Solution
Part a
p=x/n = 519/1003= 0.517
Part b
The margin of error is given by:
![ME=z_{(\alpha)/(2)}\cdot\sqrt[]{(p(1-p))/(n)}](https://img.qammunity.org/2023/formulas/mathematics/college/imfqaxxeyutb5cm8jnacqv9ym245yxfj5v.png)
At 99 % confidence level the z value is z= 2.576
Replacing we got:
![ME=2.576\cdot\sqrt[]{(0.517(1-0.517))/(1003)}=0.0407](https://img.qammunity.org/2023/formulas/mathematics/college/yjz2rr10zb6ncrq2m7s34b1zqx2tn18sr4.png)
Rounded it would 0.041
Part c
For this case we can find the confidence interval on this way:

Part d
For this case the correct answer is:
c. One has 99% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion