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_{(\alpha)/(2)}}{E})^2`

Given the parameters;

`\begin{gathered} p=16\text{ \%=0.16} \\ q=1-p=1-0.16=0.84 \\ z_{(\alpha)/(2)}=1.96 \\ E=4\text{ \%= 0.04} \end{gathered}`

Thus, we have;

`\begin{gathered} n=(0.16)(0.84)((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