Answer:
The sample size must be of 47,059,600.
Step-by-step explanation:
We have to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]
Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].
That is z with a p-value of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
Standard deviation:
[tex]\sigma = 3.5[/tex]
If you want the error bound E of a 95% confidence interval to be less than 0.001, how large must the sample size n be?
This is n for which M = 0.001. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.001 = 1.96\frac{3.5}{\sqrt{n}}[/tex]
[tex]0.001\sqrt{n} = 1.96*3.5[/tex]
[tex]\sqrt{n} = \frac{1.96*3.5}{0.001}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*3.5}{0.001})^2[/tex]
[tex]n = 47059600[/tex]
The sample size must be of 47,059,600.
