Question

construct the requested confidence interval.
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.
370 ± 1.645√((370)(130)/500)
0.74 ± 1.96√((0.74)(0.26)/500)
0.74 ± 1.645√((0.5)(0.5)/500)
0.74 ± 1.645√((0.74)(0.26)/500)

Explanation:

Step1: Calculate the sample proportion $\hat{p}$

The sample proportion $\hat{p}=\frac{370}{500} = 0.74$. The formula for the confidence interval for a proportion is $\hat{p}\pm z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$.

Step2: Determine the z - value

For a 90% confidence interval, the z - value (from the standard normal distribution) is $z = 1.645$. The sample size $n = 500$.

Step3: Substitute into the formula

Substitute $\hat{p}=0.74$, $z = 1.645$, and $n = 500$ into the formula $\hat{p}\pm z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. We get $0.74\pm1.645\sqrt{\frac{(0.74)(0.26)}{500}}$.

Answer:

The correct option is $0.74\pm1.645\sqrt{\frac{(0.74)(0.26)}{500}}$.