Question

A media poll of 500 air travelers surveyed in national airports found that 370 favor tighter security procedures in boarding planes. Construct a 90% confidence interval for the proportion of all air travelers who are in favor of tighter security procedures.

Answer 1

Answer:

92% Confidence interval:  (0.7078,0.7722)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 500

Number of travelers who favor tight security, x = 370

$\hat{p} = \dfrac{x}{n} = \dfrac{370}{500} = 0.74$

92% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.08} = \pm 1.645$

Putting the values, we get:

$0.74\pm 1.645(\sqrt{\dfrac{0.74(1-0.74)}{500}}) = 0.74\pm 0.0322\\\\=(0.7078,0.7722)$