Answer
Confidence interval: $-0.075\lt p̂_1-p̂ _2\lt0.015$
We are 90% confident that $p̂_1-p̂ _2$ is between -0.075 and 0.015.
Work Step by Step
$p̂ _1=\frac{x_1}{n_1}=\frac{368}{541}=0.680$ and $p̂ _2=\frac{x_2}{n_2}=\frac{421}{593}=0.710$
Requirements:
$n_1p̂ _1(1-p̂ _1)=541\times0.680(1-0.680)=117.72\geq10$
$n_2p̂ _2(1-p̂ _2)=593\times0.710(1-0.710)=122.10\geq10$
$z_\frac{α}{2}=z_{0.05}$
If the area of the standard normal curve to the right of $z_{0.05}$ is 0.05, then the area of the standard normal curve to the left of $z_{0.05}$ is $1−0.05=0.95$
According to Table V, there are 2 z-scores which give the closest value to 0.95: 1.64 and 1.65. So, let's find the mean of these z-scores: $\frac{1.64+1.65}{2}=1.645$
$Lower~bound=(p̂_1-p̂ _2)-z_{\frac{α}{2}}\sqrt {\frac{p̂_1(1-p̂_1)}{n_1}+\frac{p̂ _2(1-p̂ _2)}{n_2}}=(0.680-0.710)-1.645\sqrt {\frac{0.680(1-0.680)}{541}+\frac{0.710(1-0.710)}{593}}=-0.075$
$Upper~bound=(p̂_1-p̂ _2)+z_{\frac{α}{2}}\sqrt {\frac{p̂_1(1-p̂_1)}{n_1}+\frac{p̂ _2(1-p̂ _2)}{n_2}}=(0.680-0.710)+1.645\sqrt {\frac{0.680(1-0.680)}{541}+\frac{0.710(1-0.710)}{593}}=0.015$
