Answer:
The required sample size is 39.
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].
That is z with a pvalue of [tex]1 - 0.005 = 0.995[/tex], so Z = 2.575.
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.
Minimum sample size, within 1 minute of the population mean, population standar deviation of 2.4 minutes.
This minimum sample size is given by n, which is found when [tex]M = 1[/tex], and we have that [tex]\sigma = 2.4[/tex]. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]1 = 2.575\frac{2.4}{\sqrt{n}}[/tex]
[tex]\sqrt{n} = 2.575*2.4[/tex]
[tex](\sqrt{n})^2 = (2.575*2.4)^2[/tex]
[tex]n = 38.2[/tex]
Rounding up
The required sample size is 39.
