Answer:
We must survey at least 48 people from this younger age group.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
![\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
z is the zscore that has a pvalue of
.
The margin of error is:
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
90% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
How many of this younger age group must we survey in order to estimate the proportion of non-grads to within .10 with 90% confidence? Use the value of p from the over-50 age group.
This is n when ![M = 0.1, \pi = 0.23](https://tex.z-dn.net/?f=M%20%3D%200.1%2C%20%5Cpi%20%3D%200.23)
So
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
![0.1 = 1.645\sqrt{\frac{0.23*0.77}{n}}](https://tex.z-dn.net/?f=0.1%20%3D%201.645%5Csqrt%7B%5Cfrac%7B0.23%2A0.77%7D%7Bn%7D%7D)
![0.1\sqrt{n} = 1.645\sqrt{0.23*0.77}](https://tex.z-dn.net/?f=0.1%5Csqrt%7Bn%7D%20%3D%201.645%5Csqrt%7B0.23%2A0.77%7D)
![\sqrt{n} = \frac{1.645\sqrt{0.23*0.77}}{0.1}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.645%5Csqrt%7B0.23%2A0.77%7D%7D%7B0.1%7D)
![(\sqrt{n})^{2} = (\frac{1.645\sqrt{0.23*0.77}}{0.1})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B1.645%5Csqrt%7B0.23%2A0.77%7D%7D%7B0.1%7D%29%5E%7B2%7D)
![n = 47.92](https://tex.z-dn.net/?f=n%20%3D%2047.92)
Rouding up,
We must survey at least 48 people from this younger age group.
7 because all the room contain 1//2 and Brandon has 7/1/2 i t pretty easy <span />
They're probably asking for the total payment, so add $720 to the initial amount, $2,000, so the answer is $2,720.