ / E)^2

Where:
n = sample size
Z = Z-score (corresponding to the desired confidence level)
σ = population standard deviation
E = desired margin of error

a) For a desired margin of error of $500:
Using the Z-score for a 95% confidence level, which is approximately 1.96, and the planning value for the population standard deviation of $3,750, we can calculate the sample size:
n = (1.96 * 3,750 / 500)^2 = 293.03

Therefore, a sample size of approximately 293 should be taken if the desired margin of error is $500.

b) For a desired margin of error of $200:
Using the same Z-score and standard deviation, we can calculate the sample size as follows:
n = (1.96 * 3,750 / 200)^2 = 1,814.5

Therefore, a sample size of approximately 1,815 should be taken if the desired margin of error is $200.

c) For a desired margin of error of $100:
Using the same values, we can calculate the sample size:
n = (1.96 * 3,750 / 100)^2 = 27,738

Therefore, a sample size of approximately 27,738 should be taken if the desired margin of error is $100.