document.write( "Question 278054: Quality Progress, February 2005, reports on the results achieved by Bank of America in improving customer satisfaction and customer loyalty by listening to the “voice of the customer.” A key measure of customer satisfaction is the response on a scale from 1 to 10 to the question: “Considering all the business you do with Bank of America, what is your over all satisfaction with Bank of America?” Suppose that a random sample of 350 current customers results in 195 customers with a response of 9 or 10 representing “customer delight.” Find a 95 percent confidece interval for the true proportion of all current Bank of America customers who would respond with a 9 or 10. Are we 95 percent confident that this proportion exceeds .48, the historical proportion of customer delight for Bank of America? \n" ); document.write( "

Algebra.Com's Answer #364144 by maxigirl(1) $\"\"$ $\"About$

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(a) 95% confidence interval for p:
\n" ); document.write( "n = 350
\n" ); document.write( "p = 0.5571
\n" ); document.write( "% = 95
\n" ); document.write( "Standard Error, SE = p(1 - p)/n} = 0.0266
\n" ); document.write( "z- score = 1.9600
\n" ); document.write( "Width of the confidence interval = z * SE = 0.0520
\n" ); document.write( "Lower Limit of the confidence interval = P - width = 0.5051
\n" ); document.write( "Upper Limit of the confidence interval = P + width = 0.6092
\n" ); document.write( "The confidence interval is [0.5051, 0.6092]
\n" ); document.write( "(b) Yes, we can be 95% confident that p exceeds 0.48, since the entire 95% confidence interval lies above 0.48
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