Question 523819: hello
I have some question and I need the answers today because I have an assignment
Q1: This homework question:
When a pizza restaurant's delivery process is operating effectively
When a pizza restaurant's delivery process is operating effectively, pizzas are delivered in an average of 45 minutes with a standard deviation of 6 minutes. To monitor its delivery process, the restaurant randomly selects five pizzas each night and records their delivery times.
a. For the sake of argument, assume that the population of all delivery times on a given evening is normally distributed with a mean of = 45 minutes and a standard deviation of = 6 minutes. (That is we assume that the delivery process is operating effectively)
a. Find the mean of the population of all possible sample means and the standard deviation of the population of all possible sample means and Calculate an interval containing 99.73% of all possible sample means.
b. Suppose that the mean of the five sampled delivery times on a particular evening is x� = 55 minutes. Using the interval that was calculated in a(4), what would you conclude about whether the restaurant's delivery process is operating effectively? Why?
Q2: Quality Progress, February 2005, reports on improvements in customer satisfaction and loyalty made by Bank of America. 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?” Here, a response of 9 or 10 represents “customer delight.”
Historically, the percentage of Bank of America customers expressing customer delight has been 48%. Suppose that we wish to use the results of a survey of 350 Bank of America customers to justify the claim that more than 48% of all current Bank of America customers would express customer delight. The survey finds that 189 of 350 randomly selected Bank of America customers express customer delight. If, for the sake of argument, we assume that the proportion of customer delight is p = .48, calculate the probability of observing a sample proportion greater than or equal to 189/350 = .54.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! a. For the sake of argument, assume that the population of all delivery times on a given evening is normally distributed with a mean of = 45 minutes and a standard deviation of = 6 minutes. (That is we assume that the delivery process is operating effectively)
a. Find the mean of the population of all possible sample means and the standard deviation of the population of all possible sample means and Calculate an interval containing 99.73% of all possible sample means.
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mean of the sample means = 45
std of the sample means = 6/sqrt(5) = 2.6833
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1-0.9973 = 0.0027
(1/2)0.0027 = 0.0014
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Find the t value with a left tail of 0.0014 when df = 4
t = invT(0.0014,4) = -6.56
Find the correspondiing raw score value using x = ts+u
x = -6.56*2.6833+45 = 27.41
OR x = +6.56*6833+45 = 62.60
Interval with 99.73% of sample means = (27.41,62.60)
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b. Suppose that the mean of the five sampled delivery times on a particular evening is x� = 55 minutes. Using the interval that was calculated in a(4), what would you conclude about whether the restaurant's delivery process is operating effectively? Why?
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55 is in the 99.73% confidence window
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Q2: Quality Progress, February 2005, reports on improvements in customer satisfaction and loyalty made by Bank of America. 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?” Here, a response of 9 or 10 represents “customer delight.”
Historically, the percentage of Bank of America customers expressing customer delight has been 48%. Suppose that we wish to use the results of a survey of 350 Bank of America customers to justify the claim that more than 48% of all current Bank of America customers would express customer delight. The survey finds that 189 of 350 randomly selected Bank of America customers express customer delight. If, for the sake of argument, we assume that the proportion of customer delight is p = .48, calculate the probability of observing a sample proportion greater than or equal to 189/350 = .54.
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z(0.54) = (0.54-0.48)/sqrt[0.48*0.52/350] = 2.2468
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P(p-hat > 0.54) = P(z > 2.2468) = normalcdf(2.2468,100) = 0.0123
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Cheers,
Stan H.
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