Respuesta :
Answer:
The sample size necessary is of 168.
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 Z-table as such z has a p-value 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.
Based on a previous study, arrival delay times have a standard deviation of 39.6 minutes.
This means that [tex]\sigma = 39.6[/tex]
Find the sample size necessary to estimate the mean arrival delay time for all American Airlines flights from Dallas to Sacramento to within 6 minutes with 95% confidence.
This is n for which M = 6. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]6 = 1.96\frac{39.6}{\sqrt{n}}[/tex]
[tex]6\sqrt{n} = 1.96*39.6[/tex]
[tex]\sqrt{n} = \frac{1.96*39.6}{6}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*39.6}{6})^2[/tex]
[tex]n = 167.34[/tex]
Rounding up:
The sample size necessary is of 168.
