Answer:
We need a sample size of 2,071,800 or higher.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.9}{2} = 0.05[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]
Now, we 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.
In this problem:
We need a sample size of n or higher, when [tex]M = 0.04, \sigma = 35[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.04 = 1.645*\frac{35}{\sqrt{n}}[/tex]
[tex]0.04\sqrt{n} = 1.645*35[/tex]
[tex]0.04\sqrt{n} = 57.575[/tex]
[tex]\sqrt{n} = 1439.375[/tex]
[tex]\sqrt{n}^{2} = (1439.375)^{2}[/tex]
[tex]n = 2,071,800[/tex]
We need a sample size of 2,071,800 or higher.
