Answer:
The sample size should be of at least 246.
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]
The margin of error M is
[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 have that:
[tex]\sigma = 2[/tex]
We want the margin of error to be of 3 months. However the standard deviation is in years. So the margin of error must be in years. So M = 3/12 = 0.25.
As n increases, the margin of error decreases. So we need a sample of size at least n when M = 0.25.
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.25 = 1.96*\frac{2}{\sqrt{n}}[/tex]
[tex]0.25\sqrt{n} = 3.92[/tex]
[tex]\sqrt{n} = 15.68[/tex]
[tex]\sqrt{n}^{2} = (15.68)^{2}[/tex]
[tex]n = 246[/tex]
The sample size should be of at least 246.
