Answer
Confidence interval: $-0.0389\lt p̂_1-p̂ _2\lt0.0087$
We are 95% confident that $p̂_1-p̂ _2$ is between -0.0389 and 0.0087.
Work Step by Step
$p̂ _1=\frac{x_1}{n_1}=\frac{804}{874}=0.9199$ and $p̂ _2=\frac{x_2}{n_2}=\frac{892}{954}=0.9350$
Requirements:
$n_1p̂ _1(1-p̂ _1)=874\times0.9199(1-0.9199)=64.40\geq10$
$n_2p̂ _2(1-p̂ _2)=954\times0.9350(1-0.9350)=57.98\geq10$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$z_{\frac{α}{2}}=z_{0.025}$
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.
$Lower~bound=(p̂_1-p̂ _2)-z_{\frac{α}{2}}\sqrt {\frac{p̂_1(1-p̂_1)}{n_1}+\frac{p̂ _2(1-p̂ _2)}{n_2}}=(0.9199-0.9350)-1.96\sqrt {\frac{0.9199(1-0.9199)}{874}+\frac{0.9350(1-0.9350)}{954}}=-0.0389$
$Upper~bound=(p̂_1-p̂ _2)+z_{\frac{α}{2}}\sqrt {\frac{p̂_1(1-p̂_1)}{n_1}+\frac{p̂ _2(1-p̂ _2)}{n_2}}=(0.9199-0.9350)+1.96\sqrt {\frac{0.9199(1-0.9199)}{874}+\frac{0.9350(1-0.9350)}{954}}=0.0087$
