Answer:
A sample of 18 is required.
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 - 0.92}{2} = 0.04[/tex]
Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.04 = 0.96[/tex], so Z = 1.88.
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.
A previous study indicated that the standard deviation was 2.2 days.
This means that [tex]\sigma = 2.2[/tex]
How large a sample must be selected if the company wants to be 92% confident that the true mean differs from the sample mean by no more than 1 day?
This is n for which M = 1. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]1 = 1.88\frac{2.2}{\sqrt{n}}[/tex]
[tex]\sqrt{n} = 1.88*2.2[/tex]
[tex](\sqrt{n})^2 = (1.88*2.2)^2[/tex]
[tex]n = 17.1[/tex]
Rounding up:
A sample of 18 is required.
