To calculate the margin of error for a poll, we can use the formula:
\[ \text{Margin of Error} = \frac{1}{\sqrt{n}} \times Z \]
where:
- \( n \) is the sample size,
- \( Z \) is the Z-score which corresponds to the desired confidence level (for a typical 95% confidence level, \( Z \approx 1.96 \)).
Given:
- \( n = 1,325 \)
Let's calculate the standard error first:
\[ \text{Standard Error (SE)} = \sqrt{\frac{p(1-p)}{n}} \]
where \( p \) is the proportion of support for candidate A. In this case, \( p = 0.38 \).
Now, we can calculate \( p(1-p) \):
\[ p(1 - p) = 0.38 \times (1 - 0.38) = 0.38 \times 0.62 = 0.2356 \]
Next, we calculate the standard error:
\[ SE = \sqrt{\frac{0.2356}{1,325}} \approx \sqrt{0.00017743} \approx 0.01333 \]
Now, we will find the margin of error at the 95% confidence level using the Z-score of 1.96:
\[ \text{Margin of Error} = 1.96 \times SE \approx 1.96 \times 0.01333 \approx 0.0261 \]
To express this as a percentage, we multiply by 100:
\[ \text{Margin of Error as a percentage} = 0.0261 \times 100 \approx 2.61% \]
Finally, rounding to the nearest tenth of a percent, we get:
\[ \text{Margin of Error} \approx 2.6% \]
Thus, the margin of error for this poll is 2.6%.