Final answer:
98% confident of the estimate with a 5% margin of error without prior data, at least 543 employers must be surveyed. Increasing the confidence level to 96% for the population mean would typically increase the required sample size to maintain the same margin of error.
Step-by-step explanation:
To determine the number of randomly selected employers you must contact to create an estimate in which we are 98% confident with a margin of error of 5%, we use the formula for the sample size required for estimating a proportion:
n = (Z^2 × p × (1-p)) / E^2
Where Z is the z-score corresponding to the confidence level, p is the estimated proportion (we use 0.5 if we don't have a preliminary estimate), and E is the desired margin of error.
For a 98% confidence level, the z-score is approximately 2.33. Assuming we do not have a preliminary estimate, we use 0.5 for p to maximize the sample size. Thus, the calculation for the required sample size is:
n = (2.33^2 × 0.5 × (1-0.5)) / 0.05^2
n = (5.4289 × 0.5 × 0.5) / 0.0025
n = 1.357225 / 0.0025
n = 542.89
Since we can't survey a fraction of a person, we round up to the nearest whole number. Therefore, you must survey at least 543 employers to be 98% confident with a 5% margin of error.
If the firm decided that it needed to be at least 96 percent confident of the population mean to within one hour, the number of people the firm surveys would likely increase because a higher confidence level typically requires a larger sample size to ensure the same level of precision is maintained. This principle is evident in the given example where an increased confidence level necessitates a larger sample size or a larger error bound according to the error bound formula.