Answer:
0.4384 < p < 0.5049
Step-by-step explanation:
The confidence interval for the population proportion can be calculated as:
![p^(\prime)-z_{(\alpha)/(2)}\sqrt[]{(p^(\prime)(1-p^(\prime)))/(n)}<p>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.</p><p></p><p>Now, we can calculate p' as the division of the number of voters of favor approval by the total number of voters.</p>[tex]p^(\prime)=(408)/(865)=0.4717](https://img.qammunity.org/2023/formulas/mathematics/college/np9qg8mguzoxd9m16kzdudb0kmzj2zzn4y.png)
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