Question
A university wants to survey students about the classes they took last semester. It wants the margin of error to be 1.2%. Assume the standard deviation of 1, how many students should it survey?
Answers
GPT-4o mini
To determine how many students the university should survey to achieve a margin of error of 1.2%, we can use the formula for the sample size needed for estimating a population proportion. We will apply the formula used for a normal distribution given a known standard deviation.
The formula for the sample size (n) is given by:
\[
n = \left( \frac{Z \cdot \sigma}{E} \right)^2
\]
Where:
- \( Z \) is the z-value corresponding to the desired confidence level,
- \( \sigma \) is the standard deviation,
- \( E \) is the margin of error.
Assuming a common confidence level of 95%, the z-value is approximately 1.96. Given:
- \( \sigma = 1 \)
- \( E = 0.012 \) (which is 1.2% expressed as a decimal)
We can substitute these values into the formula:
\[
n = \left( \frac{1.96 \cdot 1}{0.012} \right)^2
\]
Calculating the value in the parentheses:
\[
\frac{1.96}{0.012} \approx 163.33
\]
Now squaring that value:
\[
n \approx (163.33)^2 \approx 26667.11
\]
Since we can’t survey a fraction of a person, we round up to the nearest whole number. Therefore, the required sample size is:
\[
n \approx 26668
\]
Thus, the university should survey **26,668 students** to achieve a margin of error of 1.2%.
The formula for the sample size (n) is given by:
\[
n = \left( \frac{Z \cdot \sigma}{E} \right)^2
\]
Where:
- \( Z \) is the z-value corresponding to the desired confidence level,
- \( \sigma \) is the standard deviation,
- \( E \) is the margin of error.
Assuming a common confidence level of 95%, the z-value is approximately 1.96. Given:
- \( \sigma = 1 \)
- \( E = 0.012 \) (which is 1.2% expressed as a decimal)
We can substitute these values into the formula:
\[
n = \left( \frac{1.96 \cdot 1}{0.012} \right)^2
\]
Calculating the value in the parentheses:
\[
\frac{1.96}{0.012} \approx 163.33
\]
Now squaring that value:
\[
n \approx (163.33)^2 \approx 26667.11
\]
Since we can’t survey a fraction of a person, we round up to the nearest whole number. Therefore, the required sample size is:
\[
n \approx 26668
\]
Thus, the university should survey **26,668 students** to achieve a margin of error of 1.2%.