A technology retailer is interested in the average price of their products, so they plan to create a $99\%$ confidence interval for the true mean. Find the minimum sample size needed so the margin of error is no more than \$100 if...
(B) The standard deviation is unknown but the cheapest product is \$198 and the most expensive is \$1500.

• A. $71$
• B. $282$
• C. $9$
• D. $84$

Answer: (A)

Verified step by step guidance

1. Identify the confidence level, which is 99%, and find the corresponding z-score (critical value) for a two-tailed test. Since the standard deviation is unknown, we will approximate the standard deviation using the range rule of thumb.
2. Estimate the population standard deviation \( \sigma \) by using the range of the data. The range is the difference between the most expensive and the cheapest product: \( 1500 - 198 \). Then approximate \( \sigma \) as \( \frac{\text{range}}{4} \).
3. Recall the formula for the margin of error (ME) in a confidence interval for the mean when the population standard deviation is unknown but estimated: \n\n\[ ME = z \times \frac{\sigma}{\sqrt{n}} \]\n\nwhere \( n \) is the sample size we want to find.
4. Set the margin of error \( ME \) to be no more than \$100 and solve the inequality for \( n \): \n\n\[ 100 \geq z \times \frac{\sigma}{\sqrt{n}} \]\n\nRearranging to solve for \( n \): \n\n\[ n \geq \left( \frac{z \times \sigma}{100} \right)^2 \]
5. Calculate the minimum sample size \( n \) by plugging in the values for \( z \) and \( \sigma \) from previous steps, then round up to the next whole number since sample size must be an integer.