SOLUTION: In a random sample of 150 households with an Internet connection,33 said that they had changed their Internet service provider within the past six months. a)Finda99%confidence i

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Question 927121: In a random sample of 150 households with an Internet connection,33 said that they had changed their Internet service provider within the past six months.
a)Finda99%confidence interval for the proportion of customers who changed their Internet service provider within the past six months.
b)Find the sample size needed for a99% confidence interval to specify the proportion to within �0.03.
c)If no estimate of the proportion is available,how large should the sample be?
Found 2 solutions by stanbon, ewatrrr:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
In a random sample of 150 households with an Internet connection,33 said that they had changed their Internet service provider within the past six months.
a)Finda99%confidence interval for the proportion of customers who changed their Internet service provider within the past six months.
p-hat = 33/150 = 0.22
ME = 2.5758sqrt[0.22*0.78/150] = 0.034
99% CI:: 0.22-0.034 < p < 0.22+0.034
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b)Find the sample size needed for a99% confidence interval to specify the proportion to within �0.03.
n = [2.5758/0.03]^2*0.22*0.78 = 1266 when rounded up
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c)If no estimate of the proportion is available,how large should the sample be?
n = [2.5758/0.03]^2(1/2)^2 = 1843 when rounded up
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Cheers,
Stan H.
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Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website!
a) p = 33/150 = .22
ME = 2.576sqrt((.22*.78)/150) = .034
CI: .22 � .034
..........
b) n = .22*.78(2.576/.03)^2 = 1265.2 0r 1266 (next whole number)
......
c) If no estimate of the proportion is available, how large should the sample be.
Larger the sample, smaller the ME is. Use Your own parameters.