A political candidate has asked you to conduct a poll to determine what percentage of people support her.
If the candidate only wants a 4% margin of error at a 99% confidence level, what size of sample is needed?
To determine the sample size needed for a given margin of error and confidence level, you can use the formula:
n = (Z^2 * P * (1-P)) / E^2
Where:
n = sample size
Z = z-value, based on the desired confidence level
P = estimated proportion of support for the candidate (0.5 assumes 50% support to get the maximum sample size)
E = desired margin of error, proportion (as a decimal)
In this case, the candidate wants a 4% margin of error at a 99% confidence level. First, let's calculate the z-value for a 99% confidence level.
The z-value can be determined using a table or calculator. For a 99% confidence level, the z-value is approximately 2.576.
Next, plug the values into the formula:
n = (2.576^2 * 0.5 * (1-0.5)) / 0.04^2
Simplifying the equation:
n = (6.646176 * 0.25) / 0.0016
n = 1.661544 / 0.0016
n ≈ 1038
Therefore, a sample size of approximately 1038 people is needed to achieve a 4% margin of error at a 99% confidence level.