# Talk:False precision

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## Deleted text

Deleted text: *Original version of news story: 'The hog weighed 1000 pounds, with tusks 9 inches long.' Reported overseas: 'The hog weighed 450 kilograms, with tusks 23 centimeters long.' (Note how the conversion gives rough estimates the semblance of careful measurements.)

How was this *not* false precision? The metric conversions, giving two s.f., imply precision of 5 kg and 0.5 cm, and that's certainly not warranted by the very round numbers of the original. --Calair 21:58, 27 Jan 2005 (UTC)

Yes, they certainly are warranted. What would you suggest instead?
By your own reasoning, the original 9 in implies a precision of 0.5 in, doesn't it? So what happens when we explicitly convert (9 ± 0.5 in) to (22.86 ± 1.27) cm? We get the number range of from 21.59 cm to 24.13 cm. That 23 cm in right in the middle of that range; it's as good as we are going to get. The only other possibilities for a whole number of centimeters, if we've correctly guessed the original precision, 22 cm and 24 cm. So maybe the last digit in the number we gave might be off by one. That's not bad.
Would you have us convert it to 20 cm instead, a number that is outside the range of the original measurements? Yes, if read as ± 5 cm, it would also include 23 cm, so it isn't an unreasonable choice. But it also includes the range from 15 cm to 21.58 cm and from 24.14 cm to 25.0 cm which are outside the range of the original measurements. That's quite misleading as well—in fact, that is false precision, or to be clearer, false imprecision, in the opposite direction from what you claimed to be false precision in your example. The point is, neither of them really deserve the label "false precision"—save that terminology for those at least one more digit removed in either direction.
No, it's not misleading. When the original presentation indicates that the true value is between 21.59 cm and 24.13 cm, presenting it in a way that says "between 15 cm and 25 cm" is entirely truthful. It is certainly less informative than the original figure, but 'less informative' and 'misleading' are not the same thing, and the former is preferable to the latter.
If John Smith tells me he lives in Florida, it is in no way misleading for me to say that he lives in the USA. It's vague, it'd be nice if I was more specific, but it's in no way misleading. But if I tell people "he lives in southern Florida", that *is* false precision error, because I'm being more specific than my information warrants.
You talk about "20 cm... a number that is outside the range of the original measurements" - but much of the point of this article is that 'numbers' presented in this fashion *are*, in fact, a range. When I say John Smith is 170 cm tall, it is very well understood that I'm actually claiming his height is somewhere in a range around 170 cm. Thus, the fact that 20.0 cm lies outside the possible range is irrelevant - what's important is that the *range* implied by "20 cm" completely brackets that implied by the original measurements. In contrast, the range implied by "23 cm" - i.e. 22.5 to 23.5 cm - does *not* completely enclose the original. --Calair 01:04, 28 Jan 2005 (UTC)
That's just it; for conversions like this, you have an option—you either go a little more precise, or a little less precise.
Indeed - and while one of those two options loses some information, the other risks introducing *false* information, which is distinctly the greater of evils. --Calair 01:04, 28 Jan 2005 (UTC)
No. You are missing the most important, basic philosophy of making these conversions and expressing the numbers in dual systems in the first place. Those who ignore one system of units should get basically the same information as those who ignore the other system. It won't be perfect, but when the choice is debatable, it is by no means appropriate as an example for this article.
In this case, 23 cm gives the readers a better idea of the actual measurement than the 20 cm does, and for your argument 20 cm gives less information than 2 dm would, but using 2 dm in this context would be silly simply because those units are so little used and there is no good reason to encourag their use (IMHO, we'd be better off to throw all centimeters or centi- anything else out the window too, consigning them to the same fate as the obsolete prefix myria). So a better option than 20 cm would be 0.2 m, just so that it is clear that the 0 which appears in 20 cm is not significant, since we have no simpler way of indicating whether or not it is significant. The point is, you have to take the representation of the numbers into consideration as well; some methods give more information than others, but they might be harder to read and harder to visualize. More tradeoffs to consider.
But one important factor that also gives those of us not silly enough to ignore either set of measurement an additional clue is that the figure in inches is still there. In cases like this, I'm a firm believer that the original measurment should always be stated first, just for that reason of providing the additional clues about actual precision. Gene Nygaard 02:36, 28 Jan 2005 (UTC)
You'll never get any better without explicit error ranges attached to your numbers. However, keep in mind also that our assignment of ± 0.5 inches to the original measurement is just a guess as well.
Furthermore, if it were a whole series of measurements, we might have a better idea that the 9 inches was actually only to the nearest inch. However, in real life it is more likely that the measurement would actually be to the nearest half inch, or to the nearest quarter inch, but nobody is going to write that as 9 0/2 in or 9 0/4 in.
