Answer
Confidence interval: $-0.1286\lt p̂_{men}-p̂ _{women}\lt-0.0500$
Since $p̂_{men}-p̂ _{women}=0$, that is $p̂_{men}=p̂ _{women}$, is outside the confidence interval, we reject the null hypothesis.
There is enough evidence to conclude that there is a difference in the proportion of men and women who are normal weight.
Work Step by Step
$H_0:~p_{men}=p_{women}$ versus $H_1:~p_{men}\ne p_{women}$
$p̂ _{men}=\frac{x_{men}}{n_{men}}=\frac{203}{750}=0.2707$ and $p̂ _{women}=\frac{x_{women}}{n_{women}}=\frac{270}{750}=0.36$
Requirements:
$n_{men}p̂ _{men}(1-p̂ _{men})=750\times0.2707(1-0.2707)=148.1\geq10$
$n_{women}p̂ _{women}(1-p̂ _{women})=750\times0.36(1-036)=172.8\geq10$
$level~of~confidence=(1-α).100$%
$90$% $=(1-α).100$%
$0.90=1-α$
$α=0.10$
$z_{\frac{α}{2}}=z_{0.05}$
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: $\frac{1.64+1.65}{2}=1.645$
$Lower~bound=(p̂_{men}-p̂ _{women})-z_{\frac{α}{2}}\sqrt {\frac{p̂_{men}(1-p̂_{men})}{n_{men}}+\frac{p̂ _{women}(1-p̂ _{women})}{n_{women}}}=(0.2707-0.36)-1.645\sqrt {\frac{0.2707(1-0.2707)}{750}+\frac{0.36(1-0.36)}{750}}=-0.1286$
$Upper~bound=(p̂_{men}-p̂ _{women})+z_{\frac{α}{2}}\sqrt {\frac{p̂_{men}(1-p̂_{men})}{n_{men}}+\frac{p̂ _{women}(1-p̂ _{women})}{n_{women}}}=(0.2707-0.36)+1.645\sqrt {\frac{0.2707(1-0.2707)}{750}+\frac{0.36(1-0.36)}{750}}=-0.0500$
