Answer
n=90,p̂ = 40/90 =0.44 ,q = 1-p̂ = 0.56
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
p̂ - $z_α/_2$ $\sqrt \frac{p̂q̂}{n}$ < p < p̂ + $z_α/_2$ $\sqrt \frac{p̂q̂}{n}$
= 0.44 - 1.96$\sqrt \frac{0.44*0.56}{90}$ < p < 0.44 + 1.96$\sqrt \frac{0.44*0.56}{90}$
= 0.342 < p < 0.547
= 34.2% < p <54.7%
Hence, we can say that the true proportion of families who own at least one television set with 95% confidence level is between 34.2% and 54.7%
