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.
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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