Quality Progress, February 2005, reports on the results achieved by Bank of America in improving customer satisfaction and customer loyalty b 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 result 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?

Question

Quality Progress, February 2005, reports on the results achieved by Bank of America in improving customer satisfaction and customer loyalty b 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 result 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?

Answer 1

Find a confidence interval formula using proportions. Convert your proportions to decimals for ease in calculation. Use + or - 1.96 for the interval in the formula to represent 95 percent confidence (this is found using a z-table). You should be able to draw your conclusions from there.

Answer 2

To find a 95% confidence interval for the true proportion of all current Bank of America customers who would respond with a 9 or 10, we can use the following formula:

Confidence Interval = Sample Proportion ± Margin of Error

First, let's calculate the sample proportion:

Sample Proportion = Number of customers with a response of 9 or 10 / Total sample size

Sample Proportion = 195 / 350
Sample Proportion ≈ 0.5571

Next, let's calculate the margin of error using the formula:

Margin of Error = Z * √((Sample Proportion * (1 - Sample Proportion)) / n)

Where Z is the z-score corresponding to the desired confidence level. For a 95% confidence level, the z-score is approximately 1.96. And n is the sample size, which is 350 in this case.

Margin of Error = 1.96 * √((0.5571 * (1 - 0.5571)) / 350)
Margin of Error ≈ 0.0411

Now, we can construct the confidence interval:

Confidence Interval = Sample Proportion ± Margin of Error

Confidence Interval = 0.5571 ± 0.0411

Lower Limit = 0.5571 - 0.0411 ≈ 0.5160
Upper Limit = 0.5571 + 0.0411 ≈ 0.5982

Therefore, the 95% confidence interval for the true proportion of all current Bank of America customers who would respond with a 9 or 10 is approximately 0.5160 to 0.5982.

To determine whether we are 95% confident that this proportion exceeds 0.48, the historical proportion of customer delight for Bank of America, we can check if the lower limit of the confidence interval is greater than 0.48.

Since the lower limit of the confidence interval (0.5160) is greater than 0.48, we can conclude that we are 95% confident that the proportion of customer delight (response of 9 or 10) for Bank of America exceeds 0.48.