24. Cost of Pizras A pizza shop owner wishes to find the \( 95 \% \) confidence interval of the true mean cost of a large plain pizas. How large should the sample be if she wishes to be accurate to within \( \$ 0.15 \) ? A previous study showed that the standard deviation of the price was \( \$ 0.26 \).

To determine the required sample size for a \( 95\% \) confidence interval with a margin of error of \( \$0.15 \), we'll use the following formula: \[ n = \left( \frac{z^* \cdot \sigma}{E} \right)^2 \] Where: - \( z^* \) is the z-score corresponding to the desired confidence level. - \( \sigma \) is the population standard deviation. - \( E \) is the margin of error. **Step-by-Step Calculation:** 1. **Identify the Z-Score for \( 95\% \) Confidence:** - For a \( 95\% \) confidence level, the z-score (\( z^* \)) is approximately **1.96**. 2. **Given Values:** - \( \sigma = \$0.26 \) - \( E = \$0.15 \) 3. **Plug the Values into the Formula:** \[ n = \left( \frac{1.96 \times 0.26}{0.15} \right)^2 \] 4. **Calculate the Numerator:** \[ 1.96 \times 0.26 = 0.5096 \] 5. **Divide by the Margin of Error:** \[ \frac{0.5096}{0.15} \approx 3.3987 \] 6. **Square the Result:** \[ (3.3987)^2 \approx 11.55 \] 7. **Determine the Sample Size:** - Since the sample size must be a whole number, and we can't have a fraction of a sample, we'll round up to ensure the desired confidence level and precision. \[ n \approx 12 \] **Conclusion:** The pizza shop owner should select a sample size of **12** to achieve a \( 95\% \) confidence interval with a margin of error of \( \$0.15 \).