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.
-
Given \(E = 0.012\) (1.2% expressed as a decimal) and \(p = 0.5\): \[ 0.012 = 1.96 \cdot \sqrt{\frac{0.5 \cdot (1 - 0.5)}{n}} \]
-
Rearranging to solve for \(n\): \[ 0.012 = 1.96 \cdot \sqrt{\frac{0.5 \cdot 0.5}{n}} \implies 0.012 = 1.96 \cdot \sqrt{\frac{0.25}{n}} \] \[ 0.012 = 1.96 \cdot \frac{0.5}{\sqrt{n}} \implies \sqrt{n} = \frac{1.96 \cdot 0.5}{0.012} \] \[ \sqrt{n} = \frac{0.98}{0.012} \approx 81.67 \]
-
Squaring both sides to find \(n\): \[ n \approx (81.67)^2 \approx 6687.51 \]
-
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%.