Quite possibly - but the same considerations can be applied to the figure in cm. --Calair 01:04, 28 Jan 2005 (UTC)
No, one important difference is that you are wrongly assuming decimal precision in a number which isn't normally measured decimally. he normal measurements might be 8½, 9, 9½, 10, 10½ or they might be 8½, 8¾, 9, 9¼, 9½, etc. Note that in either case there is a fairly high likelihood that a number actually accurate to the nearest half inch or to the nearest quarter inch will be expressed as a whole number.
Note in particular that if it were in fact to the nearest quarter inch, we'd then have (9 ±1/8) in = (22.86 ± 0.3175) cm or somewhere between 22.5425 cm and 23.1775, which to the nearest centimeter is ALWAYS 23 cm.Gene Nygaard 02:36, 28 Jan 2005 (UTC)
Note that 2.54 is less than the square root of ten. That's a factor here also. But it's a lot harder to explain, so I'm not going into that right now. Maybe someone else can expand on that.
No need; I'm aware of that issue. I appreciate that such values make these considerations more awkward, but since half of all data we deal with is likely to fall in those 100... to 316... brackets (and more than half, when we have to consider both pre- and post-conversion data), ducking all such examples is an exercise in escapism. --Calair 01:04, 28 Jan 2005 (UTC)
The same is true for the measurement of 1000 lb. You claim this is a measurement of an individual hog. If they best idea somebody had was that this measurement is within 500 pounds, they would say "it's about 500 to 1500 pounds". In this case, it is reasonable to assume that at least one of the zeros is significant.
Agreed. And granting that one significant zero, the implied margin of error is ~ 50 pounds, which is still a lot more than 5 kg. A figure of "1000 lbs" makes it pretty clear to anybody that this might just be an eyeballed value; unless one's specifically watching for conversions, a figure of "450 kilograms" strongly suggests that somebody actually weighed it.
People round things off to the nearest 50 kg, or to the nearest 25 kg, even when they making the original measurements in kilograms with no conversion involved. There is absolutely no reason to assume from one isolated measurement of 450 kg that it is accurate to the nearest 10 kg. Gene Nygaard 02:36, 28 Jan 2005 (UTC)
Yes, this is a real hog: compare [US http://www.msnbc.msn.com/id/5540839/] and metric versions of the story, and see which makes it clearer that these numbers are likely to be very rough. --Calair 01:04, 28 Jan 2005 (UTC)
BTW, the photo is interesting. The guy standing behind the pig has a shadow cast behind his leg. But I could not see a shadow of the pig on the guy. Go figure.--NevilleDNZ 03:47, 18 July 2005 (UTC) (Also Canada isn't exactly "overseas" from the US :-)
If the context in a more expanded story makes it clearer that the measurements are very rough, then the conversion should be more rough. The context did not make that clear in what was posted in this particular Wikipedia article. I'm not going to look up the story, because it is totally irrelevant to the discussion at hand. Gene Nygaard 02:36, 28 Jan 2005 (UTC)
Those numbers consisting of only a one are always bad examples in any case. They might not have even one significant digit; they might, for example, only be decimal orders of magnitude. If that were the case, the proper conversion from 1000 lb would be 0 kg. Or maybe 1000 kg.
In some contexts, certainly, but this particular context indicates otherwise; nobody estimates the weight of the hog they've killed in orders of magnitude, unless they're playing Fermi estimates. There's a reason why I left the context in. --Calair 01:04, 28 Jan 2005 (UTC)
Better examples are so very abundant. There's no need to throw in examples which will get you into arguments as to whether there is any "false precision" involved or not. One more apparently significant digit in the result than in the orignial measurement is not an example of false precision, especially when the initial digit of the longer (in terms of significant digits) number is the initial digit of the shorter number. Just go look at an old slide rule, or some other logarithmic scale, and you will see why especially those numbers starting with one or two should most often have that extra digit. Gene Nygaard 23:32, 27 Jan 2005 (UTC)
I'm willing to agree that this is not the *best* possible example of false precision, and the ones currently on the page are fine. But I dispute the claim that it is *not* false precision. Note that even allowing an extra significant digit on '1000' (which I agree is an eminently reasonable thing to do), it still implies a margin of error 4.5x larger than that implied by '450 kg'. To somebody trying to evaluate the story's credibility, the difference is important, because precision also implies something about the 'measurement' process. A figure of '1000 lbs' could easily be produced by overenthusiasm and poor estimation ability (which is a very likely explanation in this story); a figure of '450 kg' could not, except by the magic of unit conversions.
In terms of the errors introduced, this certainly isn't the most dramatic of examples - a factor of 4.5x for the weight, and 2.5x for the tusk length. But in its ability to mislead the reader, it is quite significant. Somebody who sees the original "1000 lbs" has a much better chance of realistically gauging that figure's reliability than somebody who sees "450 kg". --Calair 01:04, 28 Jan 2005 (UTC)
Someone who sees 23 cm gets a much better idea of the actual length of the tusks than someone who sees 20 cm. Gene Nygaard 02:36, 28 Jan 2005 (UTC)

