Question

Question In a poll, 1,325 people are surveyed about their preferred political candidate. 38% of respondents said they support candidate A. What is the margin of error for this poll, rounded to the nearest tenth of a percent? (1 point)

Answer 1

To calculate the margin of error for a poll, you can use the formula:

\[
\text{Margin of Error} = \frac{Z \times \sqrt{p(1 - p)}}{\sqrt{n}}
\]

Where:
- \( Z \) is the Z-score corresponding to the desired confidence level (for a 95% confidence level, \( Z \) is approximately 1.96).
- \( p \) is the proportion of respondents who support candidate A (38% or 0.38).
- \( n \) is the total number of respondents (1,325 in this case).

1. First, calculate \( p(1 - p) \):

\[
p(1 - p) = 0.38(1 - 0.38) = 0.38 \times 0.62 = 0.2356
\]

2. Next, calculate \( \sqrt{p(1 - p)} \):

\[
\sqrt{p(1 - p)} = \sqrt{0.2356} \approx 0.4853
\]

3. Now calculate the margin of error:

\[
\text{Margin of Error} = \frac{1.96 \times 0.4853}{\sqrt{1325}}
\]

4. Calculate \( \sqrt{1325} \):

\[
\sqrt{1325} \approx 36.4
\]

5. Now substitute back into the margin of error formula:

\[
\text{Margin of Error} = \frac{1.96 \times 0.4853}{36.4} \approx \frac{0.9509}{36.4} \approx 0.0261
\]

6. Convert to percentage:

\[
0.0261 \times 100 \approx 2.61\%
\]

7. Round to the nearest tenth of a percent:

\[
\text{Margin of Error} \approx 2.6\%
\]

Therefore, the margin of error for this poll is approximately **2.6%**.