To determine the correct margin of error for a sample size of 480, we need to understand the relationship between sample size and margin of error. Generally, as the sample size increases, the margin of error decreases.
The margin of error (E) can be estimated using the formula:
\[ E = z \times \frac{\sigma}{\sqrt{n}} \]
Where:
- \( E \) is the margin of error,
- \( z \) is the z-score associated with the desired confidence level (for example, 1.96 for 95% confidence),
- \( \sigma \) is the standard deviation of the population,
- \( n \) is the sample size.
Unfortunately, in the question, we don't have information regarding the standard deviation or the confidence level. Thus, without these values, we cannot derive the exact margin of error for the sample size of 480.
However, if we were to rank the given options based on common expectations, larger sample sizes would result in smaller margins of error. Therefore, we might anticipate that the smaller values in the options would be more likely correct.
Based on probability and standard practices in statistics, among the provided choices:
- 0.004
- 0.002
- 0.05
- 0.04
The values of 0.004 and 0.002 are particularly small, and given that 480 is a reasonable sample size, it may suggest a smaller margin of error due to the inverse relationship.
We can't definitively say which of the provided options is the correct answer without additional information, but typically, with a sample size of 480, a smaller margin of error is anticipated. If you have further context or specifications related to the population standard deviation or the confidence interval, please provide it for a more precise calculation.
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