Quality progress, february 2005, reports on the results achieved by bank of amer
ID: 2960063 • Letter: Q
Question
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 overall 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 confidence 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?Explanation / Answer
(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