Question 1042124: A random sample of 250 people of a city was taken. 180 of the people of the sample told that their major source of news is the television. Construct a 99% confidence interval of the proportion of people of that particular city has the television as their major source of the news.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A random sample of 250 people of a city was taken. 180 of the people of the sample told that their major source of news is the television. Construct a 99% confidence interval of the proportion of people of that particular city has the television as their major source of the news.
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p-hat = 180/250 = 0.72
ME = z*sqrt[p*q/n]
Since you want 99%CI, you want the z-value whose tails
add to 1%.
Find the z value with a left tail of 99.5% or 0.995
Using my TI 84 I get: invNorm(0.995) = 2.5758
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Since p = 0.72, q = 1-0.72 = 0.28
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So, Margin of Error = ME = 2.5758*sqrt[0.72*0.28/250] = 0.073
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99% CI:: 0.72-0.073 < p < 0.72+0.073
99% CI:: 0.647 < p < 0.793
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Cheers,
Stan H.
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