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.

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?

Answer 1

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.

Answer 2

To 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, we can use the formula:

CI = p̂ ± Z * √(p̂(1-p̂)/n)

Where:
- CI represents the confidence interval
- p̂ is the sample proportion (195/350=0.5571)
- Z is the z-score corresponding to the desired confidence level (95% confidence level corresponds to a z-score of 1.96)
- n is the sample size (350)

Let's plug in the values and calculate the confidence interval:

CI = 0.5571 ± 1.96 * √((0.5571(1-0.5571))/350)

Calculating this expression gives us:

CI = 0.5571 ± 1.96 * √(0.2473/350)

Simplifying further:

CI = 0.5571 ± 1.96 * 0.02244

CI = 0.5571 ± 0.04397

Now we can calculate the lower and upper bounds of the confidence interval:

Lower bound = 0.5571 - 0.04397 = 0.5131
Upper bound = 0.5571 + 0.04397 = 0.6011

Therefore, the 95 percent confidence interval for the true proportion of all current Bank of America customers who would respond with a 9 or 10 is [0.5131, 0.6011].

To determine whether this proportion exceeds 0.48, we can compare the lower bound of the confidence interval with 0.48:

Lower bound (0.5131) > 0.48

So, we can say with 95 percent confidence that the proportion of customer delight exceeds 0.48 (the historical proportion of customer delight for Bank of America).