Question

You wish to estimate, with 90% confidence, the population proportion of U.S. adults who are confident in the stability of U.S. banking system. Your estimate must be accurate within 3% of the population proportion.

(a) No preliminary estimate is available. The minimum sample size needed is
.

(b) Using a prior study that found 43% of U.S. adults are confident in the stability of U.S. banking system, the minimum sample size needed is
.

Answer 1

To estimate the minimum sample size needed to estimate the population proportion with 90% confidence and an error margin of 3%, we can use the formula: n = (Z^2 * p * (1 - p)) / E^2. Since we don't have a preliminary estimate for the population proportion, we assume the worst-case scenario where p = 0.5, which gives us the largest possible sample size: n = (1.645^2 * 0.5 * (1 - 0.5)) / 0.03^2 = 752.852 ≈ 753. Therefore, the minimum sample size needed to estimate the population proportion with 90% confidence and an error margin of 3% is 753.

Step-by-step explanation:

To estimate the minimum sample size needed to estimate the population proportion with 90% confidence and an error margin of 3%, we can use the formula:

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

Where:

• n is the sample size
• Z is the Z-score for the desired confidence level (1.645 for 90% confidence)
• p is the hypothesized population proportion (we don't have it in this case)
• E is the desired error margin (0.03)

Since we don't have a preliminary estimate for the population proportion, we assume the worst-case scenario where p = 0.5, which gives us the largest possible sample size:

n = (1.645^2 * 0.5 * (1 - 0.5)) / 0.03^2 = 752.852 ≈ 753

Therefore, the minimum sample size needed to estimate the population proportion with 90% confidence and an error margin of 3% is 753.