18. Obesity Obesity is defined as a body mass index (BMI) of \( 3 \mathrm{~kg} / \mathrm{m}^{2} \) or more. A \( 95 \% \) confidence interval for the percentage of U.S. adults aged 20 years and over who were obese was found to be 22.4 to \( 23.5 \% \). What was the sample size?

To determine the sample size for the given 95% confidence interval for the obesity rate among U.S. adults, we'll use the formula for the confidence interval of a proportion: \[ CI = \hat{p} \pm z^* \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \] Where: - \(\hat{p}\) is the sample proportion. - \(z^*\) is the z-score corresponding to the confidence level (1.96 for 95% confidence). - \(n\) is the sample size. - \(CI\) is the confidence interval. **Step 1: Identify the Given Values** - **Confidence Interval**: 22.4% to 23.5% - **Confidence Level**: 95%, so \(z^* = 1.96\) **Step 2: Calculate the Sample Proportion (\(\hat{p}\)) and Margin of Error (E)** \[ \hat{p} = \frac{22.4\% + 23.5\%}{2} = 22.95\% \] \[ E = \frac{23.5\% - 22.4\%}{2} = 0.55\% \] **Step 3: Rearrange the Confidence Interval Formula to Solve for \(n\)** \[ E = z^* \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \\ \Rightarrow n = \frac{z^{*2} \cdot \hat{p}(1 - \hat{p})}{E^2} \] **Step 4: Plug in the Values** \[ n = \frac{(1.96)^2 \cdot 0.2295 \cdot (1 - 0.2295)}{(0.0055)^2} \] \[ n = \frac{3.8416 \cdot 0.2295 \cdot 0.7705}{0.00003025} \\ n \approx \frac{0.6793}{0.00003025} \\ n \approx 22,457 \] **Step 5: Round to a Practical Sample Size** Given that sample sizes are typically rounded to whole numbers, especially in large surveys: \[ n \approx 22,450 \] **Conclusion:** The required sample size is approximately **22,450**.