Question
Explore the inverse relationship between sample size and margin of error and answer the question: Which of the following is the correct margin of error for the sample size of 480?(1 point)
Responses
0.05
0.05
0.002
0.002
0.004
0.004
0.04
Responses
0.05
0.05
0.002
0.002
0.004
0.004
0.04
Answers
GPT-4o mini
The margin of error in statistics is inversely related to the sample size. As the sample size increases, the margin of error decreases. To determine the correct margin of error for a specific sample size, you typically would use a formula that takes into account the desired confidence level, population standard deviation (if known), and the sample size.
Without more specific information regarding the confidence level and the population standard deviation, it's impossible to calculate an exact margin of error. However, in many standard situations for large sample sizes, the margin of error can sometimes be estimated using the formula:
\[ \text{Margin of Error} = \frac{Z \cdot \sigma}{\sqrt{n}} \]
where \( Z \) is the Z-score corresponding to the confidence level, \( \sigma \) is the standard deviation, and \( n \) is the sample size.
To properly answer your question about which of the provided options is correct for a sample size of 480, we need additional context on the confidence level or standard deviation. If the options are without any context, normally you might see smaller margin of errors associated with larger sample sizes.
Given the options:
- 0.05
- 0.002
- 0.004
- 0.04
The smallest values (0.002 or 0.004) could be considered more realistic for a larger sample size of 480, depending on the context. However, without additional information, we can't definitively state which margin of error applies to your specific case.
If you are engaging with a known scenario or standard values commonly used in a problem set, please provide that context for a more precise answer.
Without more specific information regarding the confidence level and the population standard deviation, it's impossible to calculate an exact margin of error. However, in many standard situations for large sample sizes, the margin of error can sometimes be estimated using the formula:
\[ \text{Margin of Error} = \frac{Z \cdot \sigma}{\sqrt{n}} \]
where \( Z \) is the Z-score corresponding to the confidence level, \( \sigma \) is the standard deviation, and \( n \) is the sample size.
To properly answer your question about which of the provided options is correct for a sample size of 480, we need additional context on the confidence level or standard deviation. If the options are without any context, normally you might see smaller margin of errors associated with larger sample sizes.
Given the options:
- 0.05
- 0.002
- 0.004
- 0.04
The smallest values (0.002 or 0.004) could be considered more realistic for a larger sample size of 480, depending on the context. However, without additional information, we can't definitively state which margin of error applies to your specific case.
If you are engaging with a known scenario or standard values commonly used in a problem set, please provide that context for a more precise answer.