Answer:
The sample size needed is at least 17,074 students.
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{1-0.95}{2} = 0.025[/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]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
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.
Sample size:
At least n, in which N is found when [tex]M = 0.03, \sigma = 2[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.03 = 1.96*\frac{2}{\sqrt{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96*2[/tex]
[tex]\sqrt{n} = \frac{1.96*2}{0.03}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96*2}{0.03})^{2}[/tex]
[tex]n = 17074[/tex]
The sample size needed is at least 17,074 students.
