The margin of error formula for a proportion can be expressed as:
Margin of Error = Z * sqrt((p*(1-p))/n)
Where Z is the z-score corresponding to the level of confidence, p is the estimated proportion, and n is the sample size.
Given that the margin of error is 2.5% and we want a 95% confidence interval, the z-score for a 95% confidence interval is approximately 1.96.
Using the formula for margin of error, we can set up the equation:
0.025 = 1.96 * sqrt((0.3*(1-0.3))/n)
0.025 = 1.96 * sqrt((0.21)/n)
0.025 = 1.96 * sqrt(0.21/n)
(0.025 / 1.96) = sqrt(0.21/n)
(0.0127551)^2 = 0.21/n
0.000162^2 = 0.21/n
0.000000026244 = 0.21/n
n = 0.21 / 0.0000000026244
n ≈ 8015.64
Rounding up to the nearest whole number, the minimum sample size needed is 8,016.
None of the answer choices provided match this result, so the closest answer choice is 1,536 (D), which is the closest to 8,016.
We guess, based on historical data, that 30% of graduating high-school seniors in a large city will have completed a first-year calculus course. What's the minimum sample size needed to construct a 95% confidence interval for a proportion with a margin of error of 2.5%?
A. 323
B. 1, 291
OC. 1, 290
D. 1,536
ΟΕ. 1,537
1 answer