Shifting the origin: why guessing a centre cannot change the answer

0:00Here is a table of thirty students' marks, sorted into six classes. To find the mean by the direct method you multiply each class mark by its frequency, total the column, and divide by thirty. Nothing about that is hard. Look at the column it makes. Six products, most of them three figures wide, and two of them carrying a half. The complaint is not that the method is wrong. The complaint is the size of what your hand has to write.

0:33So somebody had an idea. The class marks are large because they are being measured from nought, and nought is nowhere near this data. Measure them from somewhere closer, and they get small. The whole of this video is about whether you are allowed to do that. Put the six class marks on a line, fifteen apart, starting at seventeen and a half. Now plant a flag at one of them — say forty-seven and a half — and measure everything from the flag instead of from nought.

1:07The third mark is at the flag, so it becomes zero. The two marks below it become minus fifteen and minus thirty. The three above become fifteen, thirty and forty-five. That is the deviation column, and it is what the method is named after: each value's signed gap from a chosen number. The chosen number is called the assumed mean, and it is written a. Hold on to that word, assumed. By the end of this you are going to want it back.

1:41Look at what has happened to the numbers. The largest entry was ninety-two and a half. The largest entry now is forty-five. And the halves are gone, because every class mark ends in a half and so does the flag, so they cancel in the subtraction. And two of the six entries are negative. That is not damage. That is the whole mechanism. The marks below the flag pull the answer down and the marks above pull it up, and the signs are what let them cancel.

2:15Make the deviations positive because negatives look untidy and you have not simplified anything. You have computed a different quantity, and we will see exactly what it comes to. Two negative, one nought, three positive. Keep the signs. Now do the same work as before, on the new column. Multiply each deviation by its frequency. Minus sixty. Minus forty-five. Then nought times seven, which is nothing at all — that whole row costs you no arithmetic.

2:51Then ninety, one hundred and eighty, two hundred and seventy. Total the column, minding the signs: four hundred and thirty-five. Share that among the thirty students and you get fourteen and a half. Fourteen and a half is not the mean. It is the average of the deviations — the average distance from the flag, counted with sign. So add the flag back. Forty-seven and a half plus fourteen and a half.

3:21Sixty-two. Which is exactly what the direct method gives, off a column of three-figure numbers. Now, that could be a coincidence. One table, one guess, one lucky agreement. So put it on trial. Move the flag. Take a as sixty-two and a half instead — one class mark further along. Every deviation changes, and so does every product. And the total changes sign: minus fifteen. Minus fifteen shared among thirty is minus a half.

3:57And sixty-two and a half, minus a half, is sixty-two. The columns share not one entry with the columns before. The answer is the same number. So either that is a second coincidence, or the shortfall is doing something exact. It is doing something exact, and it takes four lines to see. Start with the thing we actually computed: the average of the deviation column. Sum of f times d, over sum of f.

4:30Line two. Every d is just x minus a, so write that in. Line three. A sum of differences is the difference of the sums, so split it in two. Line four. Split the division across the subtraction, and you have two separate fractions. The first is sum of f x over sum of f. That is the mean — the thing we were trying to find in the first place.

4:57The second is sum of f times a, over sum of f. And a is a constant: it does not vary from class to class, so it comes out of the sum. Leaving a times sum of f, over sum of f. Which is the only interesting step in the argument. A times sum of f, over sum of f. Sum of f is the number of observations. So you have taken one number, written it down once for every student in the table, added them all up, and divided by how many students there are.

5:34You get a back. The average of a constant is that constant. Say it out loud and it sounds like nothing. It is the load-bearing line of the method. The sums cancel — not approximately, exactly, because they are the same number top and bottom. So the second fraction is a. Not roughly a. A. And the four lines have come out to this: the average of the deviations is the mean, minus a.

6:04The column does not fall short by some amount that depends on how good your guess was. It falls short by the guess. Add a to both sides and the method is finished: the mean is a, plus the average of the deviations. So try every class mark in the table as the flag, and watch the columns. Six completely different columns, and six completely different totals. One thousand three hundred and thirty-five. Eight hundred and eighty-five. Four hundred and thirty-five. Minus fifteen. Minus four hundred and sixty-five. Minus nine hundred and fifteen.

6:48Now look at that as a sequence rather than a list. Each one is four hundred and fifty below the one before it. Every single time. And four hundred and fifty is thirty times fifteen — thirty students, and fifteen is how far the flag moved each step. The shortfall does not drift with the guess. It tracks it exactly. That is the theorem, written as six integers. And all six answers are sixty-two.

