Answer:
We need to contact 542 employees.
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)
98% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
a) How many randomly selected employers must we contact in order to create an estimate in which we are 98% confident with a margin of error of 5%
We dont know the proportion, so we use
, which is when we are going to need the largest sample size.
We have to find n when M = 0.05. 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)
![0.05 = 2.327\sqrt{\frac{0.5*0.5)}{n}}](https://tex.z-dn.net/?f=0.05%20%3D%202.327%5Csqrt%7B%5Cfrac%7B0.5%2A0.5%29%7D%7Bn%7D%7D)
![0.05\sqrt{n} = 2.327*0.5](https://tex.z-dn.net/?f=0.05%5Csqrt%7Bn%7D%20%3D%202.327%2A0.5)
![\sqrt{n} = \frac{2.327*0.5}{0.05}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B2.327%2A0.5%7D%7B0.05%7D)
![(\sqrt{n})^{2} = (\frac{2.327*0.5}{0.05})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B2.327%2A0.5%7D%7B0.05%7D%29%5E%7B2%7D)
![n = 541.49](https://tex.z-dn.net/?f=n%20%3D%20541.49)
Rounding up
We need to contact 542 employees.
Equating the formula (d is the diameter)
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
d^2/4 = 132.7
d^2 = 132.7*4/
![\pi](https://tex.z-dn.net/?f=%20%5Cpi%20)
d =
![\sqrt{169.044586}](https://tex.z-dn.net/?f=%20%5Csqrt%7B169.044586%7D%20)
d = 13.0017147331
d= 13cm
X = 75°
Y = 105°
Z =75°
<em> </em><em><u>I Took the Assignment</u></em>
Answer:
a. The population refers to targeted area under study which is all the backpacks of concertgoers
b. The sample refers to that part of the population inspected for study which is the number of backpacks checked in this case
c. P is the percentage or proportion of all backpacks of concertgoers that contained alcoholic beverages
e Using a confidence interval of 95%, I will.
This is because the true proportion lies between 7.3% and 18.9% and the data given in the question has an estimate of 13.1% which lies within the range of values
Step-by-step explanation:
PLEASE CHECK ATTACHMENT FOR EXPLANATION AND STEP BY STEP SOLUTION