Question

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. Based on a previous study, arrival delay times have a standard deviation of 39.6 minutes.

Answer 1

Answer:

The sample size necessary is of 168.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$.

That is z with a pvalue of $1 - 0.025 = 0.975$, so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ 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 $\sigma = 39.6$

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

$M = z\frac{\sigma}{\sqrt{n}}$

$6 = 1.96\frac{39.6}{\sqrt{n}}$

$6\sqrt{n} = 1.96*39.6$

$\sqrt{n} = \frac{1.96*39.6}{6}$

$(\sqrt{n})^2 = (\frac{1.96*39.6}{6})^2$

$n = 167.34$

Rounding up:

The sample size necessary is of 168.