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 𝜇𝜇 = 𝜇

and 𝜇𝜇 = 𝜇

√𝜇 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 size 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 applies, 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 𝜇 to a Z value

Then we know that P( 𝜇 < $400) = P(Z < -0.959)

We can use the Excel function NORM.S.DIST(-0.959,TRUE)

P( 𝜇 < $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 Cost

(μ).

Expected total cost = 10, 000 * $406.27 = $4,062,711

What do the following two values represent, in terms of the sampling distribution?

𝜇 + 𝜇0.10 ∗ 𝜇

√𝜇 = $397.89

𝜇 + 𝜇0.90 ∗ 𝜇

√𝜇 = $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?

When 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 between $3,978,900 and

$4,146,500