Question

How large a sample is required to estimate the tion of Americans over 33 who smoke at the 90% confidence level with an error of at most 0.02?

a) 2304
b) 2401
c) 2500
d) 2601

Answer 1

To estimate the proportion of Americans over 33 who smoke at the 90% confidence level with an error of at most 0.02, the required sample size can be calculated using the formula N = (Z^2 * p * (1-p)) / E^2, where N is the required sample size, Z is the Z-score, p is the estimated proportion, and E is the desired error tolerance. By substituting the values into the formula, the minimum sample size required is 2401.

Step-by-step explanation:

To estimate the proportion of Americans over 33 who smoke at the 90% confidence level with an error of at most 0.02, we can use the formula:

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

Where:

1. N is the required sample size
2. Z is the Z-score corresponding to the desired confidence level
3. p is the estimated proportion of Americans over 33 who smoke
4. E is the desired error tolerance

Since the confidence interval is not provided, we can assume that it is symmetrical, and the proportion estimate is a conservative estimate. Therefore, p = 0.5. The Z-score corresponding to a 90% confidence level is approximately 1.645. Substituting these values into the formula, we get:

N = (1.645^2 * 0.5 * (1-0.5)) / 0.02^2

N = 2401

Therefore, the minimum sample size required to estimate the proportion of Americans over 33 who smoke at the 90% confidence level with an error of at most 0.02 is 2401. The correct answer is option b) 2401.