If n=500 and ˆ p (p-hat) =0.49, find the margin of error at a 99% confidence level.

Answer 1

The margin of error for a sample proportion p is ME = z√[p(1-p)/n]

p = 0.49, z = 2.576 (for 99% confidence level), and n = 500:

ME = z√[p(1-p)/n] = 2.576√[(0.49)(1-0.49)/500] ≈ 0.0576

Thus, the margin of error is about 0.0576

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