To determine the margin of error for a sample size of 480, we need to understand the inverse relationship between sample size and margin of error. Generally, as the sample size increases, the margin of error decreases. The margin of error is often calculated using the formula:
\[ ME = z \times \left( \frac{\sigma}{\sqrt{n}} \right) \]
Where:
- \( ME \) is the margin of error
- \( z \) is the z-score corresponding to the desired confidence level
- \( \sigma \) is the standard deviation of the population
- \( n \) is the sample size
Without specific information about the population standard deviation or the z-score for the confidence level, it is challenging to give a precise value for the margin of error. However, we can proceed by using the options you provided, focusing on the fact that the margin of error typically becomes smaller as the sample size increases.
The given response options are:
- 0.05
- 0.002
- 0.004
- 0.04
Assuming we are working with a situation with a standard deviation and z-score that makes calculations practical, a sample size of 480 leads us to expect the margin of error to be relatively small. Calculations for a sample of this size often yield a margin of error in line with the smaller numbers provided in the options (0.002, 0.004).
Given that larger sample sizes (like 480) usually effectuate smaller margins of error, the most likely candidates from the provided options are 0.002 or 0.004.
If you have any specific assumptions (like a known population standard deviation or confidence level), we can provide a more precise calculation. But in the absence of that information, the correct answer likely is 0.004, as it is a reasonable small margin of error indicating good precision.