To calculate the minimum size of a simple random sample assuming a sampling error of 5% and a population size of 100 elements, we can use the following formula:
n = (z^2 * p * (1 - p)) / E^2
where n is the minimum sample size, z is the z-score that corresponds to the desired confidence level, p is the estimated proportion of the population, and E is the desired margin of error.
Since the population size is small (100), we can use the finite population correction factor to adjust the formula:
n = (N * z^2 * p * (1 - p)) / (N * E^2 + z^2 * p * (1 - p))
where N is the population size.
Assuming a sampling error of 5%, we have E = 0.05. Since we don’t have any information about the population proportion, we can assume a conservative estimate of p = 0.5.
To find the z-score that corresponds to a 95% confidence level, we can use a standard normal distribution table or calculator. The z-score for a 95% confidence level is approximately 1.96 1.
Substituting the values into the formula, we get:
n = (100 * 1.96^2 * 0.5 * (1 - 0.5)) / (100 * 0.05^2 + 1.96^2 * 0.5 * (1 - 0.5))
= 74.24
Therefore, the minimum size of a simple random sample assuming a sampling error of 5% and a population size of 100 elements is 74.