Answer
$0.400\lt p\lt0.462$
Work Step by Step
We have $\hat p=\frac{329}{763}=0.431, \hat q=1-\hat p=0.569 , n=763 $
At a 92% confidence the critical z-value is $z_{\alpha/2}=1.751 $
The margin of error can be found as
$E=z_{\alpha/2}\times\sqrt {\frac{\hat p\hat q}{n}}=1.751\times\sqrt {\frac{0.431\times0.569}{763}}=0.031$
4. Thus, the interval of the true proportion can be found as
$\hat p-E\lt p\lt\hat p+E$ which gives $0.400\lt p\lt0.462$
