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?
Answer by maxigirl(1)   (Show Source):
You can put this solution on YOUR website! (a) 95% confidence interval for p:
n = 350
p = 0.5571
% = 95
Standard Error, SE = p(1 - p)/n} = 0.0266
z- score = 1.9600
Width of the confidence interval = z * SE = 0.0520
Lower Limit of the confidence interval = P - width = 0.5051
Upper Limit of the confidence interval = P + width = 0.6092
The confidence interval is [0.5051, 0.6092]
(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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