Use a proportional confidence interval formula.
CI95 = p + or - 1.96(√pq/n)
...where p = 195/350 (convert to a decimal), q = 1 - p, n = 350, and + or - 1.96 represents the 95% interval using a z-table.
I'll let you take it from here.
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?
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