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Cincinnati/NorthernKentucky International Airport hadthe secondhighest on-time arrivalrate for 2005 among the nation's busiest airports (The Cincinnati Enquirer, February $3,$$2006 ) .$ Assume the findings were based on 455 on-time arrivals out of a sample of550 flights.$$\begin{array}{l}{\text { a. Develop a point estimate of the on-time arrival rate (proportion of flights arriving on }} \\ {\text { time) for the airport. }}\end{array}$$$$\begin{array}{l}{\text { b. Construct a } 95 \% \text { confidence interval for the on-time amival rate of the population of }} \\ {\text { all flights at the airport during } 2005 .}\end{array}$$
a. 0.8273b. 0.7957 to 0.8589
Intro Stats / AP Statistics
Chapter 8
Interval Estimation
Confidence Intervals
Temple University
University of North Carolina at Chapel Hill
Oregon State University
Boston College
Lectures
0:00
01:03
On-Time Airplane Arrivals …
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On-Time Flights A flight f…
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all right, This question gives us that in a sample of 550 flights, 455 of those were on time. So the first part wants us to estimate the population proportion with the point estimate. So Rp hat equals the number of successes over our sample size, which is in this case, the 455 on time flights divided by the 5 50 total, which works out to be 0.8273 So, no, it wants us to calculate the 95% confidence interval. So confidence interval is P hat plus or minus our margin of air, which is the special Z score for the confidence level. Times the square root of P Hat Q hat over end, which in this case, P hat it's 0.8273 plus or minus. Our special Z score, which for 95% confidence is the 1.96 times the square root of P hat. Once again, 0.8273 times Q hat, which is one minus that so point 17 two seven, all divided by the sample size, which is in this case 5 50 So now plugging all this in, we get our confidence. Interval ranges from 0.7957 on the low end, two 0.85 89 on the high end. So we say, with 95% confidence that the true proportion of on time flights is between 0.7957 and 0.8589
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