## kgf -> N conversion

"the exact conversion factor is 1 kgf = 9.80665 N"

Somehow I can't believe this is exact. The number comes from the force of gravity on earth, no? As I recall from physics classes, seconds is defined as a fraction of the earth's rotation period, and meters is defined as a fraction of the earth's circumference. I can't believe the force of gravity is an exact ratio of these two constants.

Perhaps the text should read, "a more precise conversion factor..."?

--ScottJ 22:55, Mar 22, 2005 (UTC)

No. That's how kilograms-force are defined. The amount of force that a kilogram will exert due to gravity will, of course, depend on its location. Even if you limit yourself to sea level on Earth, that force will vary by 0.53%, from 0.9973 kgf at sea level at the equator, to 1.0026 kgf at sea level at the poles. The second is defined based on a certain number of tranisitions in an atom of cesium-137. The meter is based on the speed of light, defined so that the speed of light is exactly 299.792458 Mm/s. Gene Nygaard 23:33, 22 Mar 2005 (UTC)

## The Speed of light and '±'

Note the number of decimal places is only loosely "correlates" to precision. A classic example:

"According to the US National Bureau of Standards, the Speed of light is equal to: 299792.4574 ± 0.0011 km/s."

Note that the accuracy here is ± 0.0011, and NOT a nice tidy ± 0.00005 as some might suggest from the 4 decimal places. I suggest using a "±" if there is confusion.

i.e. Sometimes "true precision" is also false.

NevilleDNZ 03:25, 18 July 2005 (UTC)

## Using a wikipedia article as an example

I've removed the mention of a wikipedia article as an example of false precision. It violates avoid self-references and no original research (as the data is coming from wikipedia itself). I think the other examples are far more straightforward even though I understand the conversions in the wikipedia example. Graham87 04:39, 31 October 2006 (UTC)

Agreed; I also think using an article to highlight another editor's mistakes should generally be avoided as a matter of courtesy. By all means correct the error and explain things as appropriate on Talk pages, but I think highlighting errors that have been fixed works against Wikipedia:Be bold in updating pages.--Calair 06:55, 31 October 2006 (UTC)

