Answer
n=150,p̂ = 0.36 = ,q = 1-p̂ = 0.64
When Confidence interval = 90%
α= 1-0.90 = 0.1.
α/2 = 0.05
1-0.05 = 0.95
From the table,
$z_α/_2$= 1.65
p̂ - $z_α/_2$ $\sqrt \frac{p̂q̂}{n}$ < p < p̂ + $z_α/_2$ $\sqrt \frac{p̂q̂}{n}$
= 0.36 - 1.65$\sqrt \frac{0.36*0.64}{150}$ < p < 0.36 + 1.65$\sqrt \frac{0.36*0.64}{150}$
= 0.295 < p < 0.425
= 29.5% < p < 42.5%
Hence, we can say that the true proportion of accidents that involve children under the age of 6t with 90% confidence level is between 29.5% and 42.5%
