20. Diet Habits A federal report indicated that \( 27 \% \) of children ages 2 to 5 years had a good diet-an increase over previous years. How large a sample is needed to estimate the true proportion of children with good diets within \( 2 \% \) with \( 95 \% \) confidence?

To estimate the true proportion of children with good diets within 2% with 95% confidence, we can use the formula for the sample size required for estimating a population proportion: \[ n = \left( \frac{Z_{\alpha/2}}{E} \right)^2 \times p \times (1-p) \] where: - \( n \) is the sample size required - \( Z_{\alpha/2} \) is the Z-score corresponding to the desired confidence level (95% in this case) - \( E \) is the margin of error (2% in this case) - \( p \) is the estimated proportion of children with good diets (27% in this case) First, we need to find the Z-score corresponding to 95% confidence. The Z-score for 95% confidence is approximately 1.96. Next, we can substitute the given values into the formula to calculate the sample size required. Calculate the value by following steps: - step0: Calculate: \(\left(\frac{1.96}{0.02}\right)^{2}\times 0.27\left(1-0.27\right)\) - step1: Divide the terms: \(98^{2}\times 0.27\left(1-0.27\right)\) - step2: Subtract the numbers: \(98^{2}\times 0.27\times 0.73\) - step3: Multiply the terms: \(98^{2}\times 0.1971\) - step4: Rewrite the expression: \(9604\times 0.1971\) - step5: Multiply the numbers: \(1892.9484\) The sample size required to estimate the true proportion of children with good diets within 2% with 95% confidence is approximately 1892.95. Therefore, a sample size of at least 1893 children is needed to achieve the desired level of accuracy and confidence.