Answer
when confidence interval = 95%
α= 1-0.95 = 0.05.
α/2 = 0.025
1-0.025 = 0.975
From the table,
z_α/2 = 1.96
z_α/2 = 1.96 , E=0.02, p̂ =0.27 , q̂ = 1-p̂ = 0.73
n = p̂q̂ ($ \frac{z_α/2}{E}$^2)
= p̂q ($ \frac{z_α/2}{E}$^2)
=(0.27)(0.73)($ \frac{1.96}{0.02}$^2)
=1892.95
≈1893
Hence, the sample must have 1893 individuals in order to estimate the true proportion of children with good diets within 2% with 95% confidence interval.
