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)
The margin of error is
%
.

Answer 1

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%**.