Answer
Confidence interval: $-0.090\lt p̂_1-p̂ _2\lt0.068$
We are 99% confident that $p̂_1-p̂ _2$ is between -0.090 and 0.068.
Work Step by Step
$p̂ _1=\frac{x_1}{n_1}=\frac{109}{475}=0.229$ and $p̂ _2=\frac{x_2}{n_2}=\frac{78}{325}=0.24$
Requirements:
$n_1p̂ _1(1-p̂ _1)=475\times0.229(1-0.229)=83.87\geq10$
$n_2p̂ _2(1-p̂ _2)=325\times0.24(1-0.24)=59.28\geq10$
$level~of~confidence=(1-α).100$%
$99$% $=(1-α).100$%
$0.99=1-α$
$α=0.01$
$z_{\frac{α}{2}}=z_{0.005}$
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: $\frac{2.57+2.58}{2}=2.575$
$Lower~bound=(p̂_1-p̂ _2)-z_{\frac{α}{2}}\sqrt {\frac{p̂_1(1-p̂_1)}{n_1}+\frac{p̂ _2(1-p̂ _2)}{n_2}}=(0.229-0.24)-2.575\sqrt {\frac{0.229(1-0.229)}{475}+\frac{0.24(1-0.24)}{325}}=-0.090$
$Upper~bound=(p̂_1-p̂ _2)+z_{\frac{α}{2}}\sqrt {\frac{p̂_1(1-p̂_1)}{n_1}+\frac{p̂ _2(1-p̂ _2)}{n_2}}=(0.229-0.24)+2.575\sqrt {\frac{0.229(1-0.229)}{475}+\frac{0.24(1-0.24)}{325}}=0.068$
