Answer:
The 95% confidence interval to determine the true proportion of all patrons who would be interested in attending the showing is (0.2121, 0.4671).
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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
Of the 53 patrons, 18 said that they would come to see the film.
This means that [tex]n = 53, \pi = \frac{18}{53} = 0.3396[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3396 - 1.96\sqrt{\frac{0.3396*0.6604}{53}} = 0.2121[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3396 + 1.96\sqrt{\frac{0.3396*0.6604}{53}} = 0.4671[/tex]
The 95% confidence interval to determine the true proportion of all patrons who would be interested in attending the showing is (0.2121, 0.4671).
