Problem 7.30
On February 8, 2002, the Gallup Organization released the results of a poll concerning American attitudes toward the 19th Winter Olympic Games in Salt Lake City, Utah. The poll results were based on telephone interviews with a randomly selected national sample of 1,011 adults, 18 years and older, conducted February 4-6, 2002.
a. Suppose we wish to use the poll’s results to justify the claim that more than 30 percent of Americans (18 years or older) say that figure skating is their favorite Winter Olympic event. The poll actually found that 32 percent of respondents reported that figure skating was their favorite event. If, for the sake of argument, we assume that 30 percent of Americans (18 years or older) say figure skating is their favorite event (that is, p = .3), calculate the probability of observing a sample proportion of .32 or more: that is P(p> or - .32).
b. Based on the probability you computed in part a, would you conclude that more than 30 percent of Americans (18 years or older) say that figure skating is their favorite Winter Olympic event?
z = (0.32 - 0.3)/√(0.3(1 - 0.3)/1011) = 1.3877
P(p’ > 0.32) = P(z > 1.3877) = 0.0826
(b) The claim that more than 30% of Americans say that figure skating is their favorite Winter Olympic event can neither accepted nor refuted unless we set a level of significance for the above hypothesis test. If we set α = 0.05 or anything less than 0.01, the claim can be refuted (since 0.0826 > 0.05). However, if we set α = 0.10, the claim is true (since 0.0826 < 0.10)
Problem 8.8
Recall that a bank manager has developed a new system to reduce the time customers spend waiting to be served by tellers during peak business hours. The mean waiting time during peak business hours under the current system is roughly 9 to 10 minutes. The bank manager hopes that the new system will have a mean waiting time that is less than six minutes. The mean of the sample of 100 bank customer waiting times in Table 1.8 is xbar = 5.46. If we let mew denote the mean of all possible bank customer waiting times using the new system and assume that standard deviation equals 2.47.
a. Calculate 95 percent and 99 percent confidence intervals for mew.
b. Using the 95 percent confidence interval, can the bank manager be 95 percent confident that mew is less than six minutes? Explain.
c. Using the 99 percent confidence interval, can the bank manager be 99 percent confident that mew is less than six minutes? Explain.
d. Based on your answers to parts b and c, how convinced are you that the new mean waiting time is less than six minutes?
(i) 95% CI
n = 100
x-bar = 5.46
s = 2.47
% = 95
Standard Error, SE = σ/√n = 0.2470
z- score = 1.9600
Width of the confidence interval = z * SE = 0.4841
Lower Limit of the confidence interval = x-bar - width = 4.9759
Upper Limit of the confidence interval = x-bar + width = 5.9441
The confidence interval is [4.9765.944]
(ii) 99% CI
n = 100
x-bar = 5.46
s = 2.47
% = 99
Standard Error, SE = σ/√n = 0.2470
z- score = 2.5758
Width of the confidence interval = z * SE = 0.6362
Lower Limit of the confidence interval = x-bar - width = 4.8238
Upper Limit of the confidence interval = x-bar + width = 6.0962
The confidence interval is [4.8246.096]
(b) Yes, the entire 95% CI is less than 6. The manager can be 95% confident that μ is less than 6 minutes.
(c) No, a small part of the 99% CI is greater than 6. The manager can’t be 99% confident that μ is less than 6 minutes.
(d) At α = 0.05, we are convinced that the mean waiting time is less than 6 minutes, but not at α = 0.01.
Problem 8.38
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?
p = 0.557143
% = 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 [ 0.505, 0.609]
Yes, since the entire 95% CI is greater than 0.48, we are 95% confident that this proportion exceeds 0.48.