Answer:
The 90% confidence interval for the proportion of students attending a football game is (0.6587, 0.7049).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
A survey of an urban university showed that 750 or 1,100 students sampled attended a home football game during the season. This means that [tex]n = 1100, \pi = \frac{750}{1100} = 0.6818[/tex]
90% confidence interval
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6818 - 1.645\sqrt{\frac{0.6818*0.3182}{1100}} = 0.6587[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6818 + 1.645\sqrt{\frac{0.6818*0.3182}{1100}} = 0.7049[/tex]
The 90% confidence interval for the proportion of students attending a football game is (0.6587, 0.7049).
