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 $500?

Answer 1

The planning value for the population standard deviation is $3750 and a sample size of 116 should be taken with a desired margin of error of $500.

Step-by-step explanation:

To determine the planning value for the population standard deviation, we can use the range given in the question. The range is between $30,000 and $45,000. To calculate the standard deviation, we can use the formula: Standard Deviation = (Range / 4). So, the planning value for the population standard deviation is $3750.

To find out how large a sample should be taken with a desired margin of error of $500, we can use the formula: Sample Size = (Z^2 * Population Standard Deviation^2) / Margin of Error^2. Assuming a 95% confidence interval, the Z-value is 1.96. Plugging in the values, we get: (1.96^2 * 3750^2) / 500^2 = 115.52. Therefore, a sample size of 116 should be taken.

Answer 2

To estimate the population mean annual starting salary with a 95% confidence interval and a $500 margin of error, one would estimate the standard deviation as $3,750 based on the range of salaries. Using the z-score for a 95% confidence level (1.96) and the estimated standard deviation, the calculation indicates that a sample size of 55 is necessary.

Step-by-step explanation:

To determine the sample size needed for estimating the population mean annual starting salary with a 95% confidence interval and a margin of error of $500, we will make use of the formula for the sample size of a normal distribution. This sample size (n) is calculated using the formula:

n = (Z²×σ²) / E²

where:

• Z is the Z-value from the standard normal distribution for the desired confidence level (in this case, for 95% confidence, Z is approximately 1.96).
• σ (sigma) is the population standard deviation.
• E is the desired margin of error ($500).

Since we are not given the standard deviation, we use the range of the salaries to estimate it. The planning value for the population standard deviation (σ) is often estimated as (Maximum value - Minimum value)/4. For salaries ranging from $30,000 to $45,000, this would be ($45,000 - $30,000)/4 = $3,750.

Plugging in the values:

n = (1.96²×$3,750²) / $500²

n = (3.8416×$14,062,500) / $250,000

n = 54.85

Since we can't have a fraction of a person, we would round up to the nearest whole number, which means a sample size of 55 is needed to estimate the mean starting salary with the desired margin of error.