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 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:

1. Calculate \( Z^2 \):
\[
(1.96)^2 = 3.8416
\]

2. Calculate \( p \cdot (1 - p) \):
\[
0.5 \cdot (1 - 0.5) = 0.5 \cdot 0.5 = 0.25
\]

3. Calculate \( E^2 \):
\[
(0.012)^2 = 0.000144
\]

4. 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**.