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 a sample size of 480?
A. 0.04
B. 0.004
C. 0.05
D. 0.002
Which of the following is the correct margin of error for a sample size of 480?
A. 0.04
B. 0.004
C. 0.05
D. 0.002
Answers
GPT-4o mini
To determine the margin of error based on a sample size, you'll need to know the confidence level (which affects the z-value) and the standard deviation or proportion. Without that information, we can only discuss the general concept.
The margin of error (ME) is inversely related to the square root of the sample size (n) when calculating it based on sample proportions. It is often estimated using the formula:
\[
\text{ME} = z \times \sqrt{\frac{p(1-p)}{n}}
\]
Where:
- \( z \) is the z-score reflecting the desired confidence level,
- \( p \) is the sample proportion,
- \( n \) is the sample size.
As the sample size \( n \) increases, the margin of error decreases, reflecting the idea that larger samples provide a more accurate estimate of the population parameter.
If we assume a standard confidence level (like 95%) and a typical proportion (like 0.5 for maximum variability), let's illustrate this by assuming \( p = 0.5 \):
1. **Z-score for 95% confidence**: Approximately \( z = 1.96 \)
2. Calculate the standard error (SE):
\[
SE = \sqrt{\frac{0.5(1-0.5)}{480}} = \sqrt{\frac{0.25}{480}} \approx \sqrt{0.00052083} \approx 0.0228
\]
3. Now calculate the margin of error:
\[
ME = 1.96 \times 0.0228 \approx 0.0447 \text{ (or roughly 0.045)}
\]
Given these calculations, the closest match would be:
**A. 0.04**
Keep in mind that without specific details (like the proportion or confidence level), this is a rough estimate. However, based on the options provided, **A. 0.04** seems to be the best choice for a sample size of 480 under typical assumptions.
The margin of error (ME) is inversely related to the square root of the sample size (n) when calculating it based on sample proportions. It is often estimated using the formula:
\[
\text{ME} = z \times \sqrt{\frac{p(1-p)}{n}}
\]
Where:
- \( z \) is the z-score reflecting the desired confidence level,
- \( p \) is the sample proportion,
- \( n \) is the sample size.
As the sample size \( n \) increases, the margin of error decreases, reflecting the idea that larger samples provide a more accurate estimate of the population parameter.
If we assume a standard confidence level (like 95%) and a typical proportion (like 0.5 for maximum variability), let's illustrate this by assuming \( p = 0.5 \):
1. **Z-score for 95% confidence**: Approximately \( z = 1.96 \)
2. Calculate the standard error (SE):
\[
SE = \sqrt{\frac{0.5(1-0.5)}{480}} = \sqrt{\frac{0.25}{480}} \approx \sqrt{0.00052083} \approx 0.0228
\]
3. Now calculate the margin of error:
\[
ME = 1.96 \times 0.0228 \approx 0.0447 \text{ (or roughly 0.045)}
\]
Given these calculations, the closest match would be:
**A. 0.04**
Keep in mind that without specific details (like the proportion or confidence level), this is a rough estimate. However, based on the options provided, **A. 0.04** seems to be the best choice for a sample size of 480 under typical assumptions.