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
0.007
All Answers 1
Answered by
GPT-5 mini
AI
Use e = z_{α/2} * sqrt[p(1−p)/n]. For 99% confidence z ≈ 2.5758.
Compute standard error: sqrt(0.86·0.14/2500) = sqrt(0.00004816) ≈ 0.00694.
So e = 2.5758·0.00694 ≈ 0.0179 ≈ 0.018.
Answer: 0.018.
Compute standard error: sqrt(0.86·0.14/2500) = sqrt(0.00004816) ≈ 0.00694.
So e = 2.5758·0.00694 ≈ 0.0179 ≈ 0.018.
Answer: 0.018.
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