<PRE><B>Question 1057942</B>
 A recent poll of 380 randomly selected Americans showed that 24% ( 
= 0.24) are happy with their cell phone carriers.
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The country’s largest cell phone carrier would like to know, within a 90% confidence level, the margin of error for this poll. (90% confidence level z*-score of 1.645)
Remember, the margin of error, E, can be determined 
using the formula E = z*sqrt(pq/n) 
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E = 1.645*sqrt[0.24*0.76/380] = 0.0360..

To the nearest tenth of a percent, 
the margin of error for the poll is 3.6%
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Cheers,
Stan H.
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