Answer:
a) n = 1037.
b) n = 1026.
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 ![1 - \frac{\alpha}{2}](https://tex.z-dn.net/?f=1%20-%20%5Cfrac%7B%5Calpha%7D%7B2%7D)
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)
99% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
(a) Assume that nothing is known about the percentage to be estimated.
We need to find n when M = 0.04.
We dont know the percentage to be estimated, so we use
, which is when we are going to need the largest sample size.
![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)
![0.04 = 2.575\sqrt{\frac{0.5*0.5}{n}}](https://tex.z-dn.net/?f=0.04%20%3D%202.575%5Csqrt%7B%5Cfrac%7B0.5%2A0.5%7D%7Bn%7D%7D)
![0.04\sqrt{n} = 2.575*0.5](https://tex.z-dn.net/?f=0.04%5Csqrt%7Bn%7D%20%3D%202.575%2A0.5)
![(\sqrt{n}) = \frac{2.575*0.5}{0.04}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%20%3D%20%5Cfrac%7B2.575%2A0.5%7D%7B0.04%7D)
![(\sqrt{n})^{2} = (\frac{2.575*0.5}{0.04})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B2.575%2A0.5%7D%7B0.04%7D%29%5E%7B2%7D)
![n = 1036.03](https://tex.z-dn.net/?f=n%20%3D%201036.03)
Rounding up
n = 1037.
(b) Assume prior studies have shown that about 55% of full-time students earn bachelor's degrees in four years or less.
![\pi = 0.55](https://tex.z-dn.net/?f=%5Cpi%20%3D%200.55)
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)
![0.04 = 2.575\sqrt{\frac{0.55*0.45}{n}}](https://tex.z-dn.net/?f=0.04%20%3D%202.575%5Csqrt%7B%5Cfrac%7B0.55%2A0.45%7D%7Bn%7D%7D)
![0.04\sqrt{n} = 2.575*\sqrt{0.55*0.45}](https://tex.z-dn.net/?f=0.04%5Csqrt%7Bn%7D%20%3D%202.575%2A%5Csqrt%7B0.55%2A0.45%7D)
![(\sqrt{n}) = \frac{2.575*\sqrt{0.55*0.45}}{0.04}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%20%3D%20%5Cfrac%7B2.575%2A%5Csqrt%7B0.55%2A0.45%7D%7D%7B0.04%7D)
![(\sqrt{n})^{2} = (\frac{2.575*\sqrt{0.55*0.45}}{0.04})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B2.575%2A%5Csqrt%7B0.55%2A0.45%7D%7D%7B0.04%7D%29%5E%7B2%7D)
![n = 1025.7](https://tex.z-dn.net/?f=n%20%3D%201025.7)
Rounding up
n = 1026.
Answer:
for this we need 90 percent 5410
so we multiply 5410*0.9=4869
Hope This Helps!!!
Answer:
-3 was the low and the high temp. is 7
Step-by-step explanation:
By assuming the standard deviation of population 2.2 the confidence interval is 8.67 toys,8.94 toys.
Given sample size of 1492 children,99% confidence interval , sample mean of 8.8, population standard deviation=2.2.
This type of problems can be solved through z test and in z test we have to first find the z score and then p value from normal distribution table.
First we have to find the value of α which can be calculated as under:
α=(1-0.99)/2=0.005
p=1-0.005=0.995
corresponding z value will be 2.575 for p=0.995 .
Margin of error=z*x/d
where x is mean and d is standard deviation.
M=2.575*2.2/![\sqrt{1492}](https://tex.z-dn.net/?f=%5Csqrt%7B1492%7D)
=0.14
So the lower value will be x-M
=8.8-0.14
=8.66
=8.67 ( after rounding)
The upper value will be x+M
=8.8+0.14
=8.94
Hence the confidence interval will be 8.67 toys and 8.94 toys.
Learn more about z test at brainly.com/question/14453510
#SPJ4