Answer:
0.3876<p<0.4389
Step-by-step explanation:
-Given [tex]n=2472, \ x=1022 , \ CI=0.99[/tex]
-We calculate the proportion of surveys returned:
[tex]\hat p=\frac{1022}{2472}\\\\=0.4134[/tex]
For a 99% confidence interval:
[tex]z_{\alpha/2}=2.576[/tex]
#The margin of error is calculated as;
[tex]ME=z_{0.005}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}\\\\=2.576\times \sqrt{\frac{0.4134(1-0.4134)}{2472}}\\\\=0.0255[/tex]
The confidence interval are then:
[tex]CI=\hat p\pm ME\\\\=0.4134\pm 0.0255\\\\=[0.3876,0.4389][/tex]
Hence, the confidence interval is 0.3876<p<0.4389