## 24 hours?

Should we include cases where people talk about "24 hours", rather than a day? --Lazar Taxon (talk) 22:59, 23 January 2008 (UTC)

It used to give me a chuckle how often Star Trek mentioned multiples of 12 hours. Was there ever an episode in which some disaster was due to happen in, say, 31 hours? —Tamfang (talk) 19:56, 25 May 2012 (UTC)

## Arithmetic example

Wouldn't a more concise solution to the arithmetic example be something more like:

${\displaystyle {\frac {11}{3}}\cdot 24={\frac {11\cdot 24}{3}}={\frac {264}{3}}=88}$

Seems more straightforward to me. Lurlock (talk) 00:46, 11 September 2010 (UTC)

## Irrational numbers

Perhaps the most common cause of "false precision" is the introduction of irrational constants such as Pi for calculating areas, volumes, etc. Note: in the 1970's this was commonly termed "artificial accuracy", and was often associated with the introduction of scientific calculators.

Another common type is combining estimates with exact numbers -- an example from WP:

• World War I; Infobox -- Strength : Central Powers : Total = 25,248,321

~Eric F 184.76.225.106 (talk) 15:56, 25 May 2012 (UTC)

## 100.000% image

The image shown at the examples suggest that there is a case of false precision stating "two-three digit precision suddently[sic] turns into a six digit one". Contrary to what the image suggests this is not a case of false precision, since for an addition the rule is: "For addition and subtraction, the result should have as many decimal places as the measured number with the smallest number of decimal places ". Which is the case in this picture, each number has 3 decimal digits and so the result also has 3 decimal digits regardless of the number of significant figures (which the author of the figure suggests). Therefore I remove the image Wannesvdh (talk) 09:33, 30 August 2013 (UTC)

Agreed. Staszek Lem (talk) 22:31, 30 August 2013 (UTC)

I humbly disagree: in this photo, one would expect a three digit precision, that is 100%. Instead, we have 100.000% which looks to me like 6 digit precision. Not only that, but the ingredient list is "padded" with a 5 digit precision item to arrive at the expected 100%. Even if we assume 99.303% figure is exactly correct, I would expect the resulting figure to be 100.00%, not 100.000%. — Preceding unsigned comment added by Andy.goryachev (talkcontribs) 21:03, 26 September 2013 (UTC)

This is not about what you expect, this is a simple case of applying the rules, and they are simple: for addition the number of decimals in the result is the lowest number of decimals in the terms of the sum. 0.480% has 3 decimal digits, and so have 0.185%, 0.032% and 99.303% . Thus the result should have 3 decimal digits, being 100.000%. Unless you come with some founded arguments this image remains removed. Wannesvdh (talk) 21:39, 26 September 2013 (UTC)

I did not mean "the number of decimals". I believe this article discusses precision, (significant digits) http://en.wikipedia.org/wiki/Significant_figures or http://en.wikipedia.org/wiki/Precision_%28computer_science%29 or http://reference.wolfram.com/mathematica/tutorial/NumericalPrecision.html The number 0.032 has two significant digits. — Preceding unsigned comment added by Andy.goryachev (talkcontribs) 03:47, 28 September 2013 (UTC)

I see you don't mean the number of decimals, but the rules for addition are stated in the first link you give: "When adding numbers, the number of decimal places in the result is the smallest of the number decimal places in any term." And the meaning of decimal places is also stated there: " decimal places (the number of digits following the decimal point)." So in 0.032 there are 2 significant figures, that is correct, but for the addition this doesn't have any meaning since the only thing you have to take into account is the number of decmal places, not the significant numbers. Wannesvdh (talk) 10:28, 6 October 2013 (UTC)

