Question

In a pole 1,325 people are surveyed about their preferred political candidate 38% of the respondents said they support candidate a what is the margin of error for this pole rounded to the nearest tenth of a percent

Answer 1

To calculate the margin of error for a proportion in a survey, we can use the following formula for a confidence level of approximately 95%:

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

Where:
- \( z \) is the z-score for the desired confidence level (approximately 1.96 for 95% confidence),
- \( p \) is the proportion of respondents supporting the candidate (in decimal form),
- \( n \) is the total number of respondents.

In this case:

- \( n = 1325 \)
- \( p = 0.38 \)

Now, plugging these values into the formula:

1. Calculate \( p(1-p) \):

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

2. Calculate the standard error (SE):

\[
\text{SE} = \sqrt{\frac{0.2356}{1325}} \approx \sqrt{0.0001776} \approx 0.01333
\]

3. Now, calculate the margin of error:

\[
\text{ME} = 1.96 \times 0.01333 \approx 0.0261
\]

4. Convert the margin of error to a percentage:

\[
\text{ME} \approx 0.0261 \times 100 \approx 2.61\%
\]

Finally, rounded to the nearest tenth of a percent, the margin of error is approximately:

\[
\boxed{2.6\%}
\]