Answer
Confidence interval: $−0.0086\lt p̂_{male}-p̂ _{female}\lt0.0482$
Since $p̂_{male}-p̂ _{female}=0$, that is $p̂_{male}=p̂ _{female}$, is inside the confidence interval, we do not reject the null hypothesys.
There is not enough evidence to conclude that the proportion of males that have at least one tattoo differs significantly from the proportion of females that have at least one tattoo.
Work Step by Step
$H_0:~p_{male}=p_{female}$ versus $H_1:~p_{male}\ne p{female}$
$p̂ _{male}=\frac{x_{male}}{n_{male}}=\frac{181}{1205}=0.1502$ and $p̂ _{female}=\frac{x_{female}}{n_{female}}=\frac{143}{1097}=0.1304$
Requirements:
$n_{male}p̂ _{male}(1-p̂ _{male})=1205\times0.1502(1-0.1502)=153.8\geq10$
$n_{female}p̂ _{female}(1-p̂ _{female})=1097\times0.1304(1-0.1304)=124.4\geq10$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$z_{\frac{α}{2}}=z_{0.025}$
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.
$Lower~bound=(p̂_{male}-p̂ _{female})-z_{\frac{α}{2}}\sqrt {\frac{p̂_{male}(1-p̂_{male})}{n_{male}}+\frac{p̂ _{female}(1-p̂ _{female})}{n_{female}}}=(0.1502-0.1304)-1.96\sqrt {\frac{0.1502(1-0.1502)}{1205}+\frac{0.1304(1-0.1304)}{1097}}=-0.0086$
$Upper~bound=(p̂_{male}-p̂ _{female})+z_{\frac{α}{2}}\sqrt {\frac{p̂_{male}(1-p̂_{male})}{n_{male}}+\frac{p̂ _{female}(1-p̂ _{female})}{n_{female}}}=(0.1502-0.1304)+1.96\sqrt {\frac{0.1502(1-0.1502)}{1205}+\frac{0.1304(1-0.1304)}{1097}}=0.0482$
