Asked by Fakie
given n=2500 and p=0.86 find the margin of e that corresponds to a 99% confidence level
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GPT-5 mini
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Use the normal approximation for a proportion. Margin of error (ME) at confidence level 1 − α is
ME = z_{α/2} * sqrt(p(1−p)/n).
For 99% confidence z_{α/2} ≈ 2.5758. With p = 0.86 and n = 2500:
SE = sqrt(0.86·0.14 / 2500) = sqrt(0.00004816) ≈ 0.00694
ME = 2.5758 × 0.00694 ≈ 0.0179.
So the margin of error is about 0.0179 (≈1.79 percentage points). The 99% CI would be 0.86 ± 0.0179 → about [0.8421, 0.8779].
ME = z_{α/2} * sqrt(p(1−p)/n).
For 99% confidence z_{α/2} ≈ 2.5758. With p = 0.86 and n = 2500:
SE = sqrt(0.86·0.14 / 2500) = sqrt(0.00004816) ≈ 0.00694
ME = 2.5758 × 0.00694 ≈ 0.0179.
So the margin of error is about 0.0179 (≈1.79 percentage points). The 99% CI would be 0.86 ± 0.0179 → about [0.8421, 0.8779].
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