Respuesta :
Answer:
A sample size of 551 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 - 0.95}{2} = 0.025[/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 - 0.025 = 0.975[/tex], so Z = 1.96.
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.
Population standard deviation of 7 hours.
This means that [tex]\sigma = 7[/tex]
Determine the sample size required to have a margin of error of 0.585 hours.
This is n for which M = 0.585. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.585 = 1.96\frac{7}{\sqrt{n}}[/tex]
[tex]0.585\sqrt{n} = 1.96*7[/tex]
[tex]\sqrt{n} = \frac{1.96*7}{0.585}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*7}{0.585})^2[/tex]
[tex]n = 550.04[/tex]
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.
