Answer:
The sample proportion is 0.88.
The margin of error is of 0.04.
The interval likely to contain the true population proportion is (0.84,0.92).
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 z-score that has a p-value of
.
Margin of error:
The margin of error is of:
![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)
Sample of 230 teenagers, 202 would like to have smaller class sizes at their school.
This means that
, and that the sample proportion is:
![\pi = \frac{202}{230} = 0.88](https://tex.z-dn.net/?f=%5Cpi%20%3D%20%5Cfrac%7B202%7D%7B230%7D%20%3D%200.88)
The sample proportion is 0.88.
Standard 95% confidence level
So
, z is the value of Z that has a p-value of
, so
.
Margin of error:
![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)
![M = 1.96\sqrt{\frac{0.88*0.12}{230}}](https://tex.z-dn.net/?f=M%20%3D%201.96%5Csqrt%7B%5Cfrac%7B0.88%2A0.12%7D%7B230%7D%7D)
![M = 0.04](https://tex.z-dn.net/?f=M%20%3D%200.04)
The margin of error is of 0.04.
Confidence interval:
Sample proportion plus/minus margin of error. So
0.88 - 0.04 = 0.84
0.88 + 0.04 = 0.92
The interval likely to contain the true population proportion is (0.84,0.92).