Answer:
A sample size of 551 is required.
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1 - 0.95}{2} = 0.025](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1%20-%200.95%7D%7B2%7D%20%3D%200.025)
Now, we have to find z in the Ztable as such z has a pvalue of
.
That is z with a pvalue of
, so Z = 1.96.
Now, find the margin of error M as such
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which
is the standard deviation of the population and n is the size of the sample.
Population standard deviation of 7 hours.
This means that ![\sigma = 7](https://tex.z-dn.net/?f=%5Csigma%20%3D%207)
Determine the sample size required to have a margin of error of 0.585 hours.
This is n for which M = 0.585. So
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![0.585 = 1.96\frac{7}{\sqrt{n}}](https://tex.z-dn.net/?f=0.585%20%3D%201.96%5Cfrac%7B7%7D%7B%5Csqrt%7Bn%7D%7D)
![0.585\sqrt{n} = 1.96*7](https://tex.z-dn.net/?f=0.585%5Csqrt%7Bn%7D%20%3D%201.96%2A7)
![\sqrt{n} = \frac{1.96*7}{0.585}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.96%2A7%7D%7B0.585%7D)
![(\sqrt{n})^2 = (\frac{1.96*7}{0.585})^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%28%5Cfrac%7B1.96%2A7%7D%7B0.585%7D%29%5E2)
![n = 550.04](https://tex.z-dn.net/?f=n%20%3D%20550.04)
Rounding up(as for a sample of 550 the margir of error will be a bit above the desired target):
A sample size of 551 is required.