Answer:
0.4384 < p < 0.5049
Explanation:
The confidence interval for the population proportion can be calculated as:
[tex]p^{\prime}-z_{\frac{\alpha}{2}}\sqrt[]{\frac{p^{\prime}(1-p^{\prime})}{n}}
Where p' is the sample proportion, z is the z-score related to the 95% level of confidence, n is the size of the sample and p is the population proportion.
Now, we can calculate p' as the division of the number of voters of favor approval by the total number of voters.
[tex]p^{\prime}=\frac{408}{865}=0.4717[/tex]
Additionally, n = 865 and z = 1.96 for a 95% level of confidence. So, replacing the values, we get:
[tex]\begin{gathered} 0.4717-1.96\sqrt[]{\frac{0.4717(1-0.4717)_{}}{865}}
Therefore, the confidence interval for the true proportion is:
0.4384 < p < 0.5049
