Suppose you want to estimate the difference between two population proportions correct to within 0.03 with probability 0.90. If prior information suggests that p1 ≈ 0.4 and p2 ≈ 0.8, and you want to
Select independent random samples of equal size from the populations, how large should the
Sample sizes be?
A) $\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 1203$
B) $n _ { 1 } = n _ { 2 } = 1925$
C) $\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 1708$
D) $\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 963$

Question

Suppose you want to estimate the difference between two population proportions correct to within 0.03 with probability 0.90. If prior information suggests that p1 ≈ 0.4 and p2 ≈ 0.8, and you want to
Select independent random samples of equal size from the populations, how large should the
Sample sizes be? 
A) $\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 1203$
B) $n _ { 1 } = n _ { 2 } = 1925$
C) $\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 1708$
D) $\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 963$

Quizplus · Accepted Answer

To estimate the difference between two population proportions with a specific margin of error (E) and confidence level, the sample size for each population can be calculated using the formula: $n = \left( \frac{Z^2 \cdot (p_1 \cdot (1 - p_1) + p_2 \cdot (1 - p_2))}{E^2} \right)$. Given p1 ≈ 0.4, p2 ≈ 0.8, E = 0.03, and a Z value corresponding to a 90% confidence level (approximately 1.645), plugging these values into the formula gives a sample size of approximately 1203 for each population.

Suppose You Want to Estimate the Difference Between Two Population n1=n2=1203\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 1203n1​=n2​=1203

Suppose You Want to Estimate the Difference Between Two Population $\mathrm { n } _ { 1 } = \mathrm { n } _ { 2 } = 1203$