Answer:
2/3 of 15 is 10
Step-by-step explanation:
Answer:
A sample size of 285 is needed.
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1 - 0.95}{2} = 0.025](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1%20-%200.95%7D%7B2%7D%20%3D%200.025)
Now, we have to find z in the Ztable as such z has a pvalue of
.
That is z with a pvalue of
, so Z = 1.96.
Now, find the margin of error M as such
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which
is the standard deviation of the population and n is the size of the sample.
Standard deviation is $4300
This means that ![\sigma = 4300](https://tex.z-dn.net/?f=%5Csigma%20%3D%204300)
What sample size do you need to have a margin of error equal to $500 with 95% confidence?
A sample size of n is needed. n is found when M = 500. So
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![500 = 1.96\frac{4300}{\sqrt{n}}](https://tex.z-dn.net/?f=500%20%3D%201.96%5Cfrac%7B4300%7D%7B%5Csqrt%7Bn%7D%7D)
![500\sqrt{n} = 1.96*4300](https://tex.z-dn.net/?f=500%5Csqrt%7Bn%7D%20%3D%201.96%2A4300)
![\sqrt{n} = \frac{1.96*4300}{500}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.96%2A4300%7D%7B500%7D)
![n = 284.1](https://tex.z-dn.net/?f=n%20%3D%20284.1)
Rounding up
A sample size of 285 is needed.
Answer:
your answer is option B hope ur help and mark me brainlist and dont worry about problems