Answer
Confidence interval: $-0.063\lt p̂_1-p̂ _2\lt0.044$
We are 95% confident that $p̂_1-p̂ _2$ is between -0.063 and 0.044.
Work Step by Step
$p̂ _1=\frac{x_1}{n_1}=\frac{28}{254}=0.1102$ and $p̂ _2=\frac{x_2}{n_2}=\frac{36}{301}=0.1196$
Requirements:
$n_1p̂ _1(1-p̂ _1)=254\times0.1102(1-0.1102)=24.91\geq10$
$n_2p̂ _2(1-p̂ _2)=301\times0.1196(1-0.1196)=31.69\geq10$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$z_{\frac{α}{2}}=z_{0.025}$
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.
$Lower~bound=(p̂_1-p̂ _2)-z_{\frac{α}{2}}\sqrt {\frac{p̂_1(1-p̂_1)}{n_1}+\frac{p̂ _2(1-p̂ _2)}{n_2}}=(0.1102-0.1196)-1.96\sqrt {\frac{0.1102(1-0.1102)}{254}+\frac{0.1196(1-0.1196)}{301}}=-0.063$
$Upper~bound=(p̂_1-p̂ _2)+z_{\frac{α}{2}}\sqrt {\frac{p̂_1(1-p̂_1)}{n_1}+\frac{p̂ _2(1-p̂ _2)}{n_2}}=(0.1102-0.1196)+1.96\sqrt {\frac{0.1102(1-0.1102)}{254}+\frac{0.1196(1-0.1196)}{301}}=0.044$
