Answer: [tex](0.644,\ 0.716)[/tex]
Step-by-step explanation:
Given : Sample size = 1100
The number of students attended a home football game during the season = 750
Then, the proportion of students attending a football game =[tex]\dfrac{750}{1100}=0.681818181818\approx0.68[/tex]
Critical value = [tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.576[/tex]
Now, the confidence interval for the population proportion is given by :-
[tex]p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}\\\\=0.68\pm (2.576)\sqrt{\dfrac{0.68(0.32)}{1100}}\\\\\approx0.68\pm0.036\\\\=(0.644,0.716)[/tex]
Hence, the 99% confidence interval for the proportion of students attending a football game = [tex](0.644,\ 0.716)[/tex]
