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Unformatted text preview: ISOM 111 L7L8, Fall 2010 1 Homework 4 Solutions I . 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 μ = 45 minutes and a standard deviation of σ = 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 99.7 percent 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 you calculated in (a.4), what would you conclude about whether the restaurant’s delivery process is operating effectively? Why? Solution : (a). (1). Because the population of all delivery times is normal, the population of all sample means is also normally distributed. (2). The mean of all sample means μ ¯ X = μ = 45 minutes . (3). The standard deviation of all sample means σ ¯ X = σ √ n = 6 √ 5 ≈ 2 . 68 . (4). The interval is [ μ ¯ X ± 3 σ ¯ X ] = [45 ± 3 × 2 . 68] = [36 . 96 , 53 . 04] . (b). If the delivery process is operating effectively, then the delivery time will be smaller than 53.04 with a very high probability ( ≥ 99 . 7% ). Now the sample mean delivery time is bigger than 53.04, so the delivery process is unlikely to be operating effectively. II . In an article in the Journal of Marketing , Bayus studied the differences between “early replacement buyers” and “late replacement buyers” in making consumer durable good replacement purchases. Early replacement buyers are consumers who replace a product during the early part of its lifetime, while late replacement buyers make replacement purchases late in the product’s lifetime. In particular, Bayus studied automobile replacement purchases. Consumers who traded in cars with ages of zero to three years and mileages of no more than 35,000 miles were classified as early replacement buyers. Consumers who traded in car with ages of seven or more years and mileages of more than 73,000 miles were classified as late replacement buyers. Bayus compared themileages of more than 73,000 miles were classified as late replacement buyers....
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 Fall '10
 YingYingLi
 Business, Standard Deviation, ISOM, replacement buyers, early replacement buyers

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