Question

A federal report indicated that 16% of children ages 2 to 5 years had a good diet. How large a sample is needed to estimate the true proportion of children with good diets within 4% with 95% confidence?

Answer 1

number of sample with confidence interval can be found with the formula;

$n=pq(\frac{z_{\frac{\alpha}{2}}}{E})^2$

Given the parameters;

$\begin{gathered} p=16\text{ \%=0.16} \\ q=1-p=1-0.16=0.84 \\ z_{\frac{\alpha}{2}}=1.96 \\ E=4\text{ \%= 0.04} \end{gathered}$

Thus, we have;

$\begin{gathered} n=(0.16)(0.84)(\frac{1.96}{0.04})^2 \\ n=322.6944 \\ n\approx323 \end{gathered}$

The number of sample needed to estimate the true proportion of children with good diets is 323