Question 420668: A survey of 4,581 U.S. households that owned a mobile phone found that 58 percent are satisfied with the coverage of their cellular phone provider. (a) Assuming that this was a random sample, construct a 90 percent confidence interval for the true proportion of satisfied U.S. mobile phone owners. (b) Why is the confidence interval so narrow?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A survey of 4,581 U.S. households that owned a mobile phone found that 58 percent are satisfied with the coverage of their cellular phone provider.
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(a) Assuming that this was a random sample, construct a 90 percent confidence interval for the true proportion of satisfied U.S. mobile phone owners.
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p-hat = 0.58
ME = 1.645*sqrt[0.58*0.42/4581] = 0.0987
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90%CI: 0.58-0.0987< p < 0.58+0.0987
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90%CI: 0.4813 < p < 0.6787
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(b) Why is the confidence interval so narrow?
Its width is 2ME and ME is small because the sample size is large.
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Cheers,
Stan H.
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