Answer:
We need a sample size of 2,071,800 or higher.
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{1-0.9}{2} = 0.05](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1-0.9%7D%7B2%7D%20%3D%200.05)
Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so ![z = 1.645](https://tex.z-dn.net/?f=z%20%3D%201.645)
Now, we find the margin of error M as such
![M = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%2A%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.
In this problem:
We need a sample size of n or higher, when
. So
![M = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![0.04 = 1.645*\frac{35}{\sqrt{n}}](https://tex.z-dn.net/?f=0.04%20%3D%201.645%2A%5Cfrac%7B35%7D%7B%5Csqrt%7Bn%7D%7D)
![0.04\sqrt{n} = 1.645*35](https://tex.z-dn.net/?f=0.04%5Csqrt%7Bn%7D%20%3D%201.645%2A35)
![0.04\sqrt{n} = 57.575](https://tex.z-dn.net/?f=0.04%5Csqrt%7Bn%7D%20%3D%2057.575)
![\sqrt{n} = 1439.375](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%201439.375)
![\sqrt{n}^{2} = (1439.375)^{2}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%5E%7B2%7D%20%3D%20%281439.375%29%5E%7B2%7D)
![n = 2,071,800](https://tex.z-dn.net/?f=n%20%3D%202%2C071%2C800)
We need a sample size of 2,071,800 or higher.