Answer :
Answer:
We need a sample size of at least 1698.
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.
You want the margin of error to be only one minute. What sample size do you need if the confidence level is 99%?
This is at least n when [tex]M = 1, \sigma = 16[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]1 = 2.575*\frac{16}{\sqrt{n}}[/tex]
[tex]\sqrt{n} = 2.575*16[/tex]
[tex](\sqrt{n})^{2} = (2.575*16)^{2}[/tex]
[tex]n = 1697.4[/tex]
Rouding up
We need a sample size of at least 1698.