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 $\mathrm{\backslash (}100$ if...
(A) The standard deviation is known to be about $\mathrm{\backslash )}324$

• A. $8$
• B. $6966$
• C. $70$
• D. $83$

Answer: (C)

Verified step by step guidance

1. Identify the confidence level, which is 99%, and find the corresponding z-score (critical value) from the standard normal distribution table. This z-score is denoted as \(z_{\alpha/2}\), where \(\alpha = 1 - 0.99 = 0.01\).
2. Use the formula for the margin of error (E) in estimating a population mean when the population standard deviation (\(\sigma\)) is known: \(E = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}\)
3. Rearrange the margin of error formula to solve for the minimum sample size \(n\): \(n = \left( \frac{z_{\alpha/2} \times \sigma}{E} \right)^2\)
4. Substitute the known values into the formula: - \(z_{\alpha/2}\) is the z-score for 99% confidence, - \(\sigma = 324\), - \(E = 100\) (the maximum allowed margin of error).
5. Calculate the value inside the parentheses first, then square it to find the minimum sample size \(n\). Since \(n\) must be a whole number, round up to the next whole number if necessary.