Answer
when confidence interval = 90%
α= 1-0.90 = 0.10.
α/2 = 0.05
1-0.05 = 0.95
From the table,
z_α/2 = 1.65
z_α/2 = 1.65 , E=0.025, p̂ =0.5 , q̂= 1-p̂ = 0.5
n = p̂q̂ ($ \frac{z_α/2}{E}$^2)
= p̂q̂ ($ \frac{z_α/2}{E}$^2)
=(0.5)(0.5)($ \frac{1.65}{0.025}$^2)
=1089
Hence, the sample must have 1089 individuals in order to be 90% confident that nearly one-half Americans aged 25 to 29 are unmarried.
