Final answer:
The margin of error at a 95% confidence level for a sample of 324 items from a population with a variance of 529 is approximately 2.52.
Step-by-step explanation:
When determining the margin of error at a 95% confidence level from a population with a variance of 529, we first need to consider the standard deviation of the population. Since the variance is the square of the standard deviation, we take the square root of 529 to find that the standard deviation (σ) is 23.
For a sample of 324 items, to find the margin of error at a confidence level of 95%, we utilize the z-score associated with a 95% confidence interval, which is 1.96. The formula for the margin of error (ME) is ME = z * (σ/√n), where n is the sample size.
Plugging in the values we have: ME = 1.96 * (23/√324), we find that ME is approximately 2.52.
Therefore, for a sample of 324 items from a population with a variance of 529, the margin of error at a 95% confidence level is approximately 2.52.