Question
A university wants to survey students but the classes they took last semester it once the margin of error to be 1.2% how many students should it survey?
Answers
GPT-4o mini
To determine how many students the university should survey in order to achieve a margin of error of 1.2%, you can use the formula for the sample size in a proportion estimate, which is given by:
\[
n = \left(\frac{Z^2 \cdot p \cdot (1 - p)}{E^2}\right)
\]
Where:
- \( n \) = required sample size
- \( Z \) = Z-score corresponding to the desired confidence level
- \( p \) = estimated proportion of the attribute present in the population (if unknown, use 0.5 for maximum variability)
- \( E \) = margin of error (in decimal form)
1. **Choose a Confidence Level and Find the Z-score**
Assuming a confidence level of 95%, the Z-score is approximately 1.96.
2. **Estimate Proportion (p)**
If you don't have a prior estimate of the proportion of students taking classes, use \( p = 0.5 \) for maximum variability.
3. **Convert Margin of Error to Decimal**
The margin of error \( E \) is 1.2%, which we convert to decimal form:
\[ E = 0.012 \]
4. **Plug Values into the Formula**
Now, substituting the values into the formula:
\[
n = \left(\frac{(1.96)^2 \cdot 0.5 \cdot (1 - 0.5)}{(0.012)^2}\right)
\]
Calculating:
\[
n = \left(\frac{(1.96)^2 \cdot 0.5 \cdot 0.5}{(0.012)^2}\right)
\]
\[
= \left(\frac{3.8416 \cdot 0.25}{0.000144}\right)
\]
\[
= \left(\frac{0.9604}{0.000144}\right)
\]
\[
= 6676.39
\]
Since the number of surveyed students must be a whole number, you would round up to 6677.
Therefore, the university should survey **6677 students** to achieve a margin of error of 1.2%.
\[
n = \left(\frac{Z^2 \cdot p \cdot (1 - p)}{E^2}\right)
\]
Where:
- \( n \) = required sample size
- \( Z \) = Z-score corresponding to the desired confidence level
- \( p \) = estimated proportion of the attribute present in the population (if unknown, use 0.5 for maximum variability)
- \( E \) = margin of error (in decimal form)
1. **Choose a Confidence Level and Find the Z-score**
Assuming a confidence level of 95%, the Z-score is approximately 1.96.
2. **Estimate Proportion (p)**
If you don't have a prior estimate of the proportion of students taking classes, use \( p = 0.5 \) for maximum variability.
3. **Convert Margin of Error to Decimal**
The margin of error \( E \) is 1.2%, which we convert to decimal form:
\[ E = 0.012 \]
4. **Plug Values into the Formula**
Now, substituting the values into the formula:
\[
n = \left(\frac{(1.96)^2 \cdot 0.5 \cdot (1 - 0.5)}{(0.012)^2}\right)
\]
Calculating:
\[
n = \left(\frac{(1.96)^2 \cdot 0.5 \cdot 0.5}{(0.012)^2}\right)
\]
\[
= \left(\frac{3.8416 \cdot 0.25}{0.000144}\right)
\]
\[
= \left(\frac{0.9604}{0.000144}\right)
\]
\[
= 6676.39
\]
Since the number of surveyed students must be a whole number, you would round up to 6677.
Therefore, the university should survey **6677 students** to achieve a margin of error of 1.2%.