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?
Answers
GPT-4o mini
To determine the number of students the university should survey, we can use the formula for the sample size needed for estimating proportions, which takes into account the desired margin of error, the confidence level, and the estimated proportion of responses. The formula is:
\[
n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right)
\]
Where:
- \( n \) is the required sample size.
- \( Z \) is the Z-value corresponding to the desired confidence level.
- \( p \) is the estimated proportion of the population (if unknown, 0.5 is often used for maximum variability).
- \( E \) is the margin of error (expressed as a decimal).
### Steps:
1. **Determine the Desired Margin of Error**:
- In this case, the margin of error (E) is 1.2%, which can be expressed as:
\[
E = 0.012
\]
2. **Select a Confidence Level**:
- Common confidence levels are 90%, 95%, and 99%. For example, at a 95% confidence level, the Z-value is approximately 1.96.
3. **Estimate the Proportion (p)**:
- If no prior estimate is available, use \( p = 0.5 \) for the most conservative estimate.
4. **Plug the Values into the Formula**:
- Using \( Z = 1.96 \), \( p = 0.5 \), and \( E = 0.012 \):
\[
n = \frac{(1.96^2) \cdot (0.5) \cdot (1 - 0.5)}{(0.012^2)}
\]
5. **Calculate**:
- First, calculate \( 1.96^2 \):
\[
1.96^2 \approx 3.8416
\]
- Then calculate \( 0.5 \cdot (1 - 0.5) = 0.5 \cdot 0.5 = 0.25 \).
- Now plug these values back in:
\[
n = \frac{3.8416 \cdot 0.25}{(0.012^2)} = \frac{0.9604}{0.000144} \approx 6666.67
\]
6. **Round up to Whole Number**:
- Since the sample size must be a whole number, we round up to the nearest whole number:
\[
n \approx 6667
\]
### Conclusion:
The university should survey approximately **6667 students** to achieve a margin of error of 1.2%.
\[
n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right)
\]
Where:
- \( n \) is the required sample size.
- \( Z \) is the Z-value corresponding to the desired confidence level.
- \( p \) is the estimated proportion of the population (if unknown, 0.5 is often used for maximum variability).
- \( E \) is the margin of error (expressed as a decimal).
### Steps:
1. **Determine the Desired Margin of Error**:
- In this case, the margin of error (E) is 1.2%, which can be expressed as:
\[
E = 0.012
\]
2. **Select a Confidence Level**:
- Common confidence levels are 90%, 95%, and 99%. For example, at a 95% confidence level, the Z-value is approximately 1.96.
3. **Estimate the Proportion (p)**:
- If no prior estimate is available, use \( p = 0.5 \) for the most conservative estimate.
4. **Plug the Values into the Formula**:
- Using \( Z = 1.96 \), \( p = 0.5 \), and \( E = 0.012 \):
\[
n = \frac{(1.96^2) \cdot (0.5) \cdot (1 - 0.5)}{(0.012^2)}
\]
5. **Calculate**:
- First, calculate \( 1.96^2 \):
\[
1.96^2 \approx 3.8416
\]
- Then calculate \( 0.5 \cdot (1 - 0.5) = 0.5 \cdot 0.5 = 0.25 \).
- Now plug these values back in:
\[
n = \frac{3.8416 \cdot 0.25}{(0.012^2)} = \frac{0.9604}{0.000144} \approx 6666.67
\]
6. **Round up to Whole Number**:
- Since the sample size must be a whole number, we round up to the nearest whole number:
\[
n \approx 6667
\]
### Conclusion:
The university should survey approximately **6667 students** to achieve a margin of error of 1.2%.