Question
A political candidate has asked you to conduct a poll to determine what percentage of people support her.
If the candidate only wants a 10% margin of error at a 95% confidence level, what size of sample is needed?
Give your answer in whole people.
If the candidate only wants a 10% margin of error at a 95% confidence level, what size of sample is needed?
Give your answer in whole people.
Answers
Answered by
GPT-4o mini
To determine the sample size needed for a poll with a specified margin of error and confidence level, we can use the formula for sample size in proportion studies. The formula is given by:
\[
n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right)
\]
Where:
- \( n \) is the sample size,
- \( Z \) is the Z-score 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.
1. **Identify the values**:
- For a 95% confidence level, the Z-score (for a two-tailed test) is approximately 1.96.
- The margin of error \( E \) is 10%, or 0.10.
- Since we do not know the population proportion, we will use \( p = 0.5 \) for maximum sample size.
2. **Plug values into the formula**:
\[
n = \left( \frac{(1.96)^2 \cdot 0.5 \cdot (1 - 0.5)}{(0.10)^2} \right)
\]
Simplifying this:
\[
n = \left( \frac{(1.96)^2 \cdot 0.5 \cdot 0.5}{0.01} \right)
\]
\[
n = \left( \frac{3.8416 \cdot 0.25}{0.01} \right)
\]
\[
n = \left( \frac{0.9604}{0.01} \right)
\]
\[
n = 96.04
\]
3. **Round up to the nearest whole person**:
Since sample size must be a whole number, we round up \( 96.04 \) to \( 97 \).
Therefore, the required sample size for the poll, rounded to whole people, is **97**.
\[
n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right)
\]
Where:
- \( n \) is the sample size,
- \( Z \) is the Z-score 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.
1. **Identify the values**:
- For a 95% confidence level, the Z-score (for a two-tailed test) is approximately 1.96.
- The margin of error \( E \) is 10%, or 0.10.
- Since we do not know the population proportion, we will use \( p = 0.5 \) for maximum sample size.
2. **Plug values into the formula**:
\[
n = \left( \frac{(1.96)^2 \cdot 0.5 \cdot (1 - 0.5)}{(0.10)^2} \right)
\]
Simplifying this:
\[
n = \left( \frac{(1.96)^2 \cdot 0.5 \cdot 0.5}{0.01} \right)
\]
\[
n = \left( \frac{3.8416 \cdot 0.25}{0.01} \right)
\]
\[
n = \left( \frac{0.9604}{0.01} \right)
\]
\[
n = 96.04
\]
3. **Round up to the nearest whole person**:
Since sample size must be a whole number, we round up \( 96.04 \) to \( 97 \).
Therefore, the required sample size for the poll, rounded to whole people, is **97**.