Answer:
The minimum sample size required to create the specified confidence interval is 2229.
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{1-0.99}{2} = 0.005[/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]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/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.
What is the minimum sample size required to create the specified confidence interval
This is n when [tex]M = 0.06, \sigma = 1.1[/tex]
[tex]0.06 = 2.575*\frac{1.1}{\sqrt{n}}[/tex]
[tex]0.06\sqrt{n} = 2.575*1.1[/tex]
[tex]\sqrt{n} = \frac{2.575*1.1}{0.06}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.575*1.1}{0.06})^{2}[/tex]
[tex]n = 2228.6[/tex]
Rounding up
The minimum sample size required to create the specified confidence interval is 2229.
