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Is it wrong to conclude from the following boxplot that a passenger is more likely to experience flight delays on Thursdays and Fridays? If so, why?

Note: negative delays mean early departure.

Source: Mathematical Statistics with Resampling and R, second edition

Image source: Mathematical Statistics with Resampling and R, second edition

Mainul Islam
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  • This is a very strange diagram. Why are there delays below zero on some days? Are there planes taking off before the scheduled time (and how do they appease the passengers who came on time, only to hear that their flight has already departed?) If not, what is the definition of negative delay? – rumtscho May 06 '21 at 13:29
  • Yes, negative delays mean early departure. – Mainul Islam May 06 '21 at 13:30
  • And how do you define "a passenger is more likely to experience flight delays on Thursdays and Fridays"? – rumtscho May 06 '21 at 13:38
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    I note that for example that 168 hours means departure is delayed for a week, and I guess that happens very occasionally. I would rather guess that hours should be minutes on your vertical axis. – Nick Cox May 06 '21 at 14:21
  • Box plots don't work well for very skewed distributions. The points shown individually are just those more than 1.5 IQR from the nearer quartile if the box plot is following the most common single convention -- which, here as almost everywhere, is not explained. But what the boxplot doesn't convey clearly is what fraction of the total they are. – Nick Cox May 06 '21 at 14:24
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    @NickCox this is a very good catch! I was stumped by the idea of planes leaving 7-8 hours early (e.g. the whiskers on Monday). 7-8 minutes early is indeed quite common. – rumtscho May 06 '21 at 14:26
  • To visualize well distributions that are variously negative, zero, or positive -- but with one tail or both very stretched -- you need a transformation such as cube root, neglog or asinh. Note that median and quartiles on such a scale are one-to-one with those on the original scale, but the 1.5 IQR convention is better set on one side. – Nick Cox May 06 '21 at 14:28

1 Answers1

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I will assume that your question means:

  • We define seven populations, "passengers who fly on Monday", "passengers who fly on Tuesday" etc.
  • We divide these populations in two sets, one with the passengers from Thursday and Friday, and another with the passengers from the other days
  • We consider the claim that, for any combination of a passenger sampled from a population of the first set (flying on Thu or Fri) and a passenger of a population of the second set, the first passenger has more likelihood of experiencing a flight with a positive delay.

We cannot make any conclusion about the claim from your data

The first important consideration is: the data is insufficient for conclusions of this type anyway. We are missing the information of passengers per flight. Imagine that we have a company which flies with one full Dreamliner and two Cessnas on Monday, and the Dreamliner is always delayed, and sends out three Cessnas on Thursday, only one of which is delayed. In the picture, we will see an equal distributions of delays for both days, but many more passengers will experience a delay on Monday than on Thursday.

We cannot support your claim with additional assumptions either

Let's make the assumption that the company has flights with roughly equal number of passengers on all days of the week. If this were a real-life situation, I would certainly want to check this kind of assumption first, but since you got the picture from a textbook, it may be more related to a point the author is trying to make.

With this assumption, the claim is not correct according to the graph you posted. It was difficult to recognize, but when I placed a line on the horizontal axis, with a thickness comparable to the median lines, I found that all medians fall on it, except maybe the Sunday median:

Graph with horizontal zero line

This means that, on all days of the week, half of the flights are delayed, and half of them are not. Under the assumption of equal passenger numbers per flight, you will have the same proportion of passengers experiencing delays on all days.

There is actually a bit of a difference in the median positions, but it is on the same order of magnitude as the median line's thickness. I would consider this the visual equivalent of "the difference is not significant".

Why it might be difficult for a layperson to recognize

The boxplots are obviously extending much higher on Thursday and on Friday than on other days. So maybe people are inclined to think there must be a larger proportion of delayed passengers.

This is in fact just a skewed distribution (which is what a boxplot is intended to show). It is still half the passengers who get delayed on Thurdsay and Friday, but they are delayed for much longer times. So, if the question was "How many passengers experience a delay they care about", the claim would probably be correct - and the devious thing is that this may be the more likely interpretation of the claim (not of the graph) if made by a layperson.

rumtscho
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