Question

A university wants to survey students about the classes they took last semester. It wants the margin of error to be 1.2%. How many students should it survey?(1 point) students

Answer 1

To determine the sample size needed for a survey with a specified margin of error, we can use the formula for the margin of error in relation to a proportion:

\[
E = z \cdot \sqrt{\frac{p(1-p)}{n}}
\]

Where:
- \(E\) is the margin of error,
- \(z\) is the z-score corresponding to the desired confidence level,
- \(p\) is the estimated proportion of the population, and
- \(n\) is the sample size.

For a general case where we do not have prior information about \(p\) (the proportion of students who took a class), we typically use \(p = 0.5\) (which maximizes the product \(p(1-p)\)), and we assume a common confidence level of 95%, where the z-score (\(z\)) is approximately 1.96.

1. Given \(E = 0.012\) (1.2% expressed as a decimal) and \(p = 0.5\):
\[
0.012 = 1.96 \cdot \sqrt{\frac{0.5 \cdot (1 - 0.5)}{n}}
\]

2. Rearranging to solve for \(n\):
\[
0.012 = 1.96 \cdot \sqrt{\frac{0.5 \cdot 0.5}{n}} \implies 0.012 = 1.96 \cdot \sqrt{\frac{0.25}{n}}
\]
\[
0.012 = 1.96 \cdot \frac{0.5}{\sqrt{n}} \implies \sqrt{n} = \frac{1.96 \cdot 0.5}{0.012}
\]
\[
\sqrt{n} = \frac{0.98}{0.012} \approx 81.67
\]

3. Squaring both sides to find \(n\):
\[
n \approx (81.67)^2 \approx 6687.51
\]

4. Since we cannot survey a fraction of a student, we round up to the next whole number:
\[
n \approx 6688
\]

Therefore, the university should survey approximately **6688 students** to achieve a margin of error of 1.2%.