Arrivals 2011 Will your flight get you to your destination on time? The U.S. Bureau of Transportation Statistics reported the percentage of flights that were late each month from 1995 through September of
2011. Here’s a histogram, along with some summary statistics:
Graph cannot copy
We can consider these data to be a representative sample of all months. There is no evidence of a time trend 1r = 0.072.
a) Check the assumptions and conditions for inference about the mean.
b) Find a 99% confidence interval for the true percentage of flights that arrive late.
c) Interpret this interval for a traveler planning to fly.
a. Conditions are met, assuming that the flights were randomly selected.
c. We are 99$\%$ confident that the true average percentage of flights arriving late each month is between 16.3959$\%$ and 17.8261$\% .$
this question. We have some information about percentage of time that flights arrive on time selected from an airport from 270 flights. And it turns out that over the time span that was selected, 19.696% of the flights were, um, on time, with a standard deviation of 4.2 percent. So we're going to be constructing a confidence interval. And before we do that, we have to check our conditions and there's two of them. The first is that we want to random independent, uh, random independent sample. So the independence assumption and we do have the information that there does not appear to be some kind of time trend. So on independent and representative, it tells us representative Sample was collected, so we're in good shape here. The second condition is that we have a nearly normal sampling distribution, so we have two things going for us. One end equals to 70. Um, and that's large. That's a very large, nice large sample, and then we also have our history, Graham and our hissed a gram is symmetric. So both of those are things that contribute to how the sampling distribution being likely normal or approximately normal. All right, so we are going to be constructing a 99% confidence interval. And it's going to be based on the two distribution because we don't know Sigma. Let me check up a condition there. So since sigma is unknown and over the population standard deviation, we only know the sample standard deviation. We're going to calculate a tea interval for one sample. So the format for that is the sample mean plus or minus the T star value times the standard error, which is the standard deviation divided by the square root of it. So first thing I need to look at on my degrees of freedom, um, and degree of freedom is equal to n minus one. So in this case, we're dealing with 2 69 So I need the t star value for 269 degrees of freedom, and I need to have that be associated with the 99% confidence interval. So if we think about a little sketch here of a normal model bad sketch, but none that nonetheless, this central piece of it is going to represent 99% So then the other 1% has to be split between the other between the two tails, So half of 1% is 10.5 percent at each tale. So if I go into my calculator and I do an inverse t the area that I'm looking for US 0.5% or 0.5 with 269 degrees of freedom and that gives me directly gives me negative two point 59 which tells me that that is this Z score boundary over here. And so, um, I could have the positive version of that over on that other end. So that is my tea star value. The why hat is given to me, as is given to me, are y bar rather and sample size. So now I have to substitute all these values into my formula. So 19 point 696 plus or minus 2.59 because we're going up and down that margin of air and the margin of error is 4.214 over the square root of 2 70 my actual sample size. And when you do that computation, you get this confidence interval 19.31 20 0.361 So what does that mean, then? What does this confidence interval means? Part C of the question then for his part b So Part C is to interpret this confidence interval so we could be 99% confident that the parameter the true mean percent of on time, not on time of delayed flights per month in our context is between 19.31%. Andi 20.361 percent.