Answer:
The minimum sample size is 1,704.
Explanation:
We have to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Z-table as such z has a p-value of
.
That is z with a p-value of
, so Z = 1.96.
Now, find the margin of error M as such
In which
is the standard deviation of the population and n is the size of the sample.
Standard deviation of 21,059,637 shares
This means that
![\sigma = 21059637](https://img.qammunity.org/2022/formulas/mathematics/college/roo107zuzz5s14oh4v6442bkwfn0gfmdyx.png)
What is the minimum required sample size if you would like your sampling error to be limited to 1,000,000 shares?
This is n for which
, so:
![M = z(\sigma)/(√(n))](https://img.qammunity.org/2022/formulas/mathematics/college/p19w5m3ctzqxc0b7ic9kz7y4ab19d7zpbv.png)
![1000000 = 1.96(21059637)/(√(n))](https://img.qammunity.org/2022/formulas/mathematics/college/22ozmm3cac2bj0b232lph732vkg29a6i6s.png)
![1000000√(n) = 1.96*21059637](https://img.qammunity.org/2022/formulas/mathematics/college/fr80zkv6uax0zv4fvlk7nk0t1rxem5a14l.png)
![√(n) = (1.96*21059637)/(1000000)](https://img.qammunity.org/2022/formulas/mathematics/college/q8e8hdde8lyrdz3uxv82378j6p6oykq06l.png)
![(√(n))^2 = ((1.96*21059637)/(1000000))^2](https://img.qammunity.org/2022/formulas/mathematics/college/xzenwo07t3a66guwl0anzftb0eywnnizab.png)
![n = 1703.8](https://img.qammunity.org/2022/formulas/mathematics/college/qvhllumck18ll8osmdzyia1pg7dkxcyfey.png)
Rounding up:
The minimum sample size is 1,704.