Answer:
A sample size of 6755 or higher would be appropriate.
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 M is given by:
![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
.
52% of Independents in the sample opposed the public option.
This means that ![p = 0.52](https://tex.z-dn.net/?f=p%20%3D%200.52)
If we wanted to estimate this number to within 1% with 90% confidence, what would be an appropriate sample size?
Sample size of size n or higher when
. 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)
![0.01 = 1.645\sqrt{\frac{0.52*0.48}{n}}](https://tex.z-dn.net/?f=0.01%20%3D%201.645%5Csqrt%7B%5Cfrac%7B0.52%2A0.48%7D%7Bn%7D%7D)
![0.01\sqrt{n} = 0.8218](https://tex.z-dn.net/?f=0.01%5Csqrt%7Bn%7D%20%3D%200.8218)
![\sqrt{n} = \frac{0.8218}{0.01}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B0.8218%7D%7B0.01%7D)
![\sqrt{n} = 82.18](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%2082.18)
![\sqrt{n}^{2} = (82.18)^{2}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%5E%7B2%7D%20%3D%20%2882.18%29%5E%7B2%7D)
![n = 6754.2](https://tex.z-dn.net/?f=n%20%3D%206754.2)
A sample size of 6755 or higher would be appropriate.
The one you selected is right... i just punched it into the calculator and that's correct.
37/10, 3 2/5, 3 1/4. u put in order because decimal= 3.7, 3.4, 3.25
Answer:
0, 5, 10, 15, 20
0, 5, 40, 405, 5,120
Step-by-step explanation:
One rule can be "add 5 to the previous term to get the next term."
0, 5, 10, 15, 20
Second rule:
For each term number 1, 2, 3, ..., n,
each term = 5(n - 1)^n
0, 5, 40, 405, 5,120
This is the answer for the math