Final answer:
The required sample size for a 90% confidence interval with a margin of error of 0.06 and an estimate of 0.18 for P is calculated using the margin of error formula. The Z value for a 90% confidence interval is 1.645, which is then used to solve for n, the sample size.
Step-by-step explanation:
To determine the required sample size for a 90% confidence interval with a margin of error of 0.06, using the estimate of 0.18 for P, we can use the formula for the margin of error in a proportion:
E = Z * sqrt[P(1-P)/n]
Where E is the margin of error, Z is the z-score corresponding to the desired confidence level, P is the estimated proportion, and n is the sample size.
For a 90% confidence level, the Z value is approximately 1.645. Plugging the given values into the formula:
0.06 = 1.645 * sqrt[0.18(1-0.18)/n]
Squaring both sides and rearranging for n, we get:
n = [1.645^2 * 0.18(0.82)] / 0.06^2
This calculation will give us the minimum sample size needed to estimate the population proportion with the given margin of error and confidence level. You would need to solve for n to find the numeric answer.
Alternatively, statistical software or a confidence interval calculator can be used to make this computation, accounting for the sample size recommendations