7:20If the choice really is free, then it is free, and the flag does not have to be sensible. Put it at one hundred. That is above every class mark in the table — there is no student anywhere near it. Now every deviation is negative, all the way down to minus seven and a half. The column totals minus one thousand one hundred and forty, which shared among thirty is minus thirty-eight.

7:49One hundred minus thirty-eight is sixty-two. So the flag does not have to sit inside the data. Try minus forty. Sixty-two. Try a thousand. Sixty-two. Try one third — not a class mark, not a whole number, not near anything at all. Sixty-two. Twenty-one centres were tried on five tables — a hundred and five runs — and every one returned the answer that table already had. One more. Put the flag at sixty-two: at the answer itself.

8:23Then the deviation column totals nought, which students read as the method failing. It is the opposite. You guessed the mean exactly, so the deviations balance, so there is nothing to add on, and the answer is the flag. Since nothing you do to the flag can break this, it is worth knowing what can. Four things. One: make the deviations positive. At forty-seven and a half that gives sixty-nine.

8:54And notice where the trap hides. If your flag sits at or below every mark, nothing is negative, so there is nothing to make positive, and you get the right answer anyway. Across the hundred and five runs it agreed forty-seven times — and every one of those forty-seven was a flag with nothing below it. That is how a wrong method survives. Two: never add the flag back. That gives fourteen and a half — short by forty-seven and a half, which is the flag, which is exactly what the derivation said.

9:33Three: divide by the number of classes rather than the number of students. Six instead of thirty gives one hundred and twenty, for a table whose largest mark is ninety-two and a half. Four: subtract the flag from the frequency column. The frequencies are the data, and nobody may relabel those. It returns a negative answer from a table with no negative number in it. So if accuracy is not at stake, what is?

10:07Arithmetic. And you can count it: write out every entry of the product column and its total, and count the digit characters your hand actually forms. For the six choices of flag that comes to nineteen, eighteen, sixteen, fifteen, seventeen and eighteen. The cheapest is fifteen, at sixty-two and a half, the class mark nearest the answer. So the rule of thumb is a real effect, not folklore. The dearest is nineteen, at the mark furthest away.

10:42The direct method — which, remember, is the flag at nought — costs twenty-two here. And the flag at one hundred costs twenty-three. So a bad enough guess costs more than not guessing at all, and still returns sixty-two. Every half unit from nought to a hundred was tried, and the cheapest of the two hundred and one is sixty-one and a half, at fourteen — neither a class mark nor the mean.

11:11Which tells you what kind of quantity this is. Pen-work is worth minimising, and it has nothing to do with whether the answer is right. Here is a table where the saving is worth having. Thirty-five territories, class marks running from twenty to eighty in tens. The direct method's column runs up to three hundred and thirty and totals one thousand three hundred and ninety. Twenty-four digit characters. Take the flag at fifty and the deviations are minus thirty up to thirty, in tens.

11:48The products total minus three hundred and sixty. Eighteen digit characters. Six characters cheaper, and every number in it small enough to hold in your head. Minus three hundred and sixty over thirty-five is minus ten point two eight, and fifty minus that is thirty-nine point seven one — which is what one thousand three hundred and ninety over thirty-five gives, to the digit. And that hands you something useful. If two people pick different flags and get different answers, one of them has slipped, because the method cannot produce that.

12:28Thirty-six single-entry slips were made in one column here, and all thirty-six showed up as a different answer. Two flags is a free check on your own arithmetic. One choice left, and it is the interesting one. Put the flag at nought. Then d is x minus nought, which is x. The deviation column IS the class mark column. The product column is the f times x column, and it totals one thousand eight hundred and sixty.

13:02Share that among thirty, and add the flag — which is nought, so add nothing. Sixty-two. That is the direct method. Not something like it — it is this method with a chosen to be nought, and it always was. So there are not two methods here. There is one method with a parameter, and the direct method is the case where nobody moved the origin. Which brings us back to the word. Assumed mean.

13:31Nothing has been assumed. You have not guessed at the answer and you are not hoping to be close. The derivation left no room for your choice to matter: it produced a, exactly, and cancelled it. It is not an assumed mean. It is a chosen origin. And what a good choice buys you is smaller numbers to write down — nothing more, and nothing less.