From the information given, we have that:
a) The sample proportion of students who are Democratic is 35%.
b) The 98% confidence interval for the proportion of all students at the university who are Democratic is (0.1018, 0.5982).
Item a:
7 out of 20 students are Democratic, hence:
[tex]\pi = \frac{7}{20} = 0.35[/tex]
Item b:
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]\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 z-score that has a p-value of [tex]\frac{1+\alpha}{2}[/tex].
From item a, we have that [tex]\pi = 0.35, n = 20[/tex]
98% confidence level, hence[tex]\alpha = 0.98[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.98}{2} = 0.99[/tex], so [tex]z = 2.327[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.35 - 2.327\sqrt{\frac{0.35(0.65)}{20}} = 0.1018[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.35 + 2.327\sqrt{\frac{0.35(0.65)}{20}} = 0.5982[/tex]
The 98% confidence interval for the proportion of all students at the university who are Democratic is (0.1018, 0.5982).
A similar problem is given at https://brainly.com/question/16807970
