Question

There are big spenders among University of Alabama football season ticket holders. This data set Roll Tide!! shows the dollar amount of the purchases made at the previous two Alabama home football games by 479 season ticket holders randomly selected from all season ticket holders. The Univ. Of Alabama marketing department is considering a special event hosted by the governor at the next home game for season ticket holders who spent $1000 or more at the previous two home football games. From these data construct a 95% confidence interval for the proportion p of all Alabama football season ticket holders who will qualify for the special event

Answer 1

Using the z-distribution, as we have a proportion, the 95% confidence interval is (0.2316, 0.3112).

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

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

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

In this problem, we have a 95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so the critical value is z = 1.96.

We also consider that 130 out of the 479 season ticket holders spent $1000 or more at the previous two home football games, hence:

$n = 479, \pi = \frac{130}{479} = 0.2714$

Hence the bounds of the interval are found as follows:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 - 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.2316$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 + 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.3112$

The 95% confidence interval is (0.2316, 0.3112).

More can be learned about the z-distribution at brainly.com/question/25890103