Answer:
a) 0.7562
b) 0.8074
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 level of [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:
[tex]n = 1100, \pi = \frac{860}{1100} = 0.7818[/tex]
96% confidence level
So [tex]\alpha = 0.04[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.04}{2} = 0.98[/tex], so [tex]Z = 2.054[/tex].
a) The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7818 - 2.054\sqrt{\frac{0.7818*0.2182}{1100}} = 0.7562[/tex]
B) The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7818 + 2.054\sqrt{\frac{0.7818*0.2182}{1100}} = 0.8074[/tex]
