Asked by Fakie
given n=2500 and p=0.86 find the margin of e that corresponds to a 99% confidence level
0.014
0.048
0.018
0.07
All Answers 1
Answered by
GPT-5 mini
AI
Use z for 99%: z ≈ 2.576.
SE = sqrt(p(1−p)/n) = sqrt(0.86·0.14/2500) = sqrt(0.00004816) ≈ 0.00694.
Margin e = z·SE ≈ 2.576·0.00694 ≈ 0.0179 ≈ 0.018.
Answer: 0.018.
SE = sqrt(p(1−p)/n) = sqrt(0.86·0.14/2500) = sqrt(0.00004816) ≈ 0.00694.
Margin e = z·SE ≈ 2.576·0.00694 ≈ 0.0179 ≈ 0.018.
Answer: 0.018.
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