Question

political candidate has asked you to conduct a poll to determine what percentage of people support her. If the candidate only wants a 6% margin of error at a 90% confidence level, what size of sample is needed?

Answer 1

To determine the sample size needed for a poll with a 6% margin of error at a 90% confidence level, we can use the formula: n = (Z^2 * p * (1 - p)) / E^2. Using the Z-score corresponding to a 90% confidence level (Z = 1.645) and a margin of error of 6% (E = 0.06), a sample size of 377 is needed.

Step-by-step explanation:

To determine the sample size needed for a poll with a 6% margin of error at a 90% confidence level, we can use the formula:

n = (Z^2 * p * (1 - p)) / E^2

Where:

• n is the sample size
• Z is the Z-score corresponding to the desired confidence level
• p is the estimated proportion of people supporting the candidate
• E is the desired margin of error

Since we need to estimate the proportion based on the given information, we don't have an estimated proportion. In this case, we use 0.5 as a conservative estimate, which gives us the largest sample size.

Using the Z-score corresponding to a 90% confidence level (Z = 1.645) and a margin of error of 6% (E = 0.06), we can calculate the sample size as follows:

n = (1.645^2 * 0.5 * (1 - 0.5)) / 0.06^2 = 376.36

Rounding up to the nearest whole number, a sample size of 377 is needed.