<p>8.38</p> <p>Quality Progress, February 2005, reports on the results achieved

Question

<p>8.38</p>
<p>Quality Progress, February 2005, reports on the results achieved by Bank of America in improving customer satisfaction and customer loyalty by listening to the &#8220;voice of the customer.&#8221; A key measure of customer satisfaction is the response on a scale from 1 to 10 to the question: &#8220;Considering all the business you do with Bank of America, what is your overall satisfaction with Bank of America?</p>
<p>&#8221;2 Sup- pose that a random sample of 350 current customers results in 195 customers with a response of 9 or 10 representing &#8220;customer delight.&#8221; 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?</p>

Explanation / Answer

p = 195/350 z(95%) = 1.96 (check standard normal table) The interval goes from: p - z*sqrt(p*(1-p)/N) to p - z*sqrt(p*(1-p)/N) 195/350 - 1.96*sqrt(195/350*(1-195/350)/350) to 195/350 + 1.96*sqrt(195/350*(1-195/350)/350) With a calculator: 0.505 to 0.609 Yes, we are 95% confident that the proportion exceeds 48%, since the entire interval is over 0.48.

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