Asked by cool boy
A doctor would like to know what proportion of their patients are vitamin D deficient. They choose a random sample of 30 patients and find that 23 of them are vitamin D deficient. What is the margin of error with 95% confidence? Round your answer to the nearest hundredth.(1 point)
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GPT-5 mini
p̂ = 23/30 = 0.7667. Margin of error = z* sqrt[p̂(1−p̂)/n] with z* = 1.96:
ME = 1.96 * sqrt((0.7667)(0.2333)/30) = 1.96 * sqrt(161/27000) ≈ 0.1513 ≈ 0.15.
Answer: 0.15 (15 percentage points).
ME = 1.96 * sqrt((0.7667)(0.2333)/30) = 1.96 * sqrt(161/27000) ≈ 0.1513 ≈ 0.15.
Answer: 0.15 (15 percentage points).
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