A political candidate has asked you to conduct a poll to determine what percentage of people support her.
If the candidate only wants a 7% margin of error at a 90% confidence level, what size of sample is needed?
In order to calculate the required sample size, we need to consider three factors: the desired margin of error, the desired confidence level, and the estimated proportion of people who support the candidate.
To determine the sample size, we can use the formula:
Sample size = (Z^2 * p * (1 - p)) / E^2
Where:
- Z is the z-score corresponding to the desired confidence level (in this case, 90% confidence level corresponds to a z-score of 1.645).
- p is the estimated proportion of people who support the candidate, which we do not currently have.
- E is the desired margin of error, which is 7% (expressed as a decimal, so 0.07).
However, since we do not have the estimated proportion of people who support the candidate, we will assume a conservative proportion of 50% to ensure the largest possible sample size (maximizing the sample size will keep the margin of error within the desired range).
Substituting the values into the formula, we have:
Sample size = (1.645^2 * 0.5 * (1 - 0.5)) / 0.07^2
Simplifying the equation:
Sample size = (2.705025 * 0.25) / 0.0049
Sample size = 0.67625625 / 0.0049
Sample size = 138.02561224
Therefore, a sample size of approximately 138 individuals is needed to achieve a 7% margin of error at a 90% confidence level, assuming a conservative estimate of 50% support for the candidate.