A political candidate has asked you to conduct a poll to determine what percentage of people support her.
If the candidate only wants a 8% margin of error at a 95% confidence level, what size of sample is needed?
To determine the sample size needed for a poll with a specified margin of error and confidence level, we can use a formula:
\[ n = \left(\frac{Z \cdot \sigma}{E}\right)^2 \]
Where:
- \( n \) is the sample size needed
- \( Z \) is the Z-score corresponding to the desired confidence level (in this case, 95% confidence level corresponds to a Z-score of approximately 1.96)
- \( \sigma \) is the standard deviation or anticipated variability in the population (since this value is unknown, we typically use 0.5 as a conservative estimate when estimating support percentages for binary questions)
- \( E \) is the desired margin of error (in this case, 8%)
Plugging in the values into the formula, we get:
\[ n = \left(\frac{1.96 \cdot 0.5}{0.08}\right)^2 \]
Simplifying the equation, we have:
\[ n = \left(\frac{1.96 \cdot 0.5}{0.08}\right)^2 = \left(\frac{0.98}{0.08}\right)^2 = \left(12.25\right)^2 \approx 150 \]
Therefore, a sample size of approximately 150 people would be needed in order to achieve an 8% margin of error at a 95% confidence level.