Again, I disagree. I think there might be confusion between the pure mathematical rules of addition (of infinitely precise mathematical concepts that the numbers represent) and addition of imprecise physical quantities. The rules are different. Let me illustrate the point using a different notation. Imagine one has two quantities, measured with an error: 0.5 ± 0.1 kg 0.0005 ± 0.0001 kg the error of the result will not be ±0.0001 kg, regardless of the "rules for addition". You still get something like 0.5 ± 0.1 kg because there is only 1 significant digit in the first quantity. — Preceding unsigned comment added by 99.108.140.108 (talk) 17:44, 9 October 2013 (UTC)

I think we need perhaps to get an authoritative comment from a physics professor on this. My M.S. in physics is clearly insufficient. — Preceding unsigned comment added by Andy.goryachev (talkcontribs) 17:48, 9 October 2013 (UTC)

Well nice to see a colleague MS in physics, but you keep neglecting to apply the rules. In your example you have to use (like I said before) the rules for addition. So the lowest number of decimals, which in this case is 1 decimal, determines the number of decimals in the result: 1 decimal like in your example. But if we use a very similar example add: 1.001 kg and 0.009 kg (we use a scale with 1 gram precision which is easily achievable) you suggest the answer should be 1kg(1 significant number form the last term), while it is clear that the correct answer is 1.010 kg which has 4 significant numbers. Both numbers are correct to the gram, and your rule would mean that by simply adding them you would go from a 1 gram precision to a 1kg precision, just by adding the two together. Wannesvdh (talk) 21:02, 9 October 2013 (UTC)

Let me try again. The article is about false precision. This is a phenomenon stemming from the fact that some people [who?] make mistake of applying certain rules when dealing with quantities measured with finite accuracy. As the article states, this "occurs when numerical data are presented in a manner that implies better precision than is actually the case". This happens somewhat more often than I personally would like, and I guess this is why this article exists in the first place. Perhaps it would help to figure the root cause of this confusion. Let me try to give another example. Suppose you have two quantities: a truckload of stuff and 0.123 grams of stuff. When one adds the two, the result is 1 truckload, the 0.123 gram quantity is simply swallowed by the variation in the measurement of the first! This is an example which intentionally uses different measurement units (truckloads vs grams) to illustrate the meaning of the phenomenon. Perhaps the confusion stems from the way people write and perceive numbers. Mathematically, 1, 1.000 and 1e+0 are the same numbers. However, different notation implies different precision, when dealing with real world quantities. I think this image clearly illustrates the phenomenon of false precision. They basically claim that "other" ingredients take up 99.303%, a doubtful five digit precision. Did they claim that their measurement is so precise? Or did they simply used decimal notation with 3 digits after decimal point for all their numbers, thus conveying a sense of false precision? — Preceding unsigned comment added by Andy.goryachev (talkcontribs) 22:48, 14 October 2013 (UTC)

Even though this appears to have died down, I'm going to weigh in, upset the apple cart and offer my POV. It is entirely correct to say that 0.480% + 0.185% + 0.032% + 99.303% is 100.000%, but in an article about false precision, one should not neglect accuracy. Assuming each of the original quantities is correct to ±0.001% (and ignoring covariance, at my own peril) the propagated error is ±0.002%, and 100.000% (±0.002%) seems quite reasonable.
But accuracy and precision are more about what the numbers mean than what they are. 0.480%, 0.185%, 0.032% and 99.303% were not added to form 100.000%. The first three were added to obtain 0.697% (±0.002%) and subtracted from 100%, in this case meaning everything or the mathematical constant unity. I say this because there can be no more or less of what is there than what is there.
Subtracting 0.03 ±0.01 from π gives 3.11 ±0.01 because the obtainable accuracy of pi is much greater than that of 0.03 ±0.01, and I could similarly defend the 99.303% malarky. However, 'OTHER INGREDIENTS' as listed on the container is not a measured quantity and whatever conventions say shouldn't be expressed in a way that gives that impression. Instead, saying 'approx. 99%' or <100% is actually much more meaningful, without trying to add up to 100% or 100.000%.
But that's just my two cent. Tomásdearg92 (talk) 22:57, 28 December 2013 (UTC)