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Ashford BUS 308 Week 3 statistics for managers

Week 3 Assignment
Chapter 7: 7.11, 7.30Chapter 8: 8.8, 8.38
Chapter 7:
7.11)Suppose that we will randomly select a sample
of 64 measurements from a population having a mean equal to 20 and a standard
deviation equal to 4.
a)Write a page paper – Describe the shape of the sampling
distribution of the sample mean . Do we need to make any assumptions about the shape of the
population? Why or why not?
b)Find the mean and the standard deviation
of the sampling distribution of the sample mean .
c)Calculate the probability that we will
obtain a sample mean greater than 21; that is calculate P(>21). Hint find the z value corresponding to 21 by usingand

d)Calculate the probability that we will
obtain a sample mean less than 19.385; that is calculate P(<19.385). 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, 18years 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 portion of .32 or more; that is calculate P(p^?.32) b) Based on the probability you computed in part a, would you conclude that more than 30 percent of Americans (18years or older) say that figure skating is their favorite Winter Olympic event? B)Chapter 8: 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 time in table 1.8 is = 5.46. If we let µ denote the mean of all possible bank customer waiting times using the new system and assume that ? equals 2.47: Calculate 95 percent and 99 percent confidence intervals for µ.75in;="" justify;="">.75in;=”” justify;=””>N=100 x=5.46 s=2.47 %=95.75in;=”” justify;=””>.75in;=”” justify;=”” lfo3;=””>Using the 95 percent confidence interval, can the
bank manager be 95 percent confident that µ is less than six minutes? Explain.75in;=”” justify;=”” lfo3;=””>Using the 99 percent confidence interval, can the
bank manager be 99 percent confident that µ is less than six minutes? Explain.75in;=”” justify;=”” lfo3;=””>Based on your answers to parts b and c, how
convinced are you that the new mean waiting time is less than six minutes?.75in;=”” justify;=””>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?” Suppose
that a random sample of 350 current customers’ results in 195 customers with a
response of 9 to 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.?N=350
P=0.557 %=95

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