Answer
Confidence interval: $0.0876\lt p\lt0.1824$
We are 95% confident that the proportion of incidence of salmonella for brand C is between 0.0876 and 0.1824
Work Step by Step
$p ̂=0.135$
Requirements:
$np ̂(1-p ̂)=200\times0.135(1-0.135)=23.355\gt10$
$n\leq0.05N$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$z_{\frac{α}{2}}=z_{0.025}$
If the area of the standard normal curve to the right of $z_{0.025}$ is 0.025, then the area of the standard normal curve to the left of $z_{0.025}$ is $1−0.025=0.975$
According to Table V, the z-score which gives the closest value to 0.975 is 1.96.
$Lower~bound=p ̂-z_{\frac{α}{2}}.\sqrt {\frac{p ̂(1-p ̂)}{n}}=0.135-1.96\times\sqrt {\frac{0.135(1-0.135)}{200}}=0.0876$
$Upper~bound=p ̂+z_{\frac{α}{2}}.\sqrt {\frac{p ̂(1-p ̂)}{n}}=0.135+1.96\times\sqrt {\frac{0.135(1-0.135)}{200}}=0.1824$
