Question 3
A political candidate has asked his/her assistant to conduct a poll to determine the percentage of people in the community that supports him/her. If the candidate wants a $6 \%$ margin of error at a $95 \%$ confidence level, what size of sample is needed? Be sure to round accordingly.
The candidate would need to survey people in the community in order to be within a $6 \%$ margin of error at a $95 \%$ confidence level.

Step 1 ：Translate the problem into a mathematical formula for sample size in a proportion: \(n = \frac{{Z^2 * p * (1-p)}}{{E^2}}\)

Step 2 ：Identify the given values: Z-score (Z) for a 95% confidence level is 1.96, estimated proportion of the population (p) is 0.5, and the margin of error (E) is 6% or 0.06

Step 3 ：Substitute the given values into the formula: \(n = \frac{{(1.96)^2 * 0.5 * (1-0.5)}}{{(0.06)^2}}\)

Step 4 ：Simplify the equation to find the sample size (n)

Step 5 ：Round the sample size (n) to the nearest whole number because you can't have a fraction of a person

Step 6 ：Final Answer: The candidate would need to survey at least \(\boxed{267}\) people in the community in order to be within a 6% margin of error at a 95% confidence level