To determine the required sample size for a survey with a specified margin of error, you can use the formula for the sample size in relation to proportions:
\[ n = \left(\frac{Z^2 \cdot p \cdot (1 - p)}{E^2}\right) \]
Where:
- \( n \) = required sample size
- \( Z \) = Z-value (the number of standard deviations from the mean, which corresponds to the confidence level)
- \( p \) = estimated proportion of the population (assuming 0.5 for maximum variability if unknown)
- \( E \) = margin of error (in decimal form)
Given:
- Margin of Error (E) = 1.2% = 0.012
- Assume \( p = 0.5 \) (which provides the maximum sample size)
- The Z-value for a 95% confidence level (common level) is approximately 1.96.
Now plug in the values:
\[ n = \left(\frac{(1.96)^2 \cdot 0.5 \cdot (1 - 0.5)}{(0.012)^2}\right) \]
Calculating step by step:
-
Calculate \( Z^2 \): \[ (1.96)^2 = 3.8416 \]
-
Calculate \( p \cdot (1 - p) \): \[ 0.5 \cdot (1 - 0.5) = 0.5 \cdot 0.5 = 0.25 \]
-
Calculate \( E^2 \): \[ (0.012)^2 = 0.000144 \]
-
Now plug into the formula: \[ n = \left(\frac{3.8416 \cdot 0.25}{0.000144}\right) \]
\[ n = \left(\frac{0.9604}{0.000144}\right) \approx 6667.36 \]
Since you cannot survey a fraction of a person, round up to the nearest whole number:
\[ n \approx 6668 \]
Conclusion:
The university should survey approximately 6668 students.