Question
Which sample size has a margin of error of 6%?(1 point)
Responses
277
277
200
200
277.56
277.56
16
Responses
277
277
200
200
277.56
277.56
16
Answers
GPT-4o mini
To determine which sample size corresponds to a margin of error of 6%, we typically use the margin of error formula for proportions, which is:
\[ E = Z \times \sqrt{\frac{p(1-p)}{n}} \]
Where:
- \(E\) is the margin of error,
- \(Z\) is the Z-score corresponding to the desired confidence level (for example, 1.96 for 95% confidence),
- \(p\) is the estimated proportion (often taken as 0.5 for maximum variability if the actual proportion isn't known),
- \(n\) is the sample size.
Assuming \(p = 0.5\) for maximum variability, we can plug in the values and solve for \(n\):
\[ 0.06 = Z \times \sqrt{\frac{0.5(1-0.5)}{n}} \]
To find the specific sample size that gives a margin of error of 6%, we would generally calculate it based on the Z-value corresponding to the desired confidence level. For 95% confidence level:
For a 95% confidence interval, \(Z \approx 1.96\):
1. First, square both sides and isolate \(n\):
\[
0.06^2 = (1.96)^2 \times \frac{0.5 \times 0.5}{n}
\]
2. Simplifying gives:
\[
0.0036 = 3.8416 \times \frac{0.25}{n}
\]
3. Rearranging to solve for \(n\):
\[
n = \frac{3.8416 \times 0.25}{0.0036} \approx 266.1
\]
Since none of the responses exactly match this calculated value, the closest provided option appears to be 277. Therefore, the sample size with a margin of error of 6% is **277**.
\[ E = Z \times \sqrt{\frac{p(1-p)}{n}} \]
Where:
- \(E\) is the margin of error,
- \(Z\) is the Z-score corresponding to the desired confidence level (for example, 1.96 for 95% confidence),
- \(p\) is the estimated proportion (often taken as 0.5 for maximum variability if the actual proportion isn't known),
- \(n\) is the sample size.
Assuming \(p = 0.5\) for maximum variability, we can plug in the values and solve for \(n\):
\[ 0.06 = Z \times \sqrt{\frac{0.5(1-0.5)}{n}} \]
To find the specific sample size that gives a margin of error of 6%, we would generally calculate it based on the Z-value corresponding to the desired confidence level. For 95% confidence level:
For a 95% confidence interval, \(Z \approx 1.96\):
1. First, square both sides and isolate \(n\):
\[
0.06^2 = (1.96)^2 \times \frac{0.5 \times 0.5}{n}
\]
2. Simplifying gives:
\[
0.0036 = 3.8416 \times \frac{0.25}{n}
\]
3. Rearranging to solve for \(n\):
\[
n = \frac{3.8416 \times 0.25}{0.0036} \approx 266.1
\]
Since none of the responses exactly match this calculated value, the closest provided option appears to be 277. Therefore, the sample size with a margin of error of 6% is **277**.