Question
What is the correct margin of error for a sample size of 480? A. 0.05 B.0.004 C. 0.04 D. 0.002
Answers
GPT-4o mini
To determine the margin of error for a given sample size, we typically need the following information:
1. The confidence level (e.g., 90%, 95%, 99%).
2. The population proportion (if applicable, usually denoted as \( p \)).
3. The formula for margin of error (ME) depending on the context.
The general formula for the margin of error for a proportion can be derived from:
\[
ME = z \sqrt{\frac{p(1-p)}{n}}
\]
Where:
- \( z \) is the z-score corresponding to the desired confidence level.
- \( p \) is the estimated proportion (between 0 and 1).
- \( n \) is the sample size.
If you do not have the proportion or the confidence level, it's impossible to calculate the exact margin of error with just the sample size.
However, if you assume a common scenario where \( p = 0.5 \) (which gives a maximum margin of error for proportions) and a confidence level (for example, 95%) with a \( z \)-score around 1.96, then you'd calculate as follows:
1. \( n = 480 \)
2. Assuming \( p = 0.5 \):
\[
ME = 1.96 \sqrt{\frac{0.5 \times 0.5}{480}} \approx 1.96 \sqrt{\frac{0.25}{480}} \approx 1.96 \sqrt{0.00052083} \approx 1.96 \cdot 0.0228 \approx 0.0447
\]
So the margin of error would be approximately 0.045.
Among your options, the closest provided option would be **C. 0.04**.
If you have specific parameters (confidence level or proportion), please provide those to get a precise answer.
1. The confidence level (e.g., 90%, 95%, 99%).
2. The population proportion (if applicable, usually denoted as \( p \)).
3. The formula for margin of error (ME) depending on the context.
The general formula for the margin of error for a proportion can be derived from:
\[
ME = z \sqrt{\frac{p(1-p)}{n}}
\]
Where:
- \( z \) is the z-score corresponding to the desired confidence level.
- \( p \) is the estimated proportion (between 0 and 1).
- \( n \) is the sample size.
If you do not have the proportion or the confidence level, it's impossible to calculate the exact margin of error with just the sample size.
However, if you assume a common scenario where \( p = 0.5 \) (which gives a maximum margin of error for proportions) and a confidence level (for example, 95%) with a \( z \)-score around 1.96, then you'd calculate as follows:
1. \( n = 480 \)
2. Assuming \( p = 0.5 \):
\[
ME = 1.96 \sqrt{\frac{0.5 \times 0.5}{480}} \approx 1.96 \sqrt{\frac{0.25}{480}} \approx 1.96 \sqrt{0.00052083} \approx 1.96 \cdot 0.0228 \approx 0.0447
\]
So the margin of error would be approximately 0.045.
Among your options, the closest provided option would be **C. 0.04**.
If you have specific parameters (confidence level or proportion), please provide those to get a precise answer.