Answer:
A sample size of 554 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 known to be $12,000
This means that ![\sigma = 12000](https://tex.z-dn.net/?f=%5Csigma%20%3D%2012000)
What sample size do you need to have a margin of error equal to $1000, with 95% confidence?
This is n for which M = 1000. 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)
![1000 = 1.96\frac{12000}{\sqrt{n}}](https://tex.z-dn.net/?f=1000%20%3D%201.96%5Cfrac%7B12000%7D%7B%5Csqrt%7Bn%7D%7D)
![1000\sqrt{n} = 1.96*12](https://tex.z-dn.net/?f=1000%5Csqrt%7Bn%7D%20%3D%201.96%2A12)
Dividing both sides by 1000:
![\sqrt{n} = 1.96*12](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%201.96%2A12)
![(\sqrt{n})^2 = (1.96*12)^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%281.96%2A12%29%5E2)
![n = 553.2](https://tex.z-dn.net/?f=n%20%3D%20553.2)
Rounding up:
A sample size of 554 is needed.