Question

In an online poll, 350 students were asked if they prefer to learn a language using a computer program or by learning from a tutor. 39% of respondents said they preferred learning from a computer program.

What is the margin of error for this poll?

(1 point)
Responses

5.3%
5.3%

61%
61%

39%
39%

18.7%

Answer 1

To calculate the margin of error for a proportion in a poll, we can use the formula:

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

where:
- \( p \) is the sample proportion (in this case, 39% or 0.39),
- \( n \) is the sample size (350 students), and
- \( z \) is the z-score for the desired confidence level (typically 1.96 for a 95% confidence level).

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

\[
p(1-p) = 0.39(1-0.39) = 0.39 \times 0.61 = 0.2379
\]

2. Calculate the standard error:

\[
\text{Standard Error} = \sqrt{\frac{0.2379}{350}} \approx \sqrt{0.000678} \approx 0.0260
\]

3. Now multiply by the z-score (assuming 95% confidence, z ≈ 1.96):

\[
\text{Margin of Error} \approx 1.96 \times 0.0260 \approx 0.0510 \text{ or } 5.1\%
\]

Since the options are presented with slight rounding, we can round this to about 5.3%.

Thus, the margin of error for this poll is **5.3%**.