According to USA Today, customers are not settling for automobiles straight off the production lines. As an example, those who purchase a $355,000 Rolls-Royce typically add $25,000 in accessories. One of the affordable automobiles to receive additions is BMW's Mini Cooper. A sample of 179 recent Mini purchasers yielded a sample mean of $5,000 for accessories above the $20,200 base sticker price. Suppose the cost of accessories purchased for all Mini Coopers has a population standard deviation of $1,500.
Determine the margin of error in estimating the average cost of accessories on Mini Coopers for a 95% confidence interval.

Question

BlackTom AI · Accepted Answer

We start by identifying what is being asked: the margin of error for estimating the average cost of accessories on Mini Coopers at a 95% confidence level, given a known population standard deviation.
Option 1: 219.75
- This value comes from the formula for the margin of error when σ is known: ME = z*σ/√n. Here, z for 95% is 1.96, σ = 1500, n = 179. Compute √n ≈ √179 ≈ 13.379, then σ/√n ≈ 1500/13.379 ≈ 112.2. Multiply by 1.96: 112.2 × 1.96 ≈ 219.8, which aligns with 219.75 after precise calculation. So this option is consistent with the standard method.
Option 2: 291.11
- This would require a larger ME than expected. It would occur if either z were much larger (e.g., using a t value inappropriate for known σ) or σ/√n were larger. Since σ/√n is about 112.2, achieving 291.11 would imply a multiplier around 2.6, which is not appropriate for a 95% confidence level with known σ. Therefore this is incorrect.
Option 3: 231.14
- This is close to the correct 219.75 but still significantly larger. It would result from a slightly larger ME, perhaps due to rounding differences or a mis-specified z-value. However, with z = 1.96 and the given σ and n, the ME should be around 219.75, not 231.14. Thus this option does not match the standard calculation.
Option 4: 214.41
- This value is in the same general vicinity as the correct ME but is slightly smaller. It would arise if σ/√n were about 109.4 (since 109.4 × 1.96 ≈ 214.8). Given σ/√n ≈ 112.2, this option underestimates the margin of error by a noticeable amount, so it is not correct for the provided data.
In summary, the margin of error for a 95% confidence interval with σ = 1500 and n = 179 is computed as ME = 1.96 × (1500/√179) ≈ 219.75, making option 219.75 the correct choice, while the other options reflect misapplications or rounding differences that do not align with the standard formula and given values.

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