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.

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

92% Confidence interval:

Putting the values, we get:

is the anwser b. (-13)?

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