Question

At the movies: A movie theater is considering a showing of The Princess Bride for a 80's thowback night. In order to ensure the success of the evening, they've asked a random sample of 53 patrons whether they would come to the showing or not. Of the 53 patrons, 18 said that they would come to see the film. Construct a 95% confidence interval to determine the true proportion of all patrons who would be interested in attending the showing.

Answer 1

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 $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

Of the 53 patrons, 18 said that they would come to see the film.

This means that $n = 53, \pi = \frac{18}{53} = 0.3396$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3396 - 1.96\sqrt{\frac{0.3396*0.6604}{53}} = 0.2121$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3396 + 1.96\sqrt{\frac{0.3396*0.6604}{53}} = 0.4671$

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).