Answer: 0.355,0.405)
Step-by-step explanation:
Given : Significance level : [tex]\alpha=1-0.99=0.01[/tex]
Number of subjects (n) = 2455
Number of surveys returned = 931
The probability of surveys get return will be :-
[tex]p=\dfrac{931}{2455}=0.379226069246\approx0.38[/tex]
The confidence interval for proportion is given by :-
[tex]p\pm z_{\alpha/2}\times\sqrt{\dfrac{p(1-p)}{n}}[/tex]
[tex]0.38\pm z_{0.005}\times\sqrt{\dfrac{0.38(1-0.38)}{2455}}\\\\=0.38\pm(2.576)0.0098\\\approx0.38\pm0.025=(0.355,0.405)[/tex]
Hence, the 99% confidence interval for the proportion of returned surveys : (0.355,0.405)
