Answer
$0.596\lt p\lt0.704$
Work Step by Step
We have $\hat p=\frac{195}{300}=0.65, \hat q=1-\hat p=0.35 , n=300 $
At a 95% confidence the critical z-value is $z_{\alpha/2}=1.96 $
The margin of error can be found as
$E=z_{\alpha/2}\times\sqrt {\frac{\hat p\hat q}{n}}=1.96\times\sqrt {\frac{0.65\times0.35}{300}}=0.054$
Thus, the interval of the true proportion can be found as
$\hat p-E\lt p\lt\hat p+E$ which gives $0.596\lt p\lt0.704$
