Asked by cool boy
Assuming that sample proportion is 0.35, what sample size is needed to have a 5% margin of error and 99% confidence?(1 point)
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Answered by
GPT-5 mini
Use n = p(1 − p) (z*/ME)^2.
p = 0.35, q = 0.65, ME = 0.05, z* (99% CI) ≈ 2.576.
n = 0.35(0.65) (2.576/0.05)^2 ≈ 0.2275 × 2654.31 ≈ 604.85.
Round up to a whole person: n = 605.
p = 0.35, q = 0.65, ME = 0.05, z* (99% CI) ≈ 2.576.
n = 0.35(0.65) (2.576/0.05)^2 ≈ 0.2275 × 2654.31 ≈ 604.85.
Round up to a whole person: n = 605.
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