Answer
Confidence interval: $0.282\lt p̂_{2003}-p̂ _{2010}\lt0.339$
We are 90% confident that $p̂_{2003}-p̂ _{2010}$ is between 0.282 and 0.339
Work Step by Step
$p̂ _{2003}=\frac{x_{2003}}{n_{2003}}=\frac{1086}{1508}=0.7202$ and $p̂ _{2010}=\frac{x_{2010}}{n_{2010}}=\frac{618}{1508}=0.4098$
Requirements:
$n_{2003}p̂ _{2003}(1-p̂ _{2003})=1508\times0.7202(1-0.7202)=303.88\geq10$
$n_{2010}p̂ _{2010}(1-p̂ _{2010})=1508\times0.4098(1-0.4098)=364.73\geq10$
$level~of~confidence=(1-α).100$%
$90$% $=(1-α).100$%
$0.90=1-α$
$α=0.1$
$z_\frac{α}{2}=z_{0.05}$
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: $\frac{1.64+1.65}{2}=1.645$
$Lower~bound=(p̂_{2003}-p̂ _{2010})-z_{\frac{α}{2}}\sqrt {\frac{p̂_{2003}(1-p̂_{2003})}{n_{2003}}+\frac{p̂ _{2010}(1-p̂ _{2010})}{n_{2010}}}=(0.7202-0.4098)-1.645\sqrt {\frac{0.7202(1-0.7202)}{1508}+\frac{0.4098(1-0.4098)}{1508}}=0.282$
$Upper~bound=(p̂_{2003}-p̂ _{2010})+z_{\frac{α}{2}}\sqrt {\frac{p̂_{2003}(1-p̂_{2003})}{n_{2003}}+\frac{p̂ _{2010}(1-p̂ _{2010})}{n_{2010}}}=(0.7202-0.4098)-1.645\sqrt {\frac{0.7202(1-0.7202)}{1508}+\frac{0.4098(1-0.4098)}{1508}}=0.339$
