Using the z-distribution, as we have a proportion, the 95% confidence interval is (0.2316, 0.3112).
<h3>What is a confidence interval of proportions?</h3>
A confidence interval of proportions is given by:
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which:
is the sample proportion.
In this problem, we have a 95% confidence level, hence
, z is the value of Z that has a p-value of
, 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](https://tex.z-dn.net/?f=n%20%3D%20479%2C%20%5Cpi%20%3D%20%5Cfrac%7B130%7D%7B479%7D%20%3D%200.2714)
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](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.2714%20-%201.96%5Csqrt%7B%5Cfrac%7B0.2714%280.7286%29%7D%7B479%7D%7D%20%3D%200.2316)
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 + 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.3112](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.2714%20%2B%201.96%5Csqrt%7B%5Cfrac%7B0.2714%280.7286%29%7D%7B479%7D%7D%20%3D%200.3112)
The 95% confidence interval is (0.2316, 0.3112).
More can be learned about the z-distribution at brainly.com/question/25890103
Six less than "b"
i think this would be the answer
good luck!!
Y=2x or y=x+2 If you plug in the ordered pairs for both, they fit.
What do you mean suhdbebejsjdudbdndndhdnnd
the equivalent fraction is 6/8 is the answer