Answer
Confidence interval: $0.069\lt p ̂\lt0.165$
Work Step by Step
$p̂ =\frac{x}{n}=\frac{35}{300}=0.117$
Required condition:
$np̂ (1-p̂ )=300\times0.117(1-0.117)=30.92\gt10$
$level~of~confidence=(1-α).100$%
$99$% $=(1-α).100$%
$0.99=1-α$
$α=0.01$
$z_{\frac{α}{2}}=z_{0.005}$
If the area of the standard normal curve to the right of $z_{0.005}$ is 0.005, then the area of the standard normal curve to the left of $z_{0.005}$ is $1−0.005=0.995$
According to Table V, there are 2 z-scores which give the closest value to 0.995: 2.57 and 2.58. So, let's find the mean of these z-scores: $\frac{2.57+2.58}{2}=2.575$
$Lower~bound=p ̂-z_{\frac{α}{2}}.\sqrt {\frac{p ̂(1-p ̂)}{n}}=0.117-2.575\times\sqrt {\frac{0.117(1-0.117)}{300}}=0.069$
$Upper~bound=p ̂+z_{\frac{α}{2}}.\sqrt {\frac{p ̂(1-p ̂)}{n}}=0.117+2.575\times\sqrt {\frac{0.117(1-0.117)}{300}}=0.165$
