Final answer:
To achieve a 0.1% margin of error at a 95% confidence level for a political poll, a sample size of approximately 960,400 is required. This uses a conservative estimate for p at 0.5, and the standard z-score for 95% confidence which is 1.96.
Step-by-step explanation:
To determine the sample size needed for a political candidate's poll with a 0.1% margin of error and a 95% confidence level, we use the sample size formula for proportions: n = (Z² * p * (1 - p)) / E². Here, 'n' is the sample size, 'Z' is the z-score corresponding to the confidence level, 'p' is the estimated population proportion (best guess of the true proportion), and 'E' is the margin of error. At a 95% confidence level, the standard z-score (Z) is approximately 1.96. If we do not know the true population proportion (p), a conservative estimate is to use p = 0.5, which is the value that provides the maximum product of p*(1-p) and, hence, the largest sample size. The margin of error (E) is given as 0.001 (0.1%). Substituting the values into the formula: n = (1.96² * 0.5 * (1 - 0.5)) / 0.001² = (3.8416 * 0.25) / 0.000001 = 0.9604 / 0.000001 = 960400. Therefore, a sample size of approximately 960,400 is needed to achieve a 0.1% margin of error at a 95% confidence level.