Cincinnati/Northern Kentucky International Airport had the second highest on- time arrival rate 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 of 550 flights. a. Develop a point estimate of the on-time arrival rate (proportion of flights arriving on time) for the airport. b. Construct a \(95 \%\) confidence interval for the on-time arrival rate of the population of all flights at the airport during 2005

Short Answer

Expert verified
The point estimate for the on-time arrival rate is 0.827 (82.7%). The 95% confidence interval for the on-time arrival rate of the population of all flights at the airport during 2005 is approximately (0.794, 0.860) or (79.4%, 86.0%).

Step by step solution

01

Calculate the point estimate (sample proportion)

The point estimate (sample proportion) is calculated by dividing the number of on-time arrivals by the total number of flights in the sample: \( p̂ = \frac{\text{number of on-time arrivals}}{\text{total number of flights}} \) Plug in the values: \( p̂ = \frac{455}{550} \) Now, solve for the point estimate: \( p̂ = 0.827 \) The point estimate for the on-time arrival rate is 0.827 (82.7%).
02

Find the standard error

The standard error for a proportion is given by: \( SE = \sqrt{\frac{p̂(1 - p̂)}{n}} \) Plug in the values: \( SE = \sqrt{\frac{0.827(1 - 0.827)}{550}} \) Now, solve for the standard error: \( SE \approx 0.017 \) The standard error for the proportion is approximately 0.017.
03

Find the critical value (z-score) for a 95% confidence interval

A 95% confidence interval corresponds to the area under the standard normal distribution curve where 95% of the data lies. Therefore, the critical value (z-score) corresponds to a 97.5% percentile, due to the symmetry of the curve. Using a standard normal distribution table or calculator, we find: \( z_{0.975} = 1.96 \) The critical value (z-score) for a 95% confidence interval is 1.96.
04

Calculate the margin of error

The margin of error is given by: \( Margin\,of\,Error = z_{0.975} \times SE \) Plug in the values: \( Margin\,of\,Error = 1.96 \times 0.017 \) Now, solve for the margin of error: \( Margin\,of\,Error \approx 0.033 \) The margin of error is approximately 0.033.
05

Construct the 95% confidence interval

The confidence interval is constructed by adding and subtracting the margin of error from the point estimate: \( CI = (p̂ - Margin\,of\,Error,\, p̂ + Margin\,of\,Error) \) Plug in the values: \( CI = (0.827 - 0.033,\, 0.827 + 0.033) \) Simplify: \( CI = (0.794, 0.860) \) The 95% confidence interval for the on-time arrival rate of the population of all flights at the airport during 2005 is approximately (0.794, 0.860) or (79.4%, 86.0%).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Point Estimate
When we want to give a single best guess about some quantity in a population, we use a point estimate. It's like taking a snapshot with a camera; it captures one aspect at a particular moment, but not everything going on around it. In the case of Cincinnati/Northern Kentucky International Airport's on-time arrival rate, this 'snapshot' is the ratio of flights that actually arrived on time out of the total flights observed. By dividing 455 on-time arrivals by 550 total flights, we got a point estimate of 0.827. This means our best initial guess is that approximately 82.7% of all flights at the airport arrive on time.
Sample Proportion
The sample proportion is essentially the point estimate when we're talking about a categorical variable—in this case, whether flights are on-time or not. It takes all the 'yes' instances (on-time flights, here) and divides them by the total number of instances we looked at (all the flights in the sample). It's a crucial piece of our puzzle because it's our real-world evidence; it tells us what we've actually observed in our sample, which we then use to infer what's true for the broader population of all flights.
Standard Error
Imagine if you took multiple samples. Each sample might give you a slightly different point estimate, right? The standard error tells us, on average, how much these sample proportions might fluctuate from one sample to another due to random sampling error. It gives us an idea of how precise our point estimate is. In the airport example, we calculated a standard error of about 0.017, which means we'd expect most of our samples' proportion estimates to fall within ±0.017 of our point estimate if we repeated the sampling.
Margin of Error
How confident can we be that the point estimate we've got falls within a certain range of the true population proportion? That's where the margin of error comes in. It accounts for the uncertainty and random sampling error in our point estimate. When we multiply the standard error by a z-score (which we'll cover next), we create this margin around our estimate, giving us a range we're confident contains the true population proportion. For the airport's on-time arrival rate, our margin of error is approximately 0.033. This means we believe the true proportion is no more than 3.3% away from our sample proportion, in either direction, with 95% confidence.
Z-score
To understand z-score, think about the average height of people in a room. Most people will be around the average, with few much shorter or taller; this distribution creates what we might call a 'bell curve' or normal distribution. A z-score tells us how many standard deviations away from the average we are. For confidence intervals and the margin of error, we look at z-scores that correspond to our level of confidence. The 1.96 z-score we used for the 95% confidence interval tells us that if we were to take many samples, about 95% of the time, the true population proportion would fall within 1.96 standard errors of our sample proportion (that is, our point estimate). We use z-scores to help us understand the spread of our data and to calculate the margin of error in estimating proportions.

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