Question 524239: 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)
1. Describe the shape of the population of all possible sample means. How do you know what the shape is?
2. Find the mean of the population of all possible sample means.
3. Find the standard deviation of the population of all possible sample means
4. 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?
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)
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1. Describe the shape of the population of all possible sample means.
Approximately normal with:
mean of the sample means = 45
std of the sample means = 6/sqrt(5)
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How do you know what the shape is?
The Central Limit Theorem
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2. Find the mean of the population of all possible sample means.
3. Find the standard deviation of the population of all possible sample means
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4. Calculate an interval containing 99.73% of all possible sample means.
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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?
Seems to be effective. 55 is in the 97% confidence interval.
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Cheers,
Stan H.
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