Statistics

Prompt : An automobile manufacturing company wants to estimate the cost of an anticipated recall. They randomly selected 64 vehicles that require the fix and found the average cost to complete the repair was $406.27, with a standard error of $6.54. Use what you have learned about Sampling Distributions to answer the following questions. Response parameters : What conditions, or assumptions, should be verified before using the sample values to estimate the population distribution? In order to assume the Sample Distribution of the Mean is a Normal distribution with = and = √ the population must be a Normal Distribution, or the sample size (n) must be large enough that the Central Limit Theorem applies to the sample distribution. A sample si ze of 64 should be large enough for the CLT to apply unless the parent population is very unusual. If these conditions are satisfied, what is the probability that you would get a sample with a sample mean of $400.00 or less? If we can assume the CTL appl ies, then we can use the Standardized Normal Distribution (Z - distribution) to find the probabilities associated with our sampling distribution. The first step is to transform our new to a Z value Then we know that P( < $400) = P(Z < -0.959) We can use the Excel function NORM.S.DIST( -0.959,TRUE) P( < $400) = 0.169 If they expect to recall 10,000 cars, what is their expected cost for all the repairs? The expected total cost would be the total number of cars recalled multiplied by the Mean Cos t (μ). Expected total cost = 10, 000 * $406.27 = $4,062,711 What do the following two values represent, in terms of the sampling distribution? + 0.10 ∗ √ = $397 .89 + 0.90 ∗ √ = $414 .65 These two values are the interval limits that contain 80% of the Sampling Distribution of the Mean. If we took every possible sample of size n = 64 from the population, 80% of the sample means would fall into this interval. What do these values mean in terms of the expected value of the total cost of the recall? Wh en estimating the expected total cost, based on a sample size of 64, we know that 80% of the time the value would be between 10,000 * 397.89 = $3,978,900 And 10,000 * 414.65 = $4,146,500 The company can be 80% certain that the cost of the recall will be be tween $3,978,900 and $4,146,500