Your company wants to conduct a salary survey for the position of Accounting Assistant II. The estimated salary range is $4,500, the margin of error should be $100 and the desired level of confidence is 95%. What is the optimal sample size for this salary survey?
You shouldn't be trying to get answers to the final here!
n = [(z*sigma)/E]² = [1.96*$4,500)/$100]² = ($77,792,400 / $10,000) = 7,779.24 = 7,780
To calculate the optimal sample size for the salary survey, we need to use a formula called the sample size formula for estimating a population mean. The formula is as follows:
n = (Z * σ / E)^2
Where:
n = sample size
Z = z-score corresponding to the desired level of confidence
σ = population standard deviation (unknown in this case)
E = margin of error
In this case, the desired level of confidence is 95%, which corresponds to a z-score of approximately 1.96. The margin of error is $100. However, we don't have information about the population standard deviation (σ), so we'll use a conservative estimate for this value. A commonly used estimate is to assume a population standard deviation equal to the range of the estimated salary, which in this case is $4,500.
n = (1.96 * 4,500 / 100)^2
n = (8820 / 100)^2
n = 88.2^2
n ≈ 7771
Therefore, the optimal sample size for this salary survey is approximately 7771.