Question 9 of 14, Step 1 of 1 A research company desires to know the mean consumption of meat per week among people over age 24. They believe that the meat consumption has a mean of 4.4 pounds, and want to construct a \( 90 \% \) confidence interval with a maximum error of 0.08 pounds. Assuming a variance of 0.36 pounds, what is the minimum number of people over age 24 they must include in their sample? Round your answer un to the next integer.

We know that the margin of error \( E \) for a confidence interval is given by \[ E = z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}, \] where - \( z_{\frac{\alpha}{2}} \) is the z-score corresponding to a \( 90\% \) confidence level, - \(\sigma\) is the standard deviation, - \( n \) is the sample size. Given: - \( E = 0.08 \), - \(\sigma^2 = 0.36 \) so that \(\sigma = \sqrt{0.36} = 0.6\), - For a \( 90\% \) confidence level, \(\alpha = 0.10\) and hence \(\frac{\alpha}{2} = 0.05\). The critical value is \( z_{0.05} \approx 1.645 \). Substitute the values into the formula: \[ 0.08 = 1.645 \frac{0.6}{\sqrt{n}}. \] Solve for \( n \): \[ \sqrt{n} = 1.645 \frac{0.6}{0.08}. \] First, calculate the numerator: \[ 1.645 \times 0.6 = 0.987. \] Then, \[ \sqrt{n} = \frac{0.987}{0.08} \approx 12.3375. \] Square both sides: \[ n \approx (12.3375)^2 \approx 152.1539. \] Since we must round up to the next whole number, the minimum sample size is \[ n = 153. \]