Automobile Manufacturing Company Wants To Estimate

Promptan Automobile Manufacturing Company Wants To Estimate The Cost

An automobile manufacturing company aims to estimate the cost of an anticipated vehicle recall. They selected a random sample of 64 vehicles requiring repairs, with the sample mean repair cost being $406.27 and a standard error of $6.54. The goal is to analyze this sample data to make inferences about the total population of vehicles needing repairs, based on sampling distribution principles.

Before using the sample to estimate the population parameters, it is essential to verify certain conditions or assumptions. The primary assumptions involve the normality of the sampling distribution of the mean. According to the Central Limit Theorem (CLT), when the sample size is sufficiently large—commonly n ≥ 30—the sampling distribution of the sample mean approaches normality, regardless of the parent population’s distribution. Since the sample size here is 64, which exceeds this threshold, the CLT generally applies, assuming the parent population is not extremely unusual.

However, if the parent population is heavily skewed or has significant outliers, the approximation might not be accurate. Therefore, it is also advisable to conduct exploratory data analysis, such as examining histograms or boxplots, to assess normality. If the data appear reasonably symmetrical and without extreme outliers, then the normality assumption for the sample distribution of the mean is justified, allowing us to proceed with normal-based inference methods.

Probability Calculation for a Sample Mean of $400 or Less

Assuming the conditions are satisfied for the CLT to hold, the sampling distribution of the mean can be modeled as a normal distribution with mean μ = $406.27 and standard error σₓ̅ = $6.54. To find the probability that the sample mean is $400 or less, we first standardize this value to a Z-score:

Z = (X̄ - μ) / σₓ̅ = ($400 - $406.27) / $6.54 ≈ -0.959

Using the standard normal distribution, the probability corresponding to Z

Expected Total Cost for 10,000 Vehicle Repairs

If the company plans to recall 10,000 cars, the expected total cost of repairs is the product of the total number of vehicles and the mean repair cost:

Expected total cost = 10,000 × $406.27 = $4,062,700

This provides a point estimate for the total recall cost based on the current sample mean, assuming this mean accurately reflects the population mean.

Confidence Interval for the Mean Repair Cost

The problem further presents two values, $397.89 and $414.65, which correspond to the bounds of an interval. These are derived by adding and subtracting the margin of error from the sample mean, designed to contain 80% of the sampling distribution of the mean. Specifically:

  • Lower bound: μ - Z_{0.90} × σₓ̅ = $406.27 - 0.84 × $6.54 ≈ $397.89
  • Upper bound: μ + Z_{0.90} × σₓ̅ = $406.27 + 0.84 × $6.54 ≈ $414.65

These bounds form an 80% confidence interval for the true mean repair cost, meaning that if every possible sample of size 64 were taken, approximately 80% of those sample means would fall within this interval.

In terms of total costs, multiplying these interval bounds by the total number of vehicles (10,000) yields the range within which the total recall cost is expected to fall with 80% certainty:

  • Lower total cost: 10,000 × $397.89 = $3,978,900
  • Upper total cost: 10,000 × $414.65 = $4,146,500

Therefore, the company can be 80% confident that the total cost of repairs for the 10,000 vehicles will be between $3,978,900 and $4,146,500 based on this sampling distribution analysis.

References

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