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shay03littletop5kx2p

15.02.2023

Mathematics

You must survey 784 air passengers.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so Z = 1.96 .

Assume that nothing is known about the percentage of passengers who prefer aisle seats.

This means that $\pi = 0.5$ , which is when the largest sample size will be needed.

Within 3.5 percentage points of the true population percentage.

We have to find n for which M = 0.035. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.035 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.035\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.035}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.035})^2$

n = 784

You must survey 784 air passengers.

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