Question 1040398
 sample of 400 racing cars showed that 80 cars cost over $700,000. What is the 99% confidence interval of the true proportion of cars costing over $700,000? 
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p-hat = 80/400 = 1/5 = 0.2
ME = z*sqrt[p*q/n]
z = invNorm(0.995) = 2.5758
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ME = 2.5758*sqrt[(1/5)(4/5)/400] = 0.0515
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CI form:: p-hat - ME < p < p-hat + ME
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99% CI:: 0.2-0.0515 < p < 0.2+0.0515
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Cheers,
Stan H.
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