Respuesta :
Answer:
[tex](\frac{4\sigma}{2.056})^2[/tex], rounded up, if needed, citizens should be included in the sample, in which [tex]\sigma[/tex] is the standard deviation of the population.
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 - 0.96}{2} = 0.02[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.02 = 0.98[/tex], so Z = 2.056.
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.
Within 4 years of the actual mean
We have to find n for which M = 4. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]4 = 2.056\frac{\sigma}{\sqrt{n}}[/tex]
[tex]2.056\sqrt{n} = 4\sigma[/tex]
[tex]\sqrt{n} = \frac{4\sigma}{2.056}[/tex]
[tex](\sqrt{n})^2 = (\frac{4\sigma}{2.056})^2[/tex]
[tex]n = (\frac{4\sigma}{2.056})^2[/tex]
[tex](\frac{4\sigma}{2.056})^2[/tex], rounded up, if needed, citizens should be included in the sample, in which [tex]\sigma[/tex] is the standard deviation of